%I #15 Nov 11 2022 04:34:15
%S 1,1023,59050,1047551,9765626,60408150,282475250,1072692223,
%T 3486843451,9990235398,25937424602,61857886550,137858491850,
%U 288972180750,576660215300,1098436836351,2015993900450,3567040850373,6131066257802,10229991281926,16680163512500,26533985367846,41426511213650,63342475768150
%N a(n) = Sum_{d|n} (-1)^(n/d+1)*d^10.
%H Seiichi Manyama, <a href="/A321555/b321555.txt">Table of n, a(n) for n = 1..10000</a>
%H J. W. L. Glaisher, <a href="https://books.google.com/books?id=bLs9AQAAMAAJ&pg=RA1-PA1">On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares</a>, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
%H <a href="/index/Ge#Glaisher">Index entries for sequences mentioned by Glaisher</a>.
%F G.f.: Sum_{k>=1} k^10*x^k/(1 + x^k). - _Seiichi Manyama_, Nov 25 2018
%F From _Amiram Eldar_, Nov 11 2022: (Start)
%F Multiplicative with a(2^e) = (511*2^(10*e+1)+1)/1023, and a(p^e) = (p^(10*e+10) - 1)/(p^10 - 1) if p > 2.
%F Sum_{k=1..n} a(k) ~ c * n^11, where c = 93*zeta(11)/1024 = 0.0908651... . (End)
%t f[p_, e_] := (p^(10*e + 10) - 1)/(p^10 - 1); f[2, e_] := (511*2^(10*e + 1) + 1)/1023; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* _Amiram Eldar_, Nov 11 2022 *)
%o (PARI) apply( A321555(n)=sumdiv(n, d, (-1)^(n\d-1)*d^10), [1..30]) \\ _M. F. Hasler_, Nov 26 2018
%Y Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
%Y Cf. A013669.
%K nonn,mult
%O 1,2
%A _N. J. A. Sloane_, Nov 23 2018