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A simple method to determine the number of true different quadratic and cubic permutation polynomial based interleavers for turbo codes

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Abstract

Interleavers are important blocks of the turbo codes, their types and dimensions having a significant influence on the performances of the mentioned codes. If appropriately chosen, the permutation polynomial (PP) based interleavers lead to remarkable performances of these codes. The most used interleavers from this category are quadratic permutation polynomial (QPP) and cubic permutation polynomial (CPP) based ones. In this paper, we determine the number of different QPPs and CPPs that cannot be reduced to linear permutation polynomials (LPPs) or to QPPs or LPPs, respectively. They are named true QPPs and true CPPs, respectively. Our analysis is based on the necessary and sufficient conditions for the coefficients of second and third degree polynomials to be QPPs and CPPs, respectively, and on the Chinese remainder theorem. This is of particular interest when we need to find QPP or CPP based interleavers for turbo codes.

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Acknowledgments

The authors would like to thank the anonymous reviewers for their helpful comments and suggestions that greatly improved this paper and the previous published paper in this journal, entitled “The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions”.

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Correspondence to Lucian Trifina.

Appendices

Appendix 1

1.1 Appendix 1.1: Proof [Theorem 4.1]

From Case 2) from Sect. 4, the number of possible combinations for the coefficient \(q_1 \) results equal to \(\prod \nolimits _{j=1}^s p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) \) and the number of coefficients \(q_2 \) is equal to \(\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} } \). The value \(q_2 =0\) results only when \(q_{2,j} =0,\forall j=\overline{1,s} \), that is for only one combination of the coefficients \(q_{2,j} ,j=\overline{1,s} \), which has to be removed. The number of QPPs will be that in (9).

1.2 Appendix 1.2: Proof [Theorem 4.2]

From the cases 1(b) and 2) from Sect. 4, the number of possible combinations for the coefficient \(q_1 \) results equal to \(2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) and the number of coefficients \(q_2 \), equal to \(2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \). Since from the equivalence condition of QPPs we must have \(q_2 <N/2\), a number of \(\frac{1}{2}\cdot 2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } =2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) coefficients remain, from which the value \(q_2 =0\) has to be removed, finaly remaining \(2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1\) values for true different QPPs. Then, the number of QPPs will be that in (10).

1.3 Appendix 1.3: Proof [Theorem 4.3]

From the cases 1(a) and 2 from Sect. 4 the number of possible combinations for the coefficient \(q_1 \) results equal to \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) and the number of coefficients \(q_2 \) equal to \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \). We mention that in this case, the \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) coefficients \(q_2 \) have \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) modulo N different values, all smaller than N / 2. Therefore, we have to only remove the value \(q_2 =0\), finally leading to \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1\) values for the coefficient \(q_2 \) of true different QPPs. Then the number of QPPs will be that in (11).

Appendix 2

1.1 Appendix 2.1: Proof [Theorem 5.1]

For \(n_2 =0\) and \(n_3 =0\), we have \(s_1 =1\), and the factors from the decomposition of N are of type 3(a) and/or 3(b) and/or 4(b). From the cases 3(a), 3(b) and 4(b) from Sect. 5, the number of possible combinations for the \(q_1 \) is equal to \(\Phi \left( N \right) =\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) and the total number of coefficients \(q_2 \) and \(q_3 \), respectively, are equal to \(\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} } \). The value \(q_3 =0\) results only when \(q_{3,j} =0,\forall j=\overline{1,s} \), that is, for a single combination of coefficients \(q_{3,j} ,j=\overline{1,s} \), that has to be removed. The number of CPPs will be equal to

$$\begin{aligned}&C_{N,CPPs} =\left( {\prod \limits _{j=1}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod \limits _{j=1}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=1}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad =\left( {\prod \limits _{j=1}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod \limits _{j=1}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\ \end{aligned}$$
(37)

i.e., the one from (20), for \(n_2 =0\) and \(n_3 =0\).

For \(n_2 =1\) and \(n_3 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 1(a) and 3(a) and/or 3(b) and/or 4(b). From the analysis for the cases 1(a) and 3(a), 3(b), 4(b) from Sect. 5, the number of possible combinations for the coefficient \(q_1 \) is equal to \(1\cdot \Phi \left( {N/2} \right) =\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) and the total number of coefficients \(q_2 \) and \(q_3 \), respectively, is equal to \(1\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } =\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \), from which one value is 0. By removing the value \(q_3 =0\) the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs} =\left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) =\left( {\prod \limits _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(38)

i.e., the one from (20), for \(n_2 =1\) and \(n_3 =0\).

For \(n_2 =0\) and \(n_3 =1\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 2(a) and 3(a) and/or 3(b) and/or 4(b). In determining the number of coefficients, we consider that for the two sets of coefficients valid for the factor of type 2(a), we have only a single value for \(q_2 \) and \(q_3 \). The number of possible combinations for the coefficient \(q_1 \) is equal to \(2\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) is equal to \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) and the number of coefficients \(q_3 \) is equal to \(\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \), from which one is 0. By removing the value \(q_3 =0\), the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs} =\left( {2\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad =\left( {2\cdot \prod \limits _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(39)

i.e., the one from (20), for \(n_2 =0\) and \(n_3 =1\).

