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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.
3
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
0,3
LINKS
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.
EXAMPLE
(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 + ...
PROG
(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; }
CROSSREFS
Cf. A378007.
Sequence in context: A328290 A182878 A221971 * A297785 A126217 A334702
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Nov 14 2024
STATUS
approved