A378008 a(n) = b(5*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 5; 5th column of A378007.

1, 0, 4, 1, 0, 0, 4, 0, 4, 0, 0, 0, 4, 0, 4, 0, 1, 0, 0, 0, 4, 0, 0, 0, 10, 0, 4, 0, 0, 0, 4, 0, 0, 0, 0, 4, 4, 0, 4, 0, 0, 0, 4, 0, 0, 0, 0, 0, 4, 0, 4, 1, 0, 0, 4, 0, 4, 0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 0, 16, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 0, 4, 0, 0, 0, 16

Offset: 0

Author

Jianing Song, Nov 14 2024

Keywords

• nonn
• easy

Examples

			(1 + 1/2^s + 1/3^s + 1/4^s + ...)*(1 + i/2^s - i/3^s - 1/4^s + ...)*(1 - 1/2^s - 1/3^s + 1/4^s + ...)*(1 - i/2^s + i/3^s - 1/4^s + ...) = 1 + 4/11^s + 1/16^s + 4/31^s + 4/41^s + ...
		

Links

• Jianing Song, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A378007.

Programs

PARI

A378008(n) = {
my(f = factor(5*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 5 == 1, res *= binomial(f[i,2]+3, 3));
if(f[i,1] % 5 == 4, if(f[i,2] % 2 == 0, res *= f[i,2]/2+1, return(0)));
if(f[i,1] % 5 == 2 || f[i,1] % 5 == 3, if(f[i,2] % 4 != 0, return(0))));
res; }

Formula

a(n) = b(5*n+1), where {b(n)} is multiplicative with:

- b(5^e) = 0;

- for p == 1 (mod 5), b(p^e) = binomial(e+3,3) = (e+3)*(e+2)*(e+1)/6;

- for p == 4 (mod 5), b(p^e) = e/2 + 1 if e is even, and 0 otherwise;

- for p == 2, 3 (mod 5), b(p^e) = 1 if 4 divides e, and 0 otherwise.