A294929 Number of proper divisors of n that are abundant (A005101).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 6

Offset: 1

Author

Antti Karttunen, Nov 14 2017

Keywords

• nonn

Examples

			The proper divisors of 24 are 1, 2, 3, 4, 6, 8, 12. Only one of these, 12, is abundant (in A005101), thus a(24) = 1.
The proper divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60. Six of these are abundant: 12, 20, 24, 30, 40, 60, thus a(120) = 6.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A005101, A032741, A080224, A294889, A294926, A294927, A294928, A294930, A294937.

Cf. also A294902.

Programs

Mathematica

a[n_] := Count[Most[Divisors[n]], ?(DivisorSigma[1, #] > 2*# &)]; Array[a, 100] (* _Amiram Eldar, Mar 14 2024 *)

PARI

A294929(n) = sumdiv(n, d, (d(2*d)));

Formula

a(n) = Sum_{d|n, dA294937(d).

a(n) = A080224(n) - A294937(n).

a(n) + A294928(n) = A032741(n).