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

0,1

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)

Links

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

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, …)

Example

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

Prog

(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

Crossrefs

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

Sequence in context: A344584 A037273 A285313 * A158924 A025426 A269244

Adjacent sequences: A231363 A231364 A231365 * A231367 A231368 A231369

Keyword

nonn,changed

Author

Jaroslav Krizek, Nov 09 2013

Extensions

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

Status

approved