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A284362
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a(n) = Sum_{d|n, d = 0, 1, or 5 mod 6} d.
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7
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1, 1, 1, 1, 6, 7, 8, 1, 1, 6, 12, 19, 14, 8, 6, 1, 18, 25, 20, 6, 8, 12, 24, 43, 31, 14, 1, 8, 30, 42, 32, 1, 12, 18, 48, 73, 38, 20, 14, 6, 42, 56, 44, 12, 6, 24, 48, 91, 57, 31, 18, 14, 54, 79, 72, 8, 20, 30, 60, 114, 62, 32, 8, 1, 84, 84, 68, 18, 24, 48, 72
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OFFSET
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1,5
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LINKS
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FORMULA
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O.g.f.: Sum_{k >= 1} ( (6*k)*x^(6*k)/(1 - x^(6*k)) + (6*k-1)*x^(6*k-1)/(1 - x^(6*k-1)) + (6*k-5)*x^(6*k-5)/(1 - x^(6*k-5)) ).
Define a(n) = 0 for n < 1. Then a(n) = e(n) + a(n-1) + a(n-5) - a(n-8) - a(n-16) + + - -, where [1, 5, 8, 16, ...] is the sequence of generalized octagonal numbers A001082, and e(n) = (-1)^(m+1)*n if n is a generalized octagonal number of the form m*(3*m+-2); otherwise e(n) = 0. Examples of this recurrence are given below. (End)
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/24 = A222171 = 0.411233... . - Amiram Eldar, Apr 12 2024
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EXAMPLE
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n = 20: n is not of the form m*(3*m +- 2), so e(n) = 0 and a(20) = a(19) + a(15) - a(12) - a(4) = 20 + 6 - 19 - 1 = 6;
n = 21: n = m*(3*m - 2) for m = 3, so e(n) = 21 and a(21) = 21 + a(20) + a(16) - a(13) - a(5) = 21 + 6 + 1 - 14 - 6 = 8;
n = 40: n = m*(3*m - 2) for m = 4, so e(n) = -40 and a(4) = -40 + a(39) + a(35) - a(32) - a(24) + a(19) + a(7) = -40 + 14 + 48 - 1 - 43 + 20 + 8 = 6. (End)
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MATHEMATICA
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Table[Sum[If[Mod[d, 6] <2 || Mod[d, 6]==5, d, 0], {d, Divisors[n]}], {n, 80}] (* Indranil Ghosh, Mar 25 2017 *)
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PROG
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(PARI) a(n) = sumdiv(n, d, ((d + 1) % 6 < 3) * d); \\ Amiram Eldar, Apr 12 2024
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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STATUS
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approved
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