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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A378011 a(n) = b(9*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 9; 9th column of A378007.

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%I A378011 #12 Nov 14 2024 23:23:10
%S A378011 1,0,6,0,6,0,0,1,6,0,0,0,6,0,6,0,0,0,6,0,6,0,6,0,0,0,0,0,0,0,6,0,3,0,
%T A378011 6,0,0,0,2,0,21,0,6,0,6,0,0,0,6,0,0,0,0,0,6,0,0,0,6,0,6,0,0,0,6,0,0,0,
%U A378011 6,0,6,0,0,0,0,0,0,0,36,0,0,0,6,0,6,0,0,0,0,0,6
%N A378011 a(n) = b(9*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 9; 9th column of A378007.
%H A378011 Jianing Song, <a href="/A378011/b378011.txt">Table of n, a(n) for n = 0..10000</a>
%F A378011 a(n) = b(9*n+1), where {b(n)} is multiplicative with:
%F A378011 - b(3^e) = 0;
%F A378011 - for p == 1 (mod 9), b(p^e) = binomial(e+5,5) = (e+5)*(e+4)*(e+3)*(e+2)*(e+1)/120;
%F A378011 - for p == 8 (mod 9), b(p^e) = binomial(e/2+2,2) = (e/2+2)*(e/2+1)/2 if e is even, and 0 otherwise;
%F A378011 - for p == 4, 7 (mod 9), b(p^e) = e/3 + 1 if 3 divides e, and 0 otherwise;
%F A378011 - for p == 2, 5 (mod 9), b(p^e) = 1 if 6 divides e, and 0 otherwise.
%e A378011 Write w = exp(2*Pi*i/3) = (-1 + sqrt(3)*i)/2, then (1 + 1/2^s + 1/4^s + 1/5^s + 1/7^s + 1/8^s + ...)*(1 + (w+1)/2^s + w/4^s - w/5^s - (w+1)/7^s - 1/8^s + ...)*(1 + w/2^s - (w+1)/4^s - (w+1)/5^s + w/7^s + 1/8^s + ...)*(1 - 1/2^s + 1/4^s - 1/5^s + 1/7^s - 1/8^s + ...)*(1 - (w+1)/2^s + w/4^s + w/5^s - (w+1)/7^s + 1/8^s + ...)*(1 - w/2^s - (w+1)/4^s + (w+1)/5^s + w/7^s - 1/8^s + ...) = 1 + 6/19^s + 6/37^s + 1/64^s + 6/73^s + ...
%o A378011 (PARI) A378011(n) = {
%o A378011 my(f = factor(9*n+1), res = 1); for(i=1, #f~,
%o A378011 if(f[i,1] % 9 == 1, res *= binomial(f[i,2]+5, 5));
%o A378011 if(f[i,1] % 9 == 8, if(f[i,2] % 2 == 0, res *= binomial(f[i,2]/2+2, 2), return(0)));
%o A378011 if(f[i,1] % 9 == 4 || f[i,1] % 9 == 7, if(f[i,2] % 3 == 0, res *= f[i,2]/3+1, return(0)));
%o A378011 if(f[i,1] % 9 == 2 || f[i,1] % 9 == 5, if(f[i,2] % 6 != 0, return(0))));
%o A378011 res; }
%Y A378011 Cf. A378007.
%K A378011 nonn,easy
%O A378011 0,3
%A A378011 _Jianing Song_, Nov 14 2024