A350743 Number of numbers k, 1 <= k <= n, such that k | sigma_k(n).
1, 1, 2, 1, 3, 3, 2, 3, 1, 5, 4, 5, 4, 5, 6, 1, 5, 4, 3, 7, 4, 5, 5, 8, 1, 7, 6, 6, 8, 9, 3, 5, 8, 7, 9, 3, 4, 10, 6, 13, 7, 8, 4, 8, 9, 7, 9, 5, 3, 5, 10, 10, 7, 13, 8, 14, 8, 12, 10, 18, 3, 10, 7, 1, 14, 10, 5, 16, 11, 12, 5, 12, 6, 9, 10, 10, 8, 14, 5, 11, 2, 13, 13, 15, 14
Offset: 1
Keywords
Examples
a(3) = 2; we have 1 | sigma_1(3) = 1 + 3 = 4 and 2 | sigma_2(3) = 1^2 + 3^2 = 10. (Note that 3 does not divide sigma_3(3) = 1^3 + 3^3 = 28.)
Programs
-
Mathematica
Table[n - Sum[Ceiling[DivisorSigma[k, n]/k] - Floor[DivisorSigma[k, n]/k], {k, n}], {n, 100}] -
PARI
a(n) = sum(k=1, n, (sigma(n,k) % k) == 0); \\ Michel Marcus, Jan 13 2022
-
Python
from math import prod from sympy import factorint def A350743(n): f = list(factorint(n).items()) return sum(1 for k in range(1,n+1) if prod(p**((q+1)*k)-1 for p,q in f)//prod(p**k-1 for p,q in f) % k == 0) # Chai Wah Wu, Jan 17 2022
Formula
a(n) = n - Sum_{k=1..n} (ceiling(sigma_k(n)/k) - floor(sigma_k(n)/k)).