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A050459
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a(n) = Sum_{d|n, d==1 mod 4} d^3 - Sum_{d|n, d==3 mod 4} d^3.
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2
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1, 1, -26, 1, 126, -26, -342, 1, 703, 126, -1330, -26, 2198, -342, -3276, 1, 4914, 703, -6858, 126, 8892, -1330, -12166, -26, 15751, 2198, -18980, -342, 24390, -3276, -29790, 1, 34580, 4914, -43092, 703, 50654, -6858, -57148, 126, 68922, 8892, -79506, -1330
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OFFSET
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1,3
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COMMENTS
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Multiplicative because it is the Inverse Möbius transform of [1 0 -3^3 0 5^3 0 -7^3 ...], which is multiplicative. - Christian G. Bower, May 18 2005
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LINKS
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FORMULA
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G.f.: Sum_{k>=1} (-1)^(k-1)*(2*k - 1)^3*x^(2*k-1)/(1 - x^(2*k-1)). - Ilya Gutkovskiy, Dec 22 2018
Multiplicative with a(2^e) = 1, and for an odd prime p, ((p^3)^(e+1)-1)/(p^3-1) if p == 1 (mod 4) and ((-p^3)^(e+1)-1)/(-p^3-1) if p == 3 (mod 4). - Amiram Eldar, Sep 27 2023
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MAPLE
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A050459 := proc(n) local a; a := 0 ; for d in numtheory[divisors](n) do if d mod 4 = 1 then a := a+d^3 ; elif d mod 4 = 3 then a := a-d^3 ; end if; end do; a ; end proc:
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MATHEMATICA
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s[n_, r_] := DivisorSum[n, #^3 &, Mod[#, 4]==r &]; a[n_] := s[n, 1] - s[n, 3]; Array[a, 30] (* Amiram Eldar, Dec 06 2018 *)
f[p_, e_] := If[Mod[p, 4] == 1, ((p^3)^(e+1)-1)/(p^3-1), ((-p^3)^(e+1)-1)/(-p^3-1)]; f[2, e_] := 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 60] (* Amiram Eldar, Sep 27 2023 *)
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CROSSREFS
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Glaisher's E_i (i=0..12): A002654, A050457, A002173, A050459, A050456, A321821, A321822, A321823, A321824, A321825, A321826, A321827, A321828.
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KEYWORD
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sign,easy,mult
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AUTHOR
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STATUS
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approved
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