%I #16 Jan 19 2025 09:26:29
%S 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,
%T 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,
%U 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
%N Number of numbers whose sum of non-divisors (A024816) is equal to n.
%C a(n) = frequency of values n in A024816(m), where A024816(m) = sum of non-divisors of m = antisigma(m).
%C From _Charles R Greathouse IV_, Nov 11 2013: (Start)
%C So far all n such that a(n) > 1 correspond to members of A067816:
%C a(0) = 2 from 1, 2;
%C a(9) = 2 from 5, 6;
%C a(36844389) = 2 from 8585, 8586;
%C a(129894940) = 2 from 16119, 16120;
%C a(446591224981504) = 2 from 29886159, 29886160.
%C I checked this, and thus Krizek's conjecture below, up to 4*10^19.
%C (End)
%H Antti Karttunen, <a href="/A231366/b231366.txt">Table of n, a(n) for n = 0..32001</a>
%F Conjecture: max a(n) = 2.
%F a(A231368(n)) >= 1, a(A231369(n)) = 0.
%F a(n) = 0 for such n that A231367(n) = 0, a(n) = 0 if A024816(m) = n has no solution.
%F a(n) >= 1 for such n that A231367(n) = 1, a(n) >= 1 if A024816(m) = n for any m.
%F Conjecture: a(n) = 2 iff n is number from A225775 (0, 9, 36844389, 129894940, 446591224981504, …)
%e a(9) = 2 because there are two numbers m (5, 6) with antisigma(m) = 9.
%o (PARI)
%o up_to = 105;
%o A024816(n) = (n*(n+1)/2-sigma(n));
%o 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); };
%o v231366 = A231366list(up_to);
%o A231366(n) = v231366[1+n]; \\ _Antti Karttunen_, Jan 19 2025
%Y Cf. A054973 (number of numbers whose divisors sum to n), A231365, A231368, A231367, A231369, A067816.
%K nonn
%O 0,1
%A _Jaroslav Krizek_, Nov 09 2013
%E Data section extended to a(105) by _Antti Karttunen_, Jan 19 2025