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Number of numbers whose sum of non-divisors (A024816) is equal to n.
6

%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