For \(n_2 =1\) and \(n_3 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(a) and 2(a) and 3(a) and/or 3(b) and/or 4(b). The number of possible combinations for the coefficient \(q_1 \) results equal to \(1\cdot 2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } =2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) and \(q_3 \) is equal to \(1\cdot 1\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } =\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } \), from which one is 0. After removing the value \(q_3 =0\) the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs} =\left( {2\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {\prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad =\left( {2\cdot \prod \limits _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(40)

i.e., the one from (20), for \(n_2 =1\) and \(n_3 =1\).

1.2 Appendix 2.2: Proof [Theorem 5.2]

For \(n_3 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 1(b) and 3(a) and/or 3(b) and/or 4(b). From the cases 1(b), 3(a), 3(b) and 4(b) from Sect. 5, the number of possible combinations for the coefficient \(q_1 \) is equal to \(\Phi \left( N \right) =2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) and the total number of coefficients \(q_2 \) and \(q_3 \), respectively, is equal to \(2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \). Because this case requires from the equivalence conditions that \(q_2 <N/2\) and \(q_3 <N/2\), a number of \(\frac{1}{2}\cdot 2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } =2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) possible coefficients \(q_2 \) and \(q_3 \), respectively, remains, from which one is zero. By removing the value \(q_3 =0\), the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2^{n_{N,2} -1}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&=2^{2\cdot n_{N,2} -3}\cdot \prod \limits _{j=2}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(41)

i.e., the one from (21), for \(n_3 =0\).

For \(n_3 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(b) and 2(a) and 3(a) and/or 3(b) and/or 4(b). In determining the number of coefficients, we consider that for the two sets of coefficients valid for the factor of type 2(a) we always have \(q_{2,1} =q_{3,1} =0\). From the two sets we have to keep only one for the coefficients \(q_2 \) and \(q_3 \). The number of possible combinations for the coefficient \(q_1 \) is equal to \(2^{n_{N,2} -1}\cdot 2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } =2^{n_{N,2} }\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) and \(q_3 \) is equal to \(2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } \) , from which one is 0. By removing the value \(q_3 =0\) the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2^{n_{N,2} }\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( {2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot \prod \limits _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(42)

i.e., the one from (21), for \(n_3 =1\).

1.3 Appendix 2.3: Proof [Theorem 5.3]

For \(n_2 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 2(b) and 3(a) and/or 3(b) and/or 4(b). The number of possible combinations for the coefficient \(q_1 \) is equal to \(2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) is equal to \(3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \) and the number of coefficients \(q_3 \) is equal to \(2\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \), for \(3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) of the \(2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) values for the coefficients \(q_1 \) and it is equal to \(2\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} } \), for the other \(3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) values for the coefficients \(q_1 \), from which one is 0. By removing the value \(q_3 =0\) the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&+\left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&= \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod \limits _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(43)

i.e., the one from (22), for \(n_2 =0\).

For \(n_2 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(a) and 2(b) and 3(a) and/or 3(b) and/or 4(b). From the analysis of the cases 1(a), 2(b) and 3(a), 3(b), 4(b) from Sect. 5, it follows that the number of possible combinations for the coefficient \(q_1 \) is equal to \(1\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } =2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) is equal to \(1\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } =3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } \) and a number of coefficients \(q_3 \) is equal to \(1\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } =2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } \), for \(3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) of the \(2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) values for the coefficient \(q_1 \) and equal to \(1\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } =2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } \), for the other \(3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) \) values for the coefficient \(q_1 \), from which one is 0. By removing the value \(q_3 =0\) the number of CPPs results equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&+\left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&=\left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod \limits _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(44)

i.e., the one from (22), for \(n_2 =1\).

1.4 Appendix 2.4: Proof [Theorem 5.4]

In this case, the factors from the decomposition of N are of type 1(b) and 2(b) and 3(a) and/or 3(b) and/or 4(b). The numbers of possible combinations for the coefficient \(q_1 \) is equal to \(2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } =2^{n_{N,2} }\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) is equal to \(2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } \) and the number of coefficients \(q_3 \) is equal to \(2^{n_{N,2} -2}\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } =2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } \), for \(2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) of the \(2^{n_{N,2} }\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) values for the coefficient \(q_1 \) and equal to \(2^{n_{N,2} -2}\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } =2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^{s} {p_j^{n_{N,p_j } -1} } \), for the other \(2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \) values of the coefficient \(q_1 \), from which one is 0. By removing the value \(q_3 =0\) the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs} =\left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \cdot \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\qquad +\left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad =\left( {2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod \limits _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(45)

i.e., the one from (23).

