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A320005
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Number of proper divisors of n of the form 6*k + 5.
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7
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0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 2
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OFFSET
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1,55
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LINKS
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FORMULA
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a(n) = A319995(n) - [+5 = n (mod 6)], where [ ] is the Iverson bracket, giving 1 only when n = -1 mod 6, and 0 otherwise.
G.f.: Sum_{k>=1} x^(12*k-2) / (1 - x^(6*k-1)). - Ilya Gutkovskiy, Apr 14 2021
Sum_{k=1..n} a(k) = n*log(n)/6 + c*n + O(n^(1/3)*log(n)), where c = gamma(5,6) - (2 - gamma)/6 = -0.387302..., gamma(5,6) = -(psi(5/6) + log(6))/6 is a generalized Euler constant, and gamma is Euler's constant (A001620) (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023
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MATHEMATICA
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a[n_] := DivisorSum[n, 1 &, # < n && Mod[#, 6] == 5 &]; Array[a, 100] (* Amiram Eldar, Nov 25 2023 *)
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PROG
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(PARI) A320005(n) = if(!n, n, sumdiv(n, d, (d<n)*(5==(d%6))));
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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