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 1-4 of 4 results.

A208274 Expansion of phi(q) / phi(q^4) in powers of q where phi() is a Ramanujan theta function.

Original entry on oeis.org

1, 2, 0, 0, 0, -4, 0, 0, 0, 10, 0, 0, 0, -20, 0, 0, 0, 36, 0, 0, 0, -64, 0, 0, 0, 110, 0, 0, 0, -180, 0, 0, 0, 288, 0, 0, 0, -452, 0, 0, 0, 692, 0, 0, 0, -1044, 0, 0, 0, 1554, 0, 0, 0, -2276, 0, 0, 0, 3296, 0, 0, 0, -4724, 0, 0, 0, 6696, 0, 0, 0, -9408, 0, 0
Offset: 0

Views

Author

Michael Somos, Mar 12 2012

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Differs from A127391 only at n=0. - R. J. Mathar, Mar 18 2012

Examples

			1 + 2*q - 4*q^5 + 10*q^9 - 20*q^13 + 36*q^17 - 64*q^21 + 110*q^25 - 180*q^29 + ...
		

Crossrefs

Programs

  • Mathematica
    a[n_]:= SeriesCoefficient[EllipticTheta[3, 0, q]/EllipticTheta[3, 0, q^4], {q, 0, n}]; Table[a[n], {n, 0, 50}] (* G. C. Greubel, Dec 04 2017 *)
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^5 * eta(x^16 + A)^2 / (eta(x + A)^2 * eta(x^8 + A)^5), n))}

Formula

Expansion of eta(q^2)^5 * eta(q^16)^2 / (eta(q)^2 * eta(q^8)^5) in powers of q.
Euler transform of period 16 sequence [ 2, -3, 2, -3, 2, -3, 2, 2, 2, -3, 2, -3, 2, -3, 2, 0, ...].
G.f. A(x) satisfies A(x)^2 - 2*A(x) + 2 = A134746(x^2), which means (phi(q) / phi(q^4) - 1)^2 + 1 = (phi(q^2) / phi(q^4))^2.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = (u^2 - 2*u + 2) * (v^2 - 2*v + 2) - v^2.
G.f. A(x) satisfies 0 = f(A(x), A(x^3)) where f(u, v) = 4 * u * (u - 1) * (2 - u) * v * (v - 1) * (2 - v) - (u - v)^4.
G.f. is a period 1 Fourier series which satisfies f(-1 / (16 t)) = g(t) where q = exp(2 Pi i t) and g() is g.f. for A112128.
a(4*n) = 0 unless n=0. a(4*n + 2) = a(4*n + 3) = 0. a(4*n + 1) = 2 * A079006(n). a(n) = (-1)^n * A208604(n). Convolution inverse is A112128.

A208933 Expansion of phi(q^4) / phi(-q) in powers of q where phi() is a Ramanujan theta function.

Original entry on oeis.org

1, 2, 4, 8, 16, 28, 48, 80, 128, 202, 312, 472, 704, 1036, 1504, 2160, 3072, 4324, 6036, 8360, 11488, 15680, 21264, 28656, 38400, 51182, 67864, 89552, 117632, 153836, 200352, 259904, 335872, 432480, 554952, 709728, 904784, 1149916, 1457136, 1841200, 2320128
Offset: 0

Views

Author

Michael Somos, Mar 13 2012

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			G.f. = 1 + 2*q + 4*q^2 + 8*q^3 + 16*q^4 + 28*q^5 + 48*q^6 + 80*q^7 + 128*q^8 + ...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, q^4] / EllipticTheta[ 4, 0, q], {q, 0, n}]; (* Michael Somos, Apr 25 2015 *)
    nmax=60; CoefficientList[Series[Product[(1-x^(2*k)) * (1-x^(8*k))^5 / ((1-x^k)^2 * (1-x^(4*k))^2 * (1-x^(16*k))^2),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Oct 14 2015 *)
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A) * eta(x^8 + A)^5 / (eta(x + A) * eta(x^4 + A) * eta(x^16 + A))^2, n))};

Formula

Expansion of eta(q^2) * eta(q^8)^5 / (eta(q) * eta(q^4) * eta(q^16))^2 in powers of q.
Euler transform of period 16 sequence [ 2, 1, 2, 3, 2, 1, 2, -2, 2, 1, 2, 3, 2, 1, 2, 0, ...].
G.f. is a period 1 Fourier series which satisfies f(-1 / (16 t)) = (1/4) * g(t) where q = exp(2 Pi i t) and g() is g.f. for A208603.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = (2*u - 1) * (2*v^2 - 2*v + 1) - u^2.
G.f. A(x) satisfies 0 = f(A(x), A(x^3)) where f(u, v) = 4 * u * (u - 1) * (2*u - 1) * v * (v - 1) * (2*v - 1) - (u - v)^4.
(-1)^n * a(n) = A112128(n). a(n) = 2 * A123655(n) unless n=0. 2 * a(n) = A007096(n) unless n=0. a(2*n) = A131126(n). a(2*n + 1) = 2 * A093160(n). Convolution inverse of A208604.
G.f.: (Sum_{k in Z} x^(4 * k^2)) / (Sum_{k in Z} (-1)^k * x^(k^2)) = theta_3(x^4) / theta_3(-x).
G.f.: Product_{k>0} ((1 + x^(2*k)) * (1 + x^(4*k)))^3 / ((1 + (-x)^k) * (1 + x^(8*k)))^2.
a(n) ~ exp(sqrt(n)*Pi) / (2^(7/2) * n^(3/4)). - Vaclav Kotesovec, Oct 14 2015

A097566 Number of partitions p of n for which Odd(p) = Odd(p') (mod 4), where p' is the conjugate of p.