1.5 Appendix 2.5: Proof [Theorem 5.5]

For \(n_2 =0\) and \(n_3 =0\), we have \(s_1 =1\), and the factors from the decomposition of N are of type 4(a) and 3(a) and/or 3(b) and/or 4(b). We consider the analysis of the case 4(a) from Sect. 5. It follows that for a group of \(n_{4a,0} \) prime factors of type 4(a), with \(n_{4a,0} <n_{4a} \), for which \(q_{3,j} =0\left( {\bmod \,\, p_j } \right) \), \(\forall j\in I_{n_{4a,0} } \), the number of possible combinations for the coefficient \(q_1 \) is equal to \(\prod \nolimits _{j\in I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot \prod \nolimits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \), the number of coefficients \(q_2 \) is equal to \(\prod \nolimits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot \prod \nolimits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \) and the number of coefficients \(q_3 \) is equal to \(\prod \nolimits _{j\in I_{n_{4a} } -I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot \prod \nolimits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \). In total, for a group of \(n_{4a,0} \) prime factors of the type 4(a), with \(n_{4a,0} \in \left\{ {1,2,\ldots ,n_{4a} -1} \right\} \), the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs,n_{4a,0} }\nonumber \\&\quad =\left( {\prod \limits _{j\in I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad =\prod \limits _{j\in I_{n_{4a} } } {\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\nonumber \\&\qquad \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \nonumber \\&\quad =\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\nonumber \\&\qquad \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \end{aligned}$$
(46)

For a group of \(n_{4a} \) prime factors of the type 4(a), we have to remove the case when \(q_3 =0\) and the number of CPPs will be:

$$\begin{aligned} C_{N,CPPs,n_{4a} }= & {} \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(47)

When the condition \(q_{3,j} =0\left( {\bmod \,\, p_j } \right) \) is not met for any of the \(n_{4a} \) prime factors of the type 4(a), according to the condition 4(a) from Sect. 5, the number of CPPs will be:

$$\begin{aligned} C_{N,CPPs,0}= & {} \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&=\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) }\nonumber \\&\cdot \prod \limits _{j=1}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \end{aligned}$$
(48)

The final number of CPPs results by summing the quantities from (46), for \(n_{4a,0} =1,2,\ldots ,n_{4a} -1\), (47) and (48):

$$\begin{aligned} C_{N,CPPs}= & {} \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) }\nonumber \\&\cdot \prod \limits _{j=1}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&+\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \right. \nonumber \\&\left. \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&+\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } p_j^{2\cdot \left( {n_{N,p_j } -1} \right) }\right. \nonumber \\&\left. \cdot \left( {p_j -1} \right) \right) \nonumber \\&\cdot \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) }\nonumber \\&\cdot \prod \limits _{j=1}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\quad +\sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \right. \nonumber \\&\left. \cdot \prod \limits _{j=1}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&+\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=1}^{s-n_{4a} } p_j^{2\cdot \left( {n_{N,p_j } -1} \right) }\right. \nonumber \\&\left. \cdot \left( {p_j -1} \right) \right) \nonumber \\&\cdot \left( {\prod \limits _{j=1}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(49)

i.e., the one from (24), for \(n_2 =0\) and \(n_3 =0\).

For \(n_2 =1\) and \(n_3 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 1(a) and 4(a) and 3(a) and/or 3(b) and/or 4(b). The number of CPPs results by considering the previous case (for \(n_2 =0\) and \(n_3 =0)\) and the case 1(a) from Sect. 5:

(50)

i.e., the one from (24), for \(n_2 =1\) and \(n_3 =0\).

For \(n_2 =0\) and \(n_3 =1\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 2(a) and 4(a) and 3(a) and/or 3(b) and/or 4(b). We consider the analysis from the proof of this theorem for \(n_2 =0\) and \(n_3 =0\) and the conditions from the case 2(a) from Sect. 5. For a group of \(n_{4a,0} \) prime factors of the type 4(a), with \(n_{4a,0} \in \left\{ {1,2,\ldots ,n_{4a} -1} \right\} \), the number of CPPs is equal to:

$$\begin{aligned}&C_{N,CPPs,n_{4a,0} } = \left( \prod \limits _{j\in I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot 2\right. \nonumber \\&\left. \quad \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \right) \nonumber \\&\quad \cdot \left( \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \right) \nonumber \\&= \prod \limits _{j\in I_{n_{4a} } } {\left( {p_j -1} \right) } \cdot 2\cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\nonumber \\&\quad \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \nonumber \\&= \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2\cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \nonumber \\&\quad \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \end{aligned}$$
(51)

For group of \(n_{4a} \) prime factors of the type 4(a), we have to remove the case when \(q_3 =0\) and the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs,n_{4a} }= & {} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {\prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2\cdot \prod \limits _{j=2}^{s-n_{4a} } p_j^{2\cdot \left( {n_{N,p_j } -1} \right) }\right. \nonumber \\ \nonumber \\&\quad \left. \cdot \left( {p_j -1} \right) \right) \cdot \left( {\prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(52)

When the condition \(q_{3,j} =0\left( {\bmod \,\, p_j } \right) \) is not met for any of the \(n_{4a} \) prime factors of the type 4(a), the number of CPPs will be:

$$\begin{aligned} C_{N,CPPs,0}= & {} \left( {2\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\= & {} \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2\nonumber \\&\cdot \prod \limits _{j=2}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \end{aligned}$$
(53)

The number of CPPs results by summing the quantities in (51), for \(n_{4a,0} =1,2,\ldots ,n_{4a} -1\), (52) and (53):

(54)

i.e., the one from (24), for \(n_2 =0\) and \(n_3 =1\).

For \(n_2 =1\) and \(n_3 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(a) and 2(a) and 4(a) and 3(a) and/or 3(b) and/or 4(b). In determining the number of CPPs, we consider the proof of this theorem for \(n_2 =0\) and \(n_3 =1\) and the analysis for the case 1(a) from Sect. 5. The number of CPPs will be:

(55)

i.e., the one from (24), for \(n_2 =1\) and \(n_3 =1\).

