A357050 Number of ways A005101(n)+1 can be written as sum of a subset of the proper divisors of A005101(n), the n-th abundant number.

2, 1, 1, 4, 4, 7, 2, 2, 10, 2, 2, 32, 2, 1, 26, 1, 6, 24, 1, 19, 20, 2, 1, 1, 20, 4, 1, 270, 11, 14, 1, 14, 116, 12, 9, 12, 3, 195, 1, 2, 719, 1, 42, 1, 8, 9, 8, 2, 148, 142, 6, 1, 8, 6, 6, 2154, 1, 534, 1, 6, 125, 108, 1, 6, 117, 1, 447, 4

Offset: 1

Author

M. F. Hasler, Dec 13 2022

Keywords

• nonn

Comments

Obviously, for non abundant numbers (including perfect numbers) N, there is no way to write N+1 as the sum of a subset of N's proper divisors. Therefore we consider only abundant N = A005101(n) here.

The first zero appears for the seventh weird and primitive weird number A006037(7) = A002975(7) = 9272 = A005101(2310) (which surprisingly is w = A100696(1), the first weird number such that the sum of its divisors less than its abundance A033880(w) is larger than that).

Crossrefs

Cf. A005101, A006037, A002975, A005835 (abundant, weird, primitive weird and pseudoperfect numbers).

Cf. A033880 (abundance), A100696.

Programs

PARI

{A357050(n)= sum(b=1, -1+2^#d=divisors(n)[^-1], vecsum(vecextract(d,b))==n+1)} \\ not very efficient, better use code as in is_A005835().