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%I #16 Nov 11 2022 04:34:20
%S 1,511,19684,261631,1953126,10058524,40353608,133955071,387440173,
%T 998047386,2357947692,5149944604,10604499374,20620693688,38445332184,
%U 68584996351,118587876498,197981928403,322687697780,510998308506,794320419872,1204911270612,1801152661464,2636771617564,3814699218751
%N a(n) = Sum_{d|n} (-1)^(n/d+1)*d^9.
%H Seiichi Manyama, <a href="/A321554/b321554.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^9*x^k/(1 + x^k). - _Seiichi Manyama_, Nov 24 2018
%F From _Amiram Eldar_, Nov 11 2022: (Start)
%F Multiplicative with a(2^e) = (255*2^(9*e+1)+1)/511, and a(p^e) = (p^(9*e+9) - 1)/(p^9 - 1) if p > 2.
%F Sum_{k=1..n} a(k) ~ c * n^10, where c = 511*zeta(10)/5120 = 0.0999039... . (End)
%t f[p_, e_] := (p^(9*e + 9) - 1)/(p^9 - 1); f[2, e_] := (255*2^(9*e + 1) + 1)/511; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* _Amiram Eldar_, Nov 11 2022 *)
%o (PARI) apply( A321554(n)=sumdiv(n, d, (-1)^(n\d-1)*d^9), [1..30]) \\ _M. F. Hasler_, Nov 26 2018
%Y Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
%Y Cf. A013668.
%K nonn,mult
%O 1,2
%A _N. J. A. Sloane_, Nov 23 2018
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