1.6 Appendix 2.6: Proof [Theorem 5.6]

For \(n_3 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 1(b) and 4(a) and 3(a) and/or 3(b) and/or 4(b). We consider the proof of Theorem 5.5 for \(n_2 =0\) and \(n_3 =0\) and the additional case 1(b) from Sect. 5. The number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs} = \left( {2^{n_{N,2} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad +\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left[ \left( \prod \limits _{j\in I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\right. \right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \cdot \nonumber \\&\quad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \left. {\cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) } \right] \nonumber \\&\quad +\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot n_{N,2} -5}\nonumber \\&\quad \cdot \prod \limits _{j=2}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\quad +\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \right. \nonumber \\&\quad \left. \cdot 2^{3\cdot n_{N,2} -5}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot n_{N,2} -3}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot n_{N,2} -5}\nonumber \\&\qquad \cdot \prod \limits _{j=2}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\qquad +\sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \cdot 2^{3\cdot n_{N,2} -5}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad +\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot n_{N,2} -3}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(56)

i.e., the one from (25), for \(n_3 =0\).

For \(n_3 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(b) and 2(a) and 4(a) and 3(a) and/or 3(b) and/or 4(b). We consider the proof of Theorem 5.5 for \(n_2 =0\) and \(n_3 =1\) and the additional case 1(b) from Sect. 5. The number of CPPs results equal to:

$$\begin{aligned}&C_{N,CPPs} = \left( {2^{n_{N,2} -1}\cdot 2\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad +\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left[ \left( \prod \limits _{j\in I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\cdot 2\right. \right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \cdot \nonumber \\&\quad \left. \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \right. \nonumber \\&\quad \left. \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \right] \nonumber \\&\quad +\left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\cdot 2\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot n_{N,2} -4}\nonumber \\&\qquad \cdot \prod \limits _{j=3}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\quad +\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 2^{3\cdot n_{N,2} -4}\right. \nonumber \\&\quad \left. \cdot \,\prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +\left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot n_{N,2} -4}\nonumber \\&\qquad \cdot \, \prod \limits _{j=3}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\qquad +\sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \cdot 2^{3\cdot n_{N,2} -4}\right. \nonumber \\&\qquad \left. \cdot \, \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +\left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(57)

i.e., the one from (25), for \(n_3 =1\).

1.7 Appendix 2.7: Proof [Theorem 5.7]

For \(n_2 =0\), we have \(s_1 =2\), and the factors from the decomposition of N are of type 2(b) and 4(a) and 3(a) and/or 3(b) and/or 4(b). We consider the analysis from the proof of the Theorem 5.5 for \(n_2 =0\) and \(n_3 =0\) and the conditions from the case 2(b) from Sect. 5. For a group of \(n_{4a,0} \) prime factors of the type 4(a), with \(n_{4a,0} \in \left\{ {1,2,\ldots ,n_{4a} -1} \right\} \), the number of CPPs is equal to:

$$\begin{aligned}&C_{N,CPPs,n_{4a,0} }\nonumber \\&\quad = 2\cdot \left( {\prod \limits _{j\in I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad = \prod \limits _{j\in I_{n_{4a} } }{\left( {p_j -1} \right) } \cdot 3^{3\cdot n_{N,3} -4}\cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j }\nonumber \\&\qquad \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \nonumber \\&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{3\cdot n_{N,3} -4}\cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \nonumber \\&\qquad \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \end{aligned}$$
(58)

For a group of \(n_{4a} \) prime factors of the type 4(a), we have to remove the case when \(q_3 =0\) and the number of CPPs will be equal to:

$$\begin{aligned}&C_{N,CPPs,n_{4a} }\nonumber \\&\quad = 2\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = 2\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(59)

When the condition \(q_{3,j} =0\left( {\bmod \,\, p_j } \right) \) is not met for any of the \(n_{4a} \) prime factors of the type 4(a), the number of CPPs will be:

$$\begin{aligned}&C_{N,CPPs,0}\nonumber \\&\quad = 2\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad = 4\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \end{aligned}$$
(60)

The number of CPPs results by summing the quantities in (58), for \(n_{4a,0} =1,2,\ldots ,n_{4a} -1\), (59) and (60):

$$\begin{aligned}&C_{N,CPPs} = 4\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +4\cdot \sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +2\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\quad \left. \cdot \, \prod \limits _{j=2}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = 4\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +4\cdot \sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +2\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=2}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(61)

i.e., the one from (26), for \(n_2 =0\).

For \(n_2 =1\), we have \(s_1 =3\), and the factors from the decomposition of N are of type 1(a) and 2(b) and 4(a) and 3(a) and/or 3(b) and/or 4(b). In determining the number of CPPs we consider the proof of this theorem for \(n_2 =0\) and the analysis for the case 1(a) from Sect. 5. The number of CPPs will be:

$$\begin{aligned}&C_{N,CPPs} = 2\cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad +2\cdot \sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left[ \left( \prod \limits _{j\in I_{n_{4a,0} } }{\left( {p_j -1} \right) } \cdot 3^{n_{N,3} -1} \right. \right. \nonumber \\&\quad \left. \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \right. \cdot \nonumber \\&\quad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\quad \left. \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \right] \nonumber \\&\quad +2\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\quad \cdot \left( {3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = 4\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad +4\cdot \sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j\in I_{n_{4a} } }{\left( {p_j -1} \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +2\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad = 4\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +4\cdot \sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad +2\cdot \left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\quad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\quad \cdot \left( {2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(62)

i.e., the one from (26), for \(n_2 =1\).

