A082451 Sum over divisors d of n of Kronecker symbol (-60, d).
1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 2, 1, 2, 1, 0, 0, 2, 1, 1, 0, 1, 0, 0, 1, 2, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 1, 2, 2, 1, 1, 1, 2, 0, 2, 1, 0, 0, 2, 0, 0, 1, 2, 2, 0, 1, 0, 0, 0, 2, 2, 0, 0, 1, 0, 0, 1, 2, 0, 0, 2, 1, 1, 0, 2, 0, 2, 0
Offset: 1
Examples
G.f. = q + q^2 + q^3 + q^4 + q^5 + q^6 + q^8 + q^9 + q^10 + q^12 + q^15 + q^16 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 1..10000
- Michael Somos, Introduction to Ramanujan theta functions, 2019.
- Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.
Programs
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Mathematica
a[ n_] := If[ n < 1, 0, DivisorSum[n, KroneckerSymbol[-60, #] &]]; (* Michael Somos, Nov 15 2015 *) a[ n_] := If[ n < 1, 0, Times @@ (Which[# < 7, 1, KroneckerSymbol[-60, #] == 1, #2 + 1, True, 1 - Mod[#2, 2]] & @@@ FactorInteger[n])]; (* Michael Somos, Nov 15 2015 *) a[ n_] := SeriesCoefficient[ q QPochhammer[ q^3] QPochhammer[ q^5] QPochhammer[ -q, q] QPochhammer[ -q^15, q^15], {q, 0, n}]; (* Michael Somos, Nov 15 2015 *)
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PARI
{a(n) = if( n<1, 0, sumdiv(n, d, kronecker(-60, d)))}; -
PARI
{a(n) = if( n<1, 0, direuler(p=2, n, 1 / ((1 - X) * (1 - kronecker(-60, p)*X))[n]))}; -
PARI
{a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( eta(x^2 + A) * eta(x^3 + A) * eta(x^5 + A) * eta(x^30 + A) / (eta(x + A) * eta(x^15 + A)), n))};
Formula
Expansion of q * f(-q^3) * f(-q^5) / (chi(-q) * chi(-q^15)) in powers of q where f(), chi() are Ramanujan theta functions.
Expansion of eta(q^2) * eta(q^3) * eta(q^5) * eta(q^30) / (eta(q) * eta(q^15)) in powers of q.
Euler transform of period 30 sequence [ 1, 0, 0, 0, 0, -1, 1, 0, 0, -1, 1, -1, 1, 0, 0, 0, 1, -1, 1, -1, 0, 0, 1, -1, 0, 0, 0, 0, 1, -2, ...].
a(n) is multiplicative with a(2^e) = a(3^e) = a(5^e) = 1, a(p^e) = e+1 if p == 1, 2, 4, 8 (mod 15), a(p^e) = (1 + (-1)^e)/2 if p == 7, 11, 13, 14 (mod 15).
G.f.: x * Product_{k>0} (1 + x^k) * (1 - x^(3*k)) * (1 - x^(5*k)) * (1 + x^(15*k)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/sqrt(15) = 0.811155... . - Amiram Eldar, Nov 16 2023
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