

A231366


Number of numbers whose sum of nondivisors (A024816) is equal to n.


5



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
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OFFSET

0,1


COMMENTS

a(n) = frequency of values n in A024816(m), where A024816(m) = sum of nondivisors 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

Table of n, a(n) for n=0..86.


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.


CROSSREFS

Cf. A054973 (number of numbers whose divisors sum to n), A231365, A231368, A231367, A231369, A067816.
Sequence in context: A154469 A037273 A285313 * A158924 A025426 A269244
Adjacent sequences: A231363 A231364 A231365 * A231367 A231368 A231369


KEYWORD

nonn


AUTHOR

Jaroslav Krizek, Nov 09 2013


STATUS

approved