1.8 Appendix 2.8: Proof [Theorem 5.8]

We consider the proof of the Theorem 5.7 for \(n_2 =0\) and the additional case 1 (b) from Sect. 5. The number of CPPs results equal to:

$$\begin{aligned}&C_{N,CPPs} = \left( {2\cdot 2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {p_j } \cdot 2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad +2\cdot \sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left[ \left( \prod \limits _{j\in I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\right. \right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \cdot \nonumber \\&\qquad \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \left. \cdot \left( {\prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -2}\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \right] \nonumber \\&\qquad +2\cdot \left( {\prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\qquad \cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } } \right) \nonumber \\&\qquad \cdot \left( {2^{n_{N,2} -2}\cdot 2\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\quad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{3\cdot n_{N,3} -4}\nonumber \\&\qquad \cdot \prod \limits _{j=3}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\qquad +\sum \limits _{n_{4a,0} =1}^{n_{4a} -1} \left( \prod \limits _{j\in I_{n_{4a} } }{\left( {p_j -1} \right) } \cdot \prod \limits _{j\in I_{n_{4a} } -I_{n_{4a,0} } } {p_j } \cdot 2^{3\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad +\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad \cdot \left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\qquad = \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j \cdot \left( {p_j -1} \right) } \right) } \cdot 2^{3\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{3\cdot n_{N,3} -4}\nonumber \\&\quad \cdot \prod \limits _{j=3}^{s-n_{4a} } {\left( {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) } \nonumber \\&\qquad +\sum \limits _{r_0 =1}^{n_{4a} -1} \left( C_{N_{\left( {4a} \right) } ,n_{4a} =r,n_{4a,0} =r_0 } \cdot 2^{3\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{3\cdot n_{N,3} -4}\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad +\left( \prod \limits _{j=s-n_{4a} +1}^s {\left( {p_j -1} \right) } \cdot 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\right. \nonumber \\&\qquad \left. \cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\qquad \cdot \left( {2^{n_{N,2} -1}\cdot 3^{n_{N,3} -2}\cdot \prod \limits _{j=3}^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$
(63)

1.9 Appendix 2.9: Proof [Theorem 5.9]

This case is a particular case of Theorem 5.5, therefore we can use equation (24) in which the products \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \), \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \) and \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \) are replaced by 1, and \(s=n_{4a} +s_1 -1\).

1.10 Appendix 2.10: Proof [Theorem 5.10]

This case is a particular case of Theorem 5.6, therefore we can use equation (25) in which the products \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \), \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \) are replaced by 1, and \(s=n_{4a} +s_1 -1\).

1.11 Appendix 2.11: Proof [Theorem 5.11]

This case is a particular case of Theorem 5.7, therefore we can use equation (26) in which the products \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \), \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \) are replaced by 1, and \(s=n_{4a} +s_1 -1\).

1.12 Appendix 2.12: Proof [Theorem 5.12]

This case is a particular case of Theorem 5.8, therefore we can use equation (27) in which the products \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \), \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=s_1 }^{s-n_{4a} } {p_j^{n_{N,p_j } -1} } \) are replaced by 1, and \(s=n_{4a} +s_1 -1\).

1.13 Appendix 2.13: Proof [Theorem 5.13]

This case is a particular case of Theorem 5.2, therefore we can use equation (21) in which the products \(\prod \nolimits _{j=s_1 }^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=s_1 }^s {p_j^{n_{N,p_j } -1} } \) are replaced by 1.

1.14 Appendix 2.14: Proof [Theorem 5.14]

This case is a particular case of Theorem 5.3, therefore we can use equation (22) in which the products \(\prod \nolimits _{j=s_1 }^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=s_1 }^s {p_j^{n_{N,p_j } -1} } \) are replaced by 1.

1.15 Appendix 2.15: Proof [Theorem 5.15]

This case is a particular case of Theorem 5.4, therefore we can use equation (23) in which the products \(\prod \nolimits _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \) and \(\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} } \) are replaced by 1.

1.16 Appendix 2.16: Proof [Theorem 5.16]

Since \(N=2\) is even, it requires that \(q_3 <N/2=1\). As \(q_3 \) can not be 0, there is no true CPP in this case, i.e. \(C_{2,CPPs} =0\).

When \(N=3\), from the equivalence conditions, it requires that \(q_3 <N/3=1\). As \(q_3 \) can not be 0, there is no true CPP in this case, i.e. \(C_{3,CPPs} =0\).

When \(N=6\), from the equivalence conditions of CPPs, it requires that \(q_3 <N/6=1\). As \(q_3 \) can not be 0, there is no true CPP, i.e. \(C_{6,CPPs} =0\).

Appendix 3

In this appendix, we give an example of checking the formula for the number of true different CPPs, for a more comprehensive case, i.e., when the prime-factor decomposition of the interleaver length contains almost all types of prime factors that appear in theorems presented in the paper. The chosen length of the interleaver corresponds to the decomposition from Theorem 5.5, when \(n_2 =1\) and \(n_3 =1\).

