cp's OEIS Frontend

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.

Showing 1-2 of 2 results.

A159818 Expansion of f(q) * f(q^5) in powers of q where f() is a Ramanujan theta function.

Original entry on oeis.org

1, 1, -1, 0, 0, 0, 1, -2, 0, 0, -2, -1, -1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 2, 0, 0, -2, 0, 2, 0, 0, 1, 1, 0, 0, 0, 0, -2, 2, 0, 0, 0, 0, 1, 0, 0, -2, 0, 0, 0, 0, 0, -2, 0, 0, 0, 0, -1, -2, 0, 0, -2, -1, 0, 0, 0, 2, 0, 2, 0, 0, -2, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -2, 0, 0, 1, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 1, -2, 0, 0
Offset: 0

Views

Author

Michael Somos, Apr 22 2009

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Number 71 of the 74 eta-quotients listed in Table I of Martin (1996).

Examples

			G.f. = 1 + x - x^2 + x^6 - 2*x^7 - 2*x^10 - x^11 - x^12 + 2*x^15 + x^20 + ...
G.f. = q + q^5 - q^9 + q^25 - 2*q^29 - 2*q^41 - q^45 - q^49 + 2*q^61 + q^81 + ...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ QPochhammer[ -x] QPochhammer[ -x^5], {x, 0, n}]; (* Michael Somos, Jun 10 2015 *)
  • PARI
    {a(n) = my(A, p, e, x, z); if(n<0, 0, n = 4*n + 1; A = factor(n); prod(k=1, matsize(A)[1], [p, e] = A[k,]; if(p==2, 0, p==5, 1, p%20>10, !(e%2), p%4==3, kronecker(-4, e+1), for(y=1, sqrtint(p\5), if(issquare(p - 5*y^2, &x), z = if(x%2, y, x)%4/2; break)); (-1)^(e*z) *(e+1))))};
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^3 * eta(x^10 + A)^3 / (eta(x + A) * eta(x^4 + A) * eta(x^5 + A) * eta(x^20 + A)), n))};

Formula

Expansion of q^(-1/4) * eta(q^2)^3 * eta(q^10)^3 / (eta(q) * eta(q^4) * eta(q^5) * eta(q^20)) in powers of q.
Euler transform of period 20 sequence [ 1, -2, 1, -1, 2, -2, 1, -1, 1, -4, 1, -1, 1, -2, 2, -1, 1, -2, 1, -2, ...].
a(n) = b(4*n + 1) where b(n) is multiplicative with b(2^e) = 0^e, b(5^e) = 1, b(p^e) = (1 + (-1)^e) / 2 if p == 11, 13, 17, 19 (mod 20), b(p^e) = (-1)^(e/2) * (1 + (-1)^e) / 2 if p == 3, 7 (mod 20), b(p^e) = (-1)^(e*z) * (e+1) if p == 1, 9 (mod 20) where p = x^2 + 5*y^2 and z = 1 if x or y == 0 (mod 4) else z = 0.
G.f. is a period 1 Fourier series which satisfies f(-1 / (320 t)) = (320)^(1/2) (t/i) f(t) where q = exp(2 Pi i t).
G.f.: Product_{k>0} (1 - (-x)^k) * (1 - (-x)^(5*k)).
a(n) = (-1)^n * A030202(n). Convolution square is A159817. a(5*n + 3) = a(5*n + 4) = a(9*n + 5) = a(9*n + 8) = 0. a(9*n + 2) = -a(n).

A318028 Expansion of Product_{k>=1} 1/((1 - x^k)*(1 - x^(5*k))).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 12, 17, 25, 35, 51, 69, 96, 129, 175, 235, 312, 410, 539, 700, 913, 1173, 1508, 1923, 2450, 3105, 3920, 4926, 6177, 7710, 9614, 11923, 14766, 18218, 22435, 27550, 33750, 41231, 50278, 61150, 74259, 89932, 108744, 131193, 158025, 189979, 227998, 273125, 326692
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 13 2018

Keywords

Comments

Convolution of A000712 and A145466.
Convolution inverse of A030202.
Number of partitions of n if there are 2 kinds of parts that are multiples of 5.

Examples

			a(5) = 8 because we have [5], [5'], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1] and [1, 1, 1, 1, 1].
		

Crossrefs

Programs

  • Maple
    a:=series(mul(1/((1-x^k)*(1-x^(5*k))),k=1..55),x=0,49): seq(coeff(a,x,n),n=0..48); # Paolo P. Lava, Apr 02 2019
  • Mathematica
    nmax = 48; CoefficientList[Series[Product[1/((1 - x^k) (1 - x^(5 k))), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 48; CoefficientList[Series[1/(QPochhammer[x] QPochhammer[x^5]), {x, 0, nmax}], x]
    nmax = 48; CoefficientList[Series[Exp[Sum[x^k (1 + x^k + x^(2 k) + x^(3 k) + 2 x^(4 k))/(k (1 - x^(5 k))), {k, 1, nmax}]], {x, 0, nmax}], x]
    Table[Sum[PartitionsP[k] PartitionsP[n - 5 k], {k, 0, n/5}], {n, 0, 48}]

Formula

G.f.: exp(Sum_{k>=1} x^k*(1 + x^k + x^(2*k) + x^(3*k) + 2*x^(4 k))/(k*(1 - x^(5*k)))).
a(n) ~ exp(2*Pi*sqrt(n/5)) / (4 * 5^(1/4) * n^(5/4)). - Vaclav Kotesovec, Aug 14 2018
Showing 1-2 of 2 results.