Original entry on oeis.org

1, 1, 0, 1, 5, 5, 1, 5, 20, 20, 6, 20, 65, 65, 25, 66, 185, 185, 85, 190, 481, 482, 250, 501, 1165, 1170, 666, 1230, 2666, 2685, 1646, 2850, 5827, 5887, 3830, 6303, 12251, 12415, 8487, 13395, 24912, 25323, 18052, 27507, 49215, 50176, 37072, 54832, 94781, 96905
Offset: 0

Views

Author

Wouter Meeussen, Aug 28 2004

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Odd(p) is the number of odd parts of a partition p. a(n) is denoted t(n) in Problem 10969.

Examples

			G.f. = 1 + x + x^3 + 5*x^4 + 5*x^5 + x^6 + 5*x^7 + 20*x^8 + 20*x^9 + 6*x^10 + ...
G.f. = 1/q + q^23 + q^71 + 5*q^95 + 5*q^119 + q^143 + 5*q^167 + 20*q^191 + 20*q^215 + ...
a(5) = 5 because only the partitions {5}, {3,2}, {3,1,1}, {2,2,1}, {1,1,1,1,1} have conjugates resp. {1,1,1,1,1}, {2,2,1}, {3,1,1}, {3,2}, {5} with matching counts of odd elements (resp. (1,5), (1,1), (3,3), (1,1), (5,1) being congruent modulo 4 ).
		

Crossrefs

Programs

  • Maple
    with(combinat); t1:=mul( (1+q^(2*n-1))/((1-q^(4*n))*(1+q^(4*n-2))^2), n=1..100): t2:=series(t1,q,100): f:=n->coeff(t2,q,n); p:=numbpart; t:=n->(p(n)+f(n))/2; # N. J. A. Sloane, Jan 25 2009
  • Mathematica
    fStanley[n_Integer]:=Product[(1+q^(2i-1))/(1-q^(4i))/(1+q^(4i-2))^2, {i, n}]; Table[PartitionsP[n]/2+1/2*Coefficient[Series[fStanley[n], {q, 0, n+1}], q^n], {n, 64}] or Table[Count[Partitions[n], q_/;Mod[Count[q, w_/;OddQ[w]]- Count[TransposePartition[q], w_/;OddQ[w]], 4]===0], {n, 24}]
    a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, x^8] / (EllipticTheta[ 3, 0, x^2] QPochhammer[ x]), {x, 0, n}]; (* Michael Somos, Jun 01 2014 *)
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^2 * eta(x^16 + A)^5 / (eta(x + A) * eta(x^4 + A)^5 * eta(x^32 + A)^2), n))}; /* Michael Somos, May 04 2011 */

Formula

From Michael Somos, May 04 2011: (Start)
Expansion of q^(1/24) * eta(q^2)^2 * eta(q^16)^5 / (eta(q) * eta(q^4)^5 * eta(q^32)^2) in powers of q.
Expansion of phi(x^8) / (phi(x^2) * f(-x)) in powers of x where phi(), f() are Ramanujan theta functions.
Euler transform of period 32 sequence [ 1, -1, 1, 4, 1, -1, 1, 4, 1, -1, 1, 4, 1, -1, 1, -1, 1, -1, 1, 4, 1, -1, 1, 4, 1, -1, 1, 4, 1, -1, 1, 1, ...].
G.f.: theta_3(x^8) / (theta_3(x^2) * Product_{k>0} (1 - x^k)) = A000041(x) * A112128(x^2).
a(n) = (A000041(n) + A085261(n)) / 2.
(End)

A260186 Expansion of (phi(q^4) / phi(q))^2 in powers of q where phi() is a Ramanujan theta function.

Original entry on oeis.org

1, -4, 12, -32, 80, -184, 400, -832, 1664, -3220, 6056, -11104, 19904, -34968, 60320, -102336, 171008, -281800, 458428, -736928, 1171552, -1843328, 2872368, -4435392, 6790656, -10313180, 15544136, -23259968, 34568576, -51042392, 74901984, -109268224, 158507008
Offset: 0

Views

Author

Michael Somos, Jul 17 2015

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Examples

			G.f. = 1 - 4*x + 12*x^2 - 32*x^3 + 80*x^4 - 184*x^5 + 400*x^6 - 832*x^7 + ...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ (EllipticTheta[ 3, 0, q^4] / EllipticTheta[ 3, 0, q])^2, {q, 0, n}];
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x + A) / eta(x^16 + A))^4 * (eta(x^8 + A) / eta(x^2 + A))^10, n))};

Formula

Expansion of (eta(q) / eta(q^16))^4 * (eta(q^8) / eta(q^2))^10 in powers of q.
Euler transform of period 16 sequence [ -4, 6, -4, 6, -4, 6, -4, -4, -4, 6, -4, 6, -4, 6, -4, 0, ...].
G.f. is a period 1 Fourier series which satisfies f(-1 / (16 t)) = (1/4) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A216060.
Convolution inverse is A216060. Convolution square of A112128.
a(n) ~ (-1)^n * exp(Pi*sqrt(2*n)) / (32 * 2^(1/4) * n^(3/4)). - Vaclav Kotesovec, Nov 15 2017
Showing 1-4 of 4 results.