Let \(N=16170=2\cdot 3\cdot 5\cdot 7^{2}\cdot 11\). According to Theorem 5.5, we have \(n_2 =1\), \(n_3 =1\), \(s_1 =3\), \(n_{4a} =2\). We mention that the two factors of type 4(a) are 5 and 11, and the only factor of type 3(b) is \(7^{2}\).

The quantity \(C_{N_{\left( {4a} \right) } ,n_{4a} =2,n_{4a,0} =1} \)used in (24), is given in (14). It is equal to

$$\begin{aligned} C_{N_{\left( {4a} \right) } ,n_{4a} =2,n_{4a,0} =1}= & {} \left( {5-1} \right) \cdot 11\cdot \left( {11-1} \right) \\&+\left( {11-1} \right) \cdot 5\cdot \left( {5-1} \right) =640 \end{aligned}$$

According to (24), the number of all true different CPPs is equal to:

$$\begin{aligned} \begin{array}{l} C_{16170,CPPs} =2\cdot 5\cdot \left( {5-1} \right) \cdot 11\cdot \left( {11-1} \right) \cdot 7^{3\cdot \left( {2-1} \right) }\\ \quad \cdot \left( {7-1} \right) +2\cdot 640\cdot 7^{3\cdot \left( {2-1} \right) }\cdot \left( {7-1} \right) + \\ \quad +2\cdot \left( {5-1} \right) \cdot \left( {11-1} \right) \cdot 7^{2\cdot \left( {2-1} \right) }\cdot \left( {7-1} \right) \\ \quad \cdot \left( {7^{2-1}-1} \right) =\hbox {11830560} \\ \end{array} \end{aligned}$$

In the following, we detail the coefficients of the \(\hbox {11830560}\) true different CPPs.

Because \(6\left| N \right. \), it requires that \(q_3 <N/6=2695\) and \(q_2 <N/2=8085\).

Because of the factor of type 3(b) from the decomposition of 16170, the coefficients \(q_3 \) and \(q_2 \) have to be multiples of 7, and the coefficient \(q_1 \) has to be relatively prime with 7. Because of the factor of type 2(a) from the decomposition of 16170, the coefficient \(q_2 \) also has to be multiple of 3. Therefore, \(q_3 \) can take values in the set \(\left\{ {7,14,21,\ldots ,2688} \right\} \) (with cardinality 384), \(q_2 \) can take values in the set \(\left\{ {0,21,42,\ldots ,8064} \right\} \) (with cardinality 385), and \(q_1 \) can take values in the set \(\left\{ 1,2,3,4,5,6,8,9,\ldots ,16169 \right\} \) (with cardinality \(16170\cdot \frac{6}{7}=13860)\).

Because of the factor of type 2(a), the condition \(\left( {q_1 +q_3 } \right) \ne 0\left( {\bmod \,\, 3} \right) \) must to be fulfilled. Because of the factor of type 1(a), the condition \(\left( {q_1 +q_2 +q_3 } \right) \ne 0\left( {\bmod \,\, 2} \right) \) must be fulfilled. Thus, when \(q_2 \) is even, \(\left( {q_1 +q_3 } \right) \) is required to be odd, and when \(q_2 \) is odd, \(\left( {q_1 +q_3 } \right) \) is required to be even.

Because of the two factors of type 4(a) (5 and 11), the analysis has to be performed for four different cases: a) \(q_3 \) is a multiple of \(5\cdot 11=55\); b) \(q_3 \) is a multiple of 5, but not a multiple of 11; c) \(q_3 \) is a multiple of 11, but not a multiple of 5; and d) \(q_3 \) is neither a multiple of 5, nor a multiple of 11.

(a) When \(q_3 \) is multiple of \(5\cdot 11=55\), then \(q_2 \) has to be multiple of \(5\cdot 11=55\), and \(q_1 \) has to be relatively prime with 5 and 11. This situation happens for six values from the set of 384 possible values for \(q_3 \) (namely, \(\left\{ {385,770,1155,1540,1925,2310} \right\} )\), for 7 values from the set of 385 possible values for \(q_2 \) (namely, \(\left\{ {0,1155,2310,3465,4620,5775,6930} \right\} )\) and for \(16170\cdot \frac{6}{7}\cdot \frac{4}{5}\cdot \frac{10}{11}=\hbox {1}00\hbox {8}0\) values from the set of 13860 possible values for \(q_1 \) (namely, \(\left\{ {1,2,3,4,6,8,9,12,13,\ldots ,16169} \right\} )\).

From the six previous values for \(q_3 \), two are multiple of 3 (namely, 1155 and 2310), two are multiple of 3 plus 1 (namely 385 and 1540) and two are multiple of 3 plus 2 (namely, 770 and 1925). From the 10080 values for \(q_1 \), there are \({10080}/3=3360\) values multiple of 3, multiple of 3 plus 1, and multiple of 3 plus 2, respectively. Therefore, the condition \(\left( {q_1 +q_3 } \right) \ne 0\left( {\bmod \,\, 3} \right) \) is fulfilled for \(2\cdot 3360\cdot 2\cdot 3=40320\) pairs of values for the coefficients \(\left( {q_3 ,q_1 } \right) \). Of the seven values for \(q_2 \), four values are even and three are odd. From each of the three groups of two values for \(q_3 \), one is even and one is odd and from each of three groups of 3360 values for \(q_1 \), 1680 are even and 1680 are odd. It follows that from the 40320 pairs of values for the coefficients \(\left( {q_3 ,q_1 } \right) \), the sum \(\left( {q_1 +q_3 } \right) \) is even for \({40320}/2=20160\) pairs of values and odd for \({40320}/2=20160\) pairs. Therefore, when \(q_3 \) is multiple of \(5\cdot 11=55\), we will have \(4\cdot 20160+3\cdot 20160=141120\) sets of coefficients \(\left( {q_3 ,q_2 ,q_1 } \right) \) leading to true different CPPs modulo 16170.

