A370050 Square array read by ascending antidiagonals: T(n,k) is the size of the group Z_p*/(Z_p*)^k, where p = prime(n), and Z_p is the ring of p-adic integers.

1, 1, 4, 1, 2, 1, 1, 2, 3, 8, 1, 2, 1, 2, 1, 1, 2, 3, 4, 1, 4, 1, 2, 1, 2, 5, 6, 1, 1, 2, 3, 2, 1, 2, 1, 16, 1, 2, 1, 4, 5, 6, 1, 2, 1, 1, 2, 3, 4, 1, 2, 7, 4, 9, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 3, 10, 1, 8, 1, 2, 3, 4, 1, 6, 1, 4, 1, 2, 1, 6, 1

Offset: 1

Author

Jianing Song, Apr 30 2024

Keywords

• nonn
• tabl
• easy

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.

Examples

			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.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)

Crossrefs

Cf. A370067. Row 1-4: A297402, A370180, A370181, A370182.

Programs

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)

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).