OFFSET
1,3
COMMENTS
We have that Z_p*/(Z_p*)^k is the inverse limit of (Z/p^iZ)*/((Z/p^iZ)*)^k as i tends to infinity. Write k = p^e * k' with k' not being divisible by p. If p is odd, then the group is cyclic of order p^e * gcd(p-1,k'). If p = 2 and k is odd, then the group is trivial. If p = 2 and k is even, then the group is the product of a cyclic group of order 2^e and a cyclic group of order 2.
Each row is multiplicative.
LINKS
Jianing Song, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)
FORMULA
Write k = p^e * k' with k' not being divisible by p, and p = prime(n). If p is odd, then T(n,k) = p^e * gcd(p-1,k'). If p = 2 and k is odd, then T(n,k) = 1. If p = 2 and k is even, then T(n,k) = 2^(e+1).
EXAMPLE
Table reads
1, 4, 1, 8, 1, 4, 1, 16, 1, 4
1, 2, 3, 2, 1, 6, 1, 2, 9, 2
1, 2, 1, 4, 5, 2, 1, 4, 1, 10
1, 2, 3, 2, 1, 6, 7, 2, 3, 2
1, 2, 1, 2, 5, 2, 1, 2, 1, 10
1, 2, 3, 4, 1, 6, 1, 4, 3, 2
1, 2, 1, 4, 1, 2, 1, 8, 1, 2
1, 2, 3, 2, 1, 6, 1, 2, 9, 2
1, 2, 1, 2, 1, 2, 1, 2, 1, 2
1, 2, 1, 4, 1, 2, 7, 4, 1, 2
For p = prime(1) = 2 and k = 2, we have Z_p*/(Z_p*)^k = Z_2*/(1+8Z_2) = (Z/8Z)*/(1+8Z) = C_2 X C_2, so T(1,2) = 4.
For p = prime(2) = 3 and k = 3, we have Z_p*/(Z_p*)^k = Z_3*/((1+9Z_3) U (8+9Z_3)) = (Z/9Z)*/((1+9Z) U (8+9Z)) = C_3, so T(2,3) = 3.
PROG
(PARI) T(n, k) = my(p = prime(n), e = valuation(k, p)); p^e*gcd(p-1, k/p^e) * if(p==2 && e>=1, 2, 1)
CROSSREFS
KEYWORD
AUTHOR
Jianing Song, Apr 30 2024
STATUS
approved