(b) When \(q_3 \) is a multiple of 5, but not a multiple of 11, \(q_2 \) has to be a multiple of 5, \(q_1 \) has to be relatively prime with 5, and the three coefficients must fulfil the condition \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 11} \right) \). \(q_3 \) is a multiple of 5, but not a multiple of 11, for 70 values from the set of 384 possible values for \(q_3 \) (namely, \(\left\{ 35,70,105,\ldots ,350,420,455,\ldots ,735,805,\ldots , 2660 \right\} )\). \(q_2 \) is a multiple of 5 for 77 values from the set of 385 possible values for \(q_2 \) (namely, \(\left\{ {0,105,210,\ldots ,7980} \right\} )\). For each two coefficients \(q_3 \) and \(q_2 \), from the sets previously mentioned, the congruence equation \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 11} \right) \) has only one solution modulo 11, in variable \(q_1 \), and thus, \({16170}/{11}=1470\) solutions modulo 16170. From these, we have to remove those that are multiple of 5 or multiple of 7. In this way, \(1470-{1470}/5-{1470}/7+{1470}/{35}=1008\) valid solutions remain. For example, for \(q_3 =35\) and \(q_2 =105\), the possible values for \(q_1 \) are from the set \(\left\{ {6,17,39,61,\ldots ,16154} \right\} \).

From the 70 possible solutions for \(q_3 \), 23 values are multiples of 3 and multiples of 3 plus 1, respectively, and 24 values are multiples of 3 plus 2. From the 1008 possible values for \(q_1 \), \({1008}/3=336\) values are multiples of 3, multiples of 3 plus 1, and multiple of 3 plus 2, respectively. Therefore, the condition \(\left( {q_1 +q_3 } \right) \ne 0\left( {\bmod \,\, 3} \right) \) is fulfilled for \(2\cdot 23\cdot 336\cdot 2+24\cdot 336\cdot 2=47040\) pairs of values of the coefficients\(\left( {q_3 ,q_1 } \right) \). From the 77 values for \(q_2 \), 39 values are even and 38 values are odd. From each of the three groups of 336 values for \(q_1 \), 168 values are even and 168 values are odd. From the two groups of 23 values for \(q_3 \), one contains 11 even values and 12 odd values and the other group contains 12 odd values and 11 odd values. The group of 24 values for \(q_3 \) contains 12 even and odd values, respectively. It follows that from the 47040 pairs of values for the coefficients \(\left( {q_3 ,q_1 } \right) \), the sum \(\left( {q_1 +q_3 } \right) \) is even and odd, respectively, for \(2\cdot 11\cdot 168\cdot 2+2\cdot 12\cdot 168\cdot 2+2\cdot 12\cdot 168\cdot 2=23520\) pairs of values. Therefore, the condition \(\left( {q_1 +q_2 +q_3 } \right) \ne 0\left( {\bmod \,\, 2} \right) \) is fulfilled for \(39\cdot 23520+38\cdot 23520=1811040\) sets of coefficients \(\left( {q_3 ,q_2 ,q_1 } \right) \). Thus, when \(q_3 \) is a multiple of 5, but not a multiple of 11, we will have 1811040 true different CPPs modulo 16170.

(c) When \(q_3 \) is a multiple of 11, but not a multiple of 5, \(q_2 \) has to be a multiple of 11, \(q_1 \) has to be relatively prime with 11, and the three coefficients must fulfil the condition \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 5} \right) \). \(q_3 \) is a multiple of 11, but not a multiple of 5, for 28 values from the set of 384 possible values for \(q_3 \) (namely, \(\left\{ 77,154,231,308,462,539,616,693,\ldots ,2387,2464,2541,2618 \right\} )\). \(q_2 \) is a multiple of 11 for 35 values from the set of 385 possible values for \(q_2 \) (namely, \(\left\{ 0,231,462,\ldots ,7854 \right\} )\). For each two coefficients \(q_3 \) and \(q_2 \), fixed in the sets previously mentioned, the congruence equation \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 5} \right) \) has only one solution modulo 5, in variable \(q_1 \), and thus, \({16170}/5=3234\) solutions modulo 16170. From these, we have to remove those that are multiple of 11 or multiple of 7. In this way, \(3234-{3234}/{11}-{3234}/7+{3234}/{77}=2520\) valid solutions remain. For example, for \(q_3 =77\) and \(q_2 =231\), the possible values for \(q_1 \) are from the set \(\left\{ {1,6,16,26,\ldots ,16166} \right\} \).

