A321829 a(n) = Sum_{d|n, n/d==1 mod 4} d^5 - Sum_{d|n, n/d==3 mod 4} d^5.
1, 32, 242, 1024, 3126, 7744, 16806, 32768, 58807, 100032, 161050, 247808, 371294, 537792, 756492, 1048576, 1419858, 1881824, 2476098, 3201024, 4067052, 5153600, 6436342, 7929856, 9768751, 11881408, 14290100, 17209344, 20511150, 24207744, 28629150, 33554432, 38974100, 45435456, 52535556
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, # ^5 &, 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^(5*e+5) - s[p]^(e+1))/(p^5 - s[p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2023 *)
-
PARI
apply( A321829(n)=factorback(apply(f->f[1]^(5*f[2]+5)\/(f[1]^5+f[1]%4-2),Col(factor(n)))), [1..40]) \\ M. F. Hasler, Nov 26 2018
Formula
G.f.: Sum_{k>=1} k^5*x^k/(1 + x^(2*k)). - Ilya Gutkovskiy, Nov 26 2018
Multiplicative with a(p^e) = round(p^(5e+5)/(p^5 + p%4 - 2)), where p%4 is the remainder of p modulo 4. (Following R. Israel in A321833.) - M. F. Hasler, Nov 26 2018
Sum_{k=1..n} a(k) ~ c * n^6 / 6, where c = A175570. - Amiram Eldar, Nov 04 2023
a(n) = Sum_{d|n} (n/d)^5*sin(d*Pi/2). - Ridouane Oudra, Sep 27 2024