A321833 a(n) = Sum_{d|n, n/d==1 mod 4} d^9 - Sum_{d|n, n/d==3 mod 4} d^9.
1, 512, 19682, 262144, 1953126, 10077184, 40353606, 134217728, 387400807, 1000000512, 2357947690, 5159518208, 10604499374, 20661046272, 38441425932, 68719476736, 118587876498, 198349213184, 322687697778, 512000262144, 794239673292
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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Maple
f:= n -> mul(piecewise(t[1]=2,2^(9*t[2]), t[1] mod 4 = 1, (t[1]^(9*(t[2]+1))-1)/(t[1]^9-1), (t[1]^(9*(t[2]+1))+(-1)^t[2])/(t[1]^9+1)), t = ifactors(n)[2]): map(f, [$1..100]); # Robert Israel, Nov 26 2018
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Mathematica
s[n_,r_] := DivisorSum[n, #^9 &, 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^(9*e+9) - s[p]^(e+1))/(p^9 - 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^9)), [1..30]) \\ M. F. Hasler, Nov 26 2018
Formula
G.f.: Sum_{k>=1} k^9*x^k/(1 + x^(2*k)). - Ilya Gutkovskiy, Nov 26 2018
From Robert Israel, Nov 26 2018: (Start) a(2^m) = 2^(9*m).
For prime p == 1 (mod 4), a(p^m) = (p^(9(m+1))-1)/(p^9-1).
For prime p == 3 (mod 4), a(p^m) = (p^(9(m+1))+(-1)^m)/(p^9+1). (End)
From Amiram Eldar, Nov 04 2023: (Start)
Sum_{k=1..n} a(k) ~ c * n^10 / 10, where c = beta(10) = 0.99998316402... and beta is the Dirichlet beta function. (End)
a(n) = Sum_{d|n} (n/d)^9*sin(d*Pi/2). - Ridouane Oudra, Sep 27 2024