From the 28 possible values for \(q_3 \), nine values are multiples of 3 and multiples of 3 plus 1, respectively, and 10 values are multiples of 3 plus 2. From the 2520 possible values for \(q_1 \), \({2520}/3=840\) values are multiples of 3, multiples of 3 plus 1, and multiples of 3 plus 2, respectively. Therefore, the condition \(\left( {q_1 +q_3 } \right) \ne 0\left( {\bmod \,\, 3} \right) \) is fulfilled by \(2\cdot 9\cdot 840\cdot 2+10\cdot 840\cdot 2=47040\) pairs of values of coefficients\(\left( {q_3 ,q_1 } \right) \). From the 35 values for \(q_2 \), 18 values are even and 17 values are odd. From each of the three groups of 840 values for \(q_1 \), 420 values are even and 420 are odd. From the two groups of nine values for \(q_3 \), one contains four even values and five odd and the other contains five even values and four odd values. The group of 10 values for \(q_3 \), contains five even and five odd values, respectively. It follows that from the 47040 pairs of values of the coefficients \(\left( {q_3 ,q_1 } \right) \), the sum \(\left( {q_1 +q_3 } \right) \) is even and odd, respectively, for \(2\cdot 4\cdot 420\cdot 2+2\cdot 5\cdot 420\cdot 2+2\cdot 5\cdot 420\cdot 2=23520\) pairs of values \(\left( {q_3 ,q_1 } \right) \). Therefore, the condition \(\left( {q_1 +q_2 +q_3 } \right) \ne 0\left( {\bmod \,\, 2} \right) \) is fulfilled for \(18\cdot 23520+17\cdot 23520=823200\) sets of coefficients . Thus, when \(q_3 \) is a multiple of 11, but not a multiple of 5, we have 823200 true different CPPs modulo 16170.

(d) When \(q_3 \) is neither a multiple of 5 nor a multiple of 11, \(q_1 \) has to be relatively prime with 11 and 5 and the three coefficients have to fulfil the conditions \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 5} \right) \) and \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 11} \right) \). \(q_3 \) is neither a multiple of 5 nor of 11, but is a multiple of 7, for 384-6-70-28=280 values from the initial set of 384 possible values for \(q_3 \) (namely, \(\left\{ {7,14,21,28,42,49,56,63,84,\ldots ,2688} \right\} )\). \(q_2 \) should only have 3 or 7 as prime factors, thus, it can take 385 values, as it was stated at the beginning of this appendix. For each two coefficients \(q_3 \) and \(q_2 \), from the sets previously mentioned, the congruence equations \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 5} \right) \) and \(q_2^2 =3q_1 q_3 \left( {\bmod \,\, 11} \right) \) have a single solution modulo 55, in variable \(q_1 \), and thus, \({16170}/{55}=294\) solutions modulo 16170. From these, we have to remove those that are multiples of 7. In this way \(294-{294}/7=252\) valid solutions remain. For example, for \(q_3 =7\) and \(q_2 =21\), the possible values for \(q_1 \) are from the set \(\left\{ {76,131,186,\ldots ,16136} \right\} \).

From the 280 possible values for \(q_3 \), 94 are multiples of 3 and multiple of 3 plus 1, respectively and 92 values are multiples of 3 plus 2. From the 252 values possible for \(q_1 \), \({252}/3=84\) values are multiples of 3, multiples of 3 plus 1, and multiples of 3 plus 2, respectively. Therefore, the condition \(\left( {q_1 +q_3 } \right) \ne 0\left( {\bmod \,\, 3} \right) \) is fulfilled by \(2\cdot 94\cdot 84\cdot 2+92\cdot 84\cdot 2=47040\) pairs of values of the coefficients \(\left( {q_3 ,q_1 } \right) \). From the 385 values for \(q_2 \), 193 are even and 192 are odd. From each of the three groups of 84 values for \(q_1 \), 42 are even and 42 are odd. From the two groups of 94 values for \(q_3 \), one contains 46 even values and 48 odd values and the other one contains 48 even values and 46 odd values. The group of 92 values for \(q_3 \) contains 46 even and odd values, respectively. It follows that from the 47040 pairs of values of the coefficients \(\left( {q_3 ,q_1 } \right) \), the sum \(\left( {q_1 +q_3 } \right) \) is even and odd, respectively, for \(2\cdot 46\cdot 42\cdot 2+2\cdot 48\cdot 42\cdot 2+2\cdot 46\cdot 42\cdot 2=23520\) pairs of values \(\left( {q_3 ,q_1 } \right) \). Therefore the condition \(\left( {q_1 +q_2 +q_3 } \right) \ne 0\left( {\bmod \,\, 2} \right) \) is fulfilled for \(193\cdot 23520+192\cdot 23520=9055200\) sets of coefficients \(\left( {q_3 ,q_2 ,q_1 } \right) \). Thus, when \(q_3 \) is neither a multiple of 11 nor of 5, we will have 9055200 true different CPPs modulo 16170.

The total number of true different CPPs modulo 16170 results by summing up the number of CPPs from the four cases. It is equal to \(141120+1811040+823200+9055200=11830560\), which is the one determined by the formula in Theorem 5.5.

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Trifina, L., Tarniceriu, D. A simple method to determine the number of true different quadratic and cubic permutation polynomial based interleavers for turbo codes. Telecommun Syst 64, 147–171 (2017). https://doi.org/10.1007/s11235-016-0166-2

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