A123864 Expansion of (eta(q^3) * eta(q^5))^2 / (eta(q) * eta(q^15)) in powers of q.
1, 1, 2, 1, 3, 1, 2, 0, 4, 1, 2, 0, 3, 0, 0, 1, 5, 2, 2, 2, 3, 0, 0, 2, 4, 1, 0, 1, 0, 0, 2, 2, 6, 0, 4, 0, 3, 0, 4, 0, 4, 0, 0, 0, 0, 1, 4, 2, 5, 1, 2, 2, 0, 2, 2, 0, 0, 2, 0, 0, 3, 2, 4, 0, 7, 0, 0, 0, 6, 2, 0, 0, 4, 0, 0, 1, 6, 0, 0, 2, 5, 1, 0, 2, 0, 2, 0, 0, 0, 0, 2, 0, 6, 2, 4, 2, 6, 0, 2, 0, 3, 0, 4, 0, 0
Offset: 0
Examples
G.f. = 1 + q + 2*q^2 + q^3 + 3*q^4 + q^5 + 2*q^6 + 4*q^8 + q^9 + 2*q^10 + ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1000
- David Broadhurst and Daniele Dorigoni, Resurgent Lambert series with characters, arXiv:2507.21352 [math.NT], 2025. See p. 37.
- Yves Martin, Multiplicative eta-quotients, Trans. Amer. Math. Soc. 348 (1996), no. 12, 4825-4856, see page 4852 Table I.
- Michael Somos, Index to Yves Martin's list of 74 multiplicative eta-quotients and their A-numbers, 2016.
Programs
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Magma
A := Basis( ModularForms( Gamma1(15), 1), 106); A[1] + A[2] + 2*A[3] + A[4] + 3*A[5] + A[6] + 2*A[7]; /* Michael Somos, Feb 10 2015 */
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Mathematica
a[ n_] := SeriesCoefficient[ (QPochhammer[ q^3] QPochhammer[ q^5])^2 / ( QPochhammer[ q] QPochhammer[ q^15]), {q, 0, n}]; (* Michael Somos, Feb 10 2015 *) a[ n_] := If[ n < 1, Boole[n == 0], Sum[ KroneckerSymbol[ -15, d], { d, Divisors[ n]}]]; (* Michael Somos, Feb 10 2015 *) -
PARI
{a(n) = if( n<1, n==0, sumdiv( n, d, kronecker( -15, d)))}; -
PARI
{a(n) = if( n<1, n==0, (qfrep( [2, 1; 1, 8],n, 1) + qfrep( [4, 1; 1, 4], n, 1))[n])}; -
PARI
{a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^3 + A) * eta(x^5 + A))^2 / (eta(x + A) * eta(x^15 + A)), n))};
Formula
Euler transform of period 15 sequence [ 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -2, ...].
Moebius transform is period 15 sequence [ 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0, ...].
G.f. A(q) satisfies 0 = f(A(q), A(q^2), A(q^4)) where f(u, v, w) = - v^3 + 4*u*v*w - 2*u*w^2 - u^2*w.
G.f.: Product_{k>0} ((1 - x^(3*k)) * (1 - x^(5*k)))^2 / ((1 - x^k) * (1 - x^(15*k))).
G.f.: (1/2) * (Sum_{n,m in Z} x^(n^2 + n*m + 4*m^2) + x^(2*n^2 + n*m + 2*m^2)).
a(15*n + 7) = a(15*n + 11) = a(15*n + 13) = a(15*n + 14) = 0. a(3*n) = a(n).
G.f. is a period 1 Fourier series which satisfies f(-1 / (15 t)) = 15^(1/2) (t/i) f(t) where q = exp(2 Pi i t). - Michael Somos, Feb 10 2015
a(n) = Sum_{d | n} Kronecker(-15, d). - Andrew Howroyd, Jul 27 2018
From Amiram Eldar, Feb 20 2024: (Start)
Multiplicative with a(p^e) = 1 if p = 3 or 5, e + 1 if Kronecker(-15, p) = 1, and 1 - (e mod 2) if Kronecker(-15, p) = -1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2*Pi/sqrt(15). (End)
Comments