A321836 a(n) = Sum_{d|n, n/d==1 mod 4} d^12 - Sum_{d|n, n/d==3 mod 4} d^12.
1, 4096, 531440, 16777216, 244140626, 2176778240, 13841287200, 68719476736, 282429005041, 1000000004096, 3138428376720, 8916083671040, 23298085122482, 56693912371200, 129746094281440, 281474976710656, 582622237229762, 1156829204647936
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
- J. W. L. Glaisher, On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
- Index entries for sequences mentioned by Glaisher.
Crossrefs
Programs
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Mathematica
s[n_,r_] := DivisorSum[n, #^12 &, Mod[n/#,4]==r &]; a[n_] := s[n,1] - s[n,3]; Array[a, 30] (* Amiram Eldar, Nov 26 2018 *) s[n_] := If[OddQ[n], (-1)^((n-1)/2), 0]; (* A101455 *) f[p_, e_] := (p^(12*e+12) - s[p]^(e+1))/(p^12 - s[p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2023 *)
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PARI
apply( a(n)=sumdiv(n, d, if(bittest(n\d,0),(2-n\d%4)*d^12)), [1..30]) \\ M. F. Hasler, Nov 26 2018
Formula
G.f.: Sum_{k>=1} k^12*x^k/(1 + x^(2*k)). - Ilya Gutkovskiy, Nov 26 2018
From Amiram Eldar, Nov 04 2023: (Start)
Sum_{k=1..n} a(k) ~ c * n^13 / 13, where c = beta(13) = 540553*Pi^13/1569592442880 = 0.999999373583..., and beta is the Dirichlet beta function. (End)
a(n) = Sum_{d|n} (n/d)^12*sin(d*Pi/2). - Ridouane Oudra, Sep 27 2024
Comments