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a(n) = Sum_{d|n} (-1)^(n/d+1)*d^8.
7

%I #16 Nov 11 2022 04:34:27

%S 1,255,6562,65279,390626,1673310,5764802,16711423,43053283,99609630,

%T 214358882,428360798,815730722,1470024510,2563287812,4278124287,

%U 6975757442,10978587165,16983563042,25499674654,37828630724,54661514910,78310985282,109660357726,152588281251,208011334110,282472589764

%N a(n) = Sum_{d|n} (-1)^(n/d+1)*d^8.

%H Seiichi Manyama, <a href="/A321553/b321553.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&amp;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^8*x^k/(1 + x^k). - _Seiichi Manyama_, Nov 23 2018

%F From _Amiram Eldar_, Nov 11 2022: (Start)

%F Multiplicative with a(2^e) = (127*2^(8*e+1)+1)/255, and a(p^e) = (p^(8*e+8) - 1)/(p^8 - 1) if p > 2.

%F Sum_{k=1..n} a(k) ~ c * n^9, where c = 85*zeta(9)/768 = 0.110899... . (End)

%t Table[Total[(-1)^(n/#+1) #^8&/@Divisors[n]],{n,30}] (* _Harvey P. Dale_, May 05 2021 *)

%t f[p_, e_] := (p^(8*e + 8) - 1)/(p^8 - 1); f[2, e_] := (127*2^(8*e + 1) + 1)/255; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* _Amiram Eldar_, Nov 11 2022 *)

%o (PARI) apply( A321553(n)=sumdiv(n, d, (-1)^(n\d-1)*d^8), [1..30]) \\ _M. F. Hasler_, Nov 26 2018

%Y Cf. A321543 - A321565, A321807 - A321836 for similar sequences.

%Y Cf. A013667.

%K nonn,mult

%O 1,2

%A _N. J. A. Sloane_, Nov 23 2018