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A306653
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a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*A008683(n/k)*n/k*[k divides m + 2^p]*A008683(k)*k, where p can be a positive integer: 1,2,3,4,5,...
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3
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1, -2, -2, 2, -2, 4, -2, 0, 0, 4, -2, -4, -2, 4, 4, 0, -2, 0, -2, -4, 4, 4, -2, 0, 0, 4, 0, -4, -2, -8, -2, 0, 4, 4, 4, 0, -2, 4, 4, 0, -2, -8, -2, -4, 0, 4, -2, 0, 0, 0, 4, -4, -2, 0, 4, 0, 4, 4, -2, 8, -2, 4, 0, 0, 4, -8, -2, -4, 4, -8, -2, 0, -2, 4, 0, -4, 4, -8, -2, 0, 0, 4, -2, 8, 4
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OFFSET
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1,2
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COMMENTS
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It appears that for p=1 and p=2, a(n) is identically the same for all n except for n equal to multiples of 16. See A306652 for comparison.
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LINKS
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FORMULA
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a(n) = 1/n * Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*[k divides m + 2^p]*A008683(n/k)*n/k*A008683(k)*k, where p can be any positive integer: 1,2,3,4,5,...
Dirichlet generating function, after Daniel Suteu in A298826 and Álvar Ibeas in A076479, appears to be: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + Sum_{n>=2} (1/2/2^(n*(s - 1)))). - Mats Granvik, Apr 07 2019
The conjectured Dirichlet generating function simplifies to: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + 2^(1 - s)/(2^s - 2)). - Steven Foster Clark, Sep 23 2022
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MATHEMATICA
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(* Dirichlet Convolution. *)
p=1;
a[n_] := 1/n*Sum[Sum[If[Mod[n, k] == 0, 1, 0]*If[Mod[m, n/k] == 0, 1, 0]*If[Mod[m + 2^p, k] == 0, 1, 0]*MoebiusMu[n/k]*n/k*MoebiusMu[k]*k, {k, 1, n}], {m, 1, n}]; a /@ Range[85]
(* conjectured faster program *)
nn = 85;
b = 4*Select[Range[1, nn, 2], SquareFreeQ];
bb = Table[DivisorSigma[0, n]*(MoebiusMu[n] + Sum[If[b[[j]] == n, LiouvilleLambda[n]*2/3, 0], {j, 1, Length[b]}]), {n, 1, nn}]
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PROG
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(PARI) A306653(n) = (1/n)*sum(m=1, n, sumdiv(n, k, if( !(m%(n/k)) && !((m+(2^1))%k), n*moebius(n/k)*moebius(k), 0))); \\ Antti Karttunen, Mar 15 2019
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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