A231366 Number of numbers whose sum of non-divisors (A024816) is equal to n.

2, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Author

Jaroslav Krizek, Nov 09 2013

Keywords

• nonn

Comments

a(n) = frequency of values n in A024816(m), where A024816(m) = sum of non-divisors of m = antisigma(m).

From Charles R Greathouse IV, Nov 11 2013: (Start)

So far all n such that a(n) > 1 correspond to members of A067816:

a(0) = 2 from 1, 2;

a(9) = 2 from 5, 6;

a(36844389) = 2 from 8585, 8586;

a(129894940) = 2 from 16119, 16120;

a(446591224981504) = 2 from 29886159, 29886160.

I checked this, and thus Krizek's conjecture below, up to 4*10^19.

(End)

Examples

			a(9) = 2 because there are two numbers m (5, 6) with antisigma(m) = 9.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..32001

Crossrefs

Cf. A054973 (number of numbers whose divisors sum to n), A231365, A231368, A231367, A231369, A067816.

Programs

PARI

up_to = 105;
A024816(n) = (n*(n+1)/2-sigma(n));
A231366list(up_to) = { my(v=vector(1+up_to), u); for(n=1, 2+up_to, if((u = A024816(n))<=up_to, v[1+u]++)); (v); };
v231366 = A231366list(up_to);
A231366(n) = v231366[1+n]; \\ Antti Karttunen, Jan 19 2025

Formula

Conjecture: max a(n) = 2.

a(A231368(n)) >= 1, a(A231369(n)) = 0.

a(n) = 0 for such n that A231367(n) = 0, a(n) = 0 if A024816(m) = n has no solution.

a(n) >= 1 for such n that A231367(n) = 1, a(n) >= 1 if A024816(m) = n for any m.

Conjecture: a(n) = 2 iff n is number from A225775 (0, 9, 36844389, 129894940, 446591224981504, …)

Extensions

Data section extended to a(105) by Antti Karttunen, Jan 19 2025