OFFSET
1,3
COMMENTS
We have Q_p* = p^Z X Z_p*, so Q_p*/(Q_p*)^k = (p^Z/p^(kZ)) X (Z_p*/(Z_p*)^k). Note that p^Z/p^(kZ) is a cyclic group of order k. For the group structure of (Z_p*/(Z_p*)^k), see A370050.
Each row is multiplicative.
LINKS
Jianing Song, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)
FORMULA
T(n,k) = k * A370050(n,k).
Write k = p^e * k' with k' not being divisible by p, and p = prime(n). If p is odd, then T(n,k) = k * p^e * gcd(p-1,k'). If p = 2 and k is odd, then T(n,k) = k. If p = 2 and k is even, then T(n,k) = k * 2^(e+1).
EXAMPLE
Table reads
1, 8, 3, 32, 5, 24, 7, 128, 9, 40
1, 4, 9, 8, 5, 36, 7, 16, 81, 20
1, 4, 3, 16, 25, 12, 7, 32, 9, 100
1, 4, 9, 8, 5, 36, 49, 16, 27, 20
1, 4, 3, 8, 25, 12, 7, 16, 9, 100
1, 4, 9, 16, 5, 36, 7, 32, 27, 20
1, 4, 3, 16, 5, 12, 7, 64, 9, 20
1, 4, 9, 8, 5, 36, 7, 16, 81, 20
1, 4, 3, 8, 5, 12, 7, 16, 9, 20
1, 4, 3, 16, 5, 12, 49, 32, 9, 20
PROG
(PARI) T(n, k) = my(p = prime(n), e = valuation(k, p)); k * 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