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%I #41 Jul 06 2024 19:45:32
%S 1,1,1,1,1,1,1,2,1,1,1,2,1,1,1,1,2,1,2,1,1,1,2,1,2,1,1,1,1,2,1,3,1,2,
%T 1,1,1,4,2,2,1,2,1,1,1,1,2,1,2,1,2,1,2,1,1,1,2,2,4,1,2,1,2,1,1,1,1,2,
%U 1,2,1,4,1,3,1,2,1,1,1,4,1,3,1,4,1,2,1,2,1,1,1
%N Square array read by ascending antidiagonals: T(m,n) is the number of non-isomorphic groups G such that G is the semidirect product of C_m and C_n, where C_m is a normal subgroup of G and C_n is a subgroup of G.
%C The semidirect product of C_m and C_n has group representation G = <x, y|x^m = y^n = 1, yxy^(-1) = x^r>, where r is any number such that r^n == 1 (mod m). Two groups G = <x, y|x^m = y^n = 1, yxy^(-1) = x^r> and G' = <x, y|x^m = y^n = 1, yxy^(-1) = x^s> are isomorphic if and only if there exists some k, gcd(k,n) = 1 such that r^k == s (mod m), in which case f(x^i*y^j) = x^i*y^(k*j) is an isomorphic mapping from G to G'.
%C Given m, T(m,n) only depends on the value of gcd(n,psi(m)), psi = A002322 (Carmichael lambda). So each row is periodic with period psi(m). See A327925 for an alternative version.
%C Every number k occurs in the table. By Dirichlet's theorem on arithmetic progressions, there exists a prime p such that p == 1 (mod 2^(k-1)), then T(p,2^(k-1)) = d(gcd(2^(k-1),p-1)) = k (see the formula below). For example, T(5,4) = 3, T(17,8) = 4, T(17,16) = 5, T(97,32) = 6, T(193,64) = 7, ...
%C Row m and Row m' are the same if and only if (Z/mZ)* = (Z/m'Z)*, where (Z/mZ)* is the multiplicative group of integers modulo m. The if part is clear; for the only if part, note that the two sequences {(number of x in (Z/mZ)* such that x^n = 1)}_{n>=1} and {T(m,n)}_{n>=1} determine each other, and the structure of a finite abelian group G is uniquely determined by the sequence {(number of x in G such that x^n = 1)}_{n>=1}. - _Jianing Song_, May 16 2022
%H Jianing Song, <a href="/A327924/b327924.txt">Table of n, a(n) for n = 1..5050</a> (the first 100 ascending diagonals)
%H Math Overflow, <a href="https://mathoverflow.net/q/455396/494268">When are two semidirect products of two cyclic groups isomorphic</a>.
%H Jianing Song, <a href="/A327924/a327924.png">An equivalent formula</a>.
%F T(m,n) = Sum_{d|n} (number of elements x such that ord(x,m) = d)/phi(d), where ord(x,m) is the multiplicative order of x modulo m, phi = A000010. There is a version to compute the terms more conveniently, see the links section. [Proof: let (Z/mZ)* denote the multiplicative group modulo m. For every d|n, the elements in (Z/mZ)* having order d are put into different equivalence classes, where each class is of the form {a^k: gcd(k,n)=1}. The size of each equivalence class is the number of different residues modulo d of the numbers that are coprime to n, which is phi(d). - _Jianing Song_, Sep 17 2022]
%F Equivalently, T(m,n) = Sum_{d|gcd(n,psi(m))} (number of elements x such that ord(x,m) = d)/phi(d). - _Jianing Song_, May 16 2022 [This is because the order of elements in (Z/mZ)* must divide psi(m). - _Jianing Song_, Sep 17 2022]
%F Let U(m,q) be the number of solutions to x^q == 1 (mod m):
%F T(m,1) = U(m,1) = 1;
%F T(m,2) = U(m,2) = A060594(m);
%F T(m,3) = (1/2)*U(m,3) + (1/2)*U(m,1) = (1/2)*A060839(m) + 1/2;
%F T(m,4) = (1/2)*U(m,4) + (1/2)*U(m,2) = (1/2)*A073103(m) + 1/2;
%F T(m,5) = (1/4)*U(m,5) + (3/4)*U(m,1) = (1/4)*A319099(m) + 3/4;
%F T(m,6) = (1/2)*U(m,6) + (1/2)*U(m,2) = (1/2)*A319100(m) + 1/2;
%F T(m,7) = (1/6)*U(m,7) + (5/6)*U(m,1) = (1/6)*A319101(m) + 5/6;
%F T(m,8) = (1/4)*U(m,8) + (1/4)*U(m,4) + (1/2)*U(m,2) = (1/4)*A247257(m) + (1/4)*A073103(m) + (1/2)*A060594(m);
%F T(m,9) = (1/6)*U(m,9) + (1/3)*U(m,3) + (1/2)*U(m,1);
%F T(m,10) = (1/4)*U(m,10) + (3/4)*U(m,2).
%F For odd primes p, T(p^e,n) = d(gcd(n,(p-1)*p^(e-1))), d = A000005; for e >= 3, T(2^e,n) = 2*(min{v2(n),e-2}+1) for even n and 1 for odd n, where v2 is the 2-adic valuation.
%e m/n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
%e 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
%e 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
%e 3 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
%e 4 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
%e 5 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3
%e 6 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
%e 7 1 2 2 2 1 4 1 2 2 2 1 4 1 2 2 2 1 4 1 2
%e 8 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4
%e 9 1 2 2 2 1 4 1 2 2 2 1 4 1 2 2 2 1 4 1 2
%e 10 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3 1 2 1 3
%e 11 1 2 1 2 2 2 1 2 1 4 1 2 1 2 2 2 1 2 1 4
%e 12 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4
%e 13 1 2 2 3 1 4 1 3 2 2 1 6 1 2 2 3 1 4 1 3
%e 14 1 2 2 2 1 4 1 2 2 2 1 4 1 2 2 2 1 4 1 2
%e 15 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6
%e 16 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6
%e 17 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3
%e 18 1 2 2 2 1 4 1 2 2 2 1 4 1 2 2 2 1 4 1 2
%e 19 1 2 2 2 1 4 1 2 3 2 1 4 1 2 2 2 1 6 1 2
%e 20 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6 1 4 1 6
%e Example shows that T(16,4) = 6: The semidirect product of C_16 and C_4 has group representation G = <x, y|x^16 = y^4 = 1, yxy^(-1) = x^r>, where r = 1, 3, 5, 7, 9, 11, 13, 15. Since 3^3 == 11 (mod 16), 5^3 == 13 (mod 16), <x, y|x^16 = y^4 = 1, yxy^(-1) = x^3> and <x, y|x^16 = y^4 = 1, yxy^(-1) = x^11> are isomorphic, <x, y|x^16 = y^4 = 1, yxy^(-1) = x^5> and <x, y|x^16 = y^4 = 1, yxy^(-1) = x^13> are isomorphic, giving a total of 6 non-isomorphic groups.
%o (PARI) numord(n,q) = my(v=divisors(q),r=znstar(n)[2]); sum(i=1,#v,prod(j=1,#r,gcd(v[i],r[j]))*moebius(q/v[i]))
%o T(m,n) = my(u=divisors(n)); sum(i=1,#u,numord(m,u[i])/eulerphi(u[i]))
%Y Cf. A327925, A002322, A000010, A000005.
%Y Cf. also A060594, A060839, A073103, A319099, A319100, A319101, A247257.
%K nonn,tabl,changed
%O 1,8
%A _Jianing Song_, Sep 30 2019
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