A327924 -id:A327924 - cp's OEIS frontend

A354057 Square array read by ascending antidiagonals: T(n,k) is the number of solutions to x^k == 1 (mod n).

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, 4, 1, 2, 1, 1, 1, 4, 3, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 3, 4, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 6, 1, 4, 1, 2, 1, 1, 1, 4, 1, 4, 1, 4, 1, 2, 1, 2, 1, 1, 1

Offset: 1

Jianing Song, May 16 2022

Row n and Row n' are the same if and only if (Z/nZ)* = (Z/n'Z)*, where (Z/nZ)* is the multiplicative group of integers modulo n.
Given n, T(n,k) only depends on gcd(k,psi(n)). For the truncated version see A354060.
Each column is multiplicative.

			  n/k  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
   1   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   2   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   3   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   4   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   5   1  2  1  4  1  2  1  4  1  2  1  4  1  2  1  4  1  2  1  4
   6   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   7   1  2  3  2  1  6  1  2  3  2  1  6  1  2  3  2  1  6  1  2
   8   1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4
   9   1  2  3  2  1  6  1  2  3  2  1  6  1  2  3  2  1  6  1  2
  10   1  2  1  4  1  2  1  4  1  2  1  4  1  2  1  4  1  2  1  4
  11   1  2  1  2  5  2  1  2  1 10  1  2  1  2  5  2  1  2  1 10
  12   1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4
  13   1  2  3  4  1  6  1  4  3  2  1 12  1  2  3  4  1  6  1  4
  14   1  2  3  2  1  6  1  2  3  2  1  6  1  2  3  2  1  6  1  2
  15   1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8
  16   1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8
  17   1  2  1  4  1  2  1  8  1  2  1  4  1  2  1 16  1  2  1  4
  18   1  2  3  2  1  6  1  2  3  2  1  6  1  2  3  2  1  6  1  2
  19   1  2  3  2  1  6  1  2  9  2  1  6  1  2  3  2  1 18  1  2
  20   1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8  1  4  1  8

Jianing Song, Table of n, a(n) for n = 1..5050 (the first 100 ascending diagonals)

k-th column: A060594 (k=2), A060839 (k=3), A073103 (k=4), A319099 (k=5), A319100 (k=6), A319101 (k=7), A247257 (k=8).
Applying Moebius transform to the rows gives A354059.
Applying Moebius transform to the columns gives A354058.
Cf. A327924.

T(n,k)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(k, Z[i]))

If (Z/nZ)* = C_{k_1} X C_{k_2} X ... X C_{k_r}, then T(n,k) = Product_{i=1..r} gcd(k,k_r).
T(p^e,k) = gcd((p-1)*p^(e-1),k) for odd primes p. T(2,k) = 1, T(2^e,k) = 2*gcd(2^(e-2),k) if k is even and 1 if k is odd.
A327924(n,k) = Sum_{q|n} T(n,k) * (Sum_{s|n/q} mu(s)/phi(s*q)).

A327925 Irregular table read by rows: 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, 1 <= n <= A002322(m).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 1, 4, 1, 4, 1, 2, 2, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 1, 2, 1, 4, 1, 4, 1, 2, 2, 3, 1, 4, 1, 3, 2, 2, 1, 6, 1, 2, 2, 2, 1, 4, 1, 4, 1, 6, 1, 4, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 2, 2, 1, 4, 1, 2, 2, 2, 1, 4, 1, 2, 3, 2, 1, 4, 1, 2, 2, 2, 1, 6

Offset: 1

Jianing Song, Sep 30 2019

The semidirect product of C_m and C_n has group representation G = , where r is any number such that r^n == 1 (mod m). Two groups G =  and G' =  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'.
Given m, T(m,n) only depends on the value of gcd(n,psi(m)), psi = A002322 (Carmichael lambda). So each row of A327924 is periodic with period psi(m), so we have this for an alternative version.
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, ...
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

			Table starts
m = 1: 1;
m = 2: 1;
m = 3: 1, 2;
m = 4: 1, 2;
m = 5: 1, 2, 1, 3;
m = 6: 1, 2;
m = 7: 1, 2, 2, 2, 1, 4;
m = 8: 1, 4;
m = 9: 1, 2, 2, 2, 1, 4;
m = 10: 1, 2, 1, 3;
m = 11: 1, 2, 1, 2, 2, 2, 1, 2, 1, 4;
m = 12: 1, 4;
m = 13: 1, 2, 2, 3, 1, 4, 1, 3, 2, 2, 1, 6;
m = 14: 1, 2, 2, 2, 1, 4;
m = 15: 1, 4, 1, 6;
m = 16: 1, 4, 1, 6;
m = 17: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5;
m = 18: 1, 2, 2, 2, 1, 4;
m = 19: 1, 2, 2, 2, 1, 4, 1, 2, 3, 2, 1, 4, 1, 2, 2, 2, 1, 6;
m = 20: 1, 4, 1, 6;
Example shows that T(21,6) = 6: The semidirect product of C_21 and C_6 has group representation G = <x, y|x^21 = y^6 = 1, yxy^(-1) = x^r>, where r = 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20. Since 2^5 == 11 (mod 21), 4^5 == 16 (mod 21), 5^5 == 17 (mod 21), 10^5 == 19 (mod 21), there are actually four pairs of isomorphic groups, giving a total of 8 non-isomorphic groups.

Jianing Song, Table of n, a(n) for n = 1..8346 (the first 200 rows)
Math Overflow, When are two semidirect products of two cyclic groups isomorphic

Cf. A327925, A002322, A000010, A000005.

Cf. also A060594, A060839, A073103, A319099, A319100, A319101, A051194.

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]))
T(m,n) = my(u=divisors(n)); sum(i=1,#u,numord(m,u[i])/eulerphi(u[i]))
Row(m) = my(l=if(m>2,znstar(m)[2][1],1), R=vector(l,n,T(m,n))); R

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.
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
For odd primes p, T(p^e,n) = d(gcd(n,(p-1)*p^(e-1))) = A051194((p-1)*p^(e-1),n), d = A000005; for e >= 3, T(2^e,n) = 2*(v2(n)+1) for even n and 1 for odd n, where v2 is the 2-adic valuation.

A354059 Square array read by ascending antidiagonals: T(n,k) is the number of elements in the multiplicative group of integers modulo n that have order k.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 2, 0, 0, 0, 0, 1, 3, 2, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Jianing Song, May 16 2022

Row n and Row n' are the same if and only if (Z/nZ)* = (Z/n'Z)*, where (Z/nZ)* is the multiplicative group of integers modulo n.

For the truncated version see A252911.

			The 7th, 9th, 14th and 18th rows of A354047 are {1,2,3,2,1,6,1,2,3,2,1,6,...}, so applying the Moebius transform gives {1,1,2,0,0,2,0,0,0,0,0,0,...}.

Jianing Song, Table of n, a(n) for n = 1..5050

Moebius transform of A354057 applied to each row.

Cf. A327924.

b(n,k)=my(Z=znstar(n)[2]); prod(i=1, #Z, gcd(k, Z[i]));
T(n,k) = sumdiv(k, d, moebius(k/d)*b(n,d))

A327924(n,k) = Sum_{d|k} T(n,k)/phi(d).

cp's OEIS Frontend

A354057 Square array read by ascending antidiagonals: T(n,k) is the number of solutions to x^k == 1 (mod n).

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A327925 Irregular table read by rows: 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, 1 <= n <= A002322(m).

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A354059 Square array read by ascending antidiagonals: T(n,k) is the number of elements in the multiplicative group of integers modulo n that have order k.

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