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.

A369671 Expansion of g.f. A(x) satisfying Sum_{n=-oo..+oo} (x^n - 4*A(x))^n = theta_4(x).

Original entry on oeis.org

1, 4, 15, 52, 177, 664, 3038, 16268, 90660, 490456, 2541387, 12819184, 64665462, 333763444, 1776226471, 9670530120, 53128162973, 291546645940, 1592977754671, 8685610041084, 47462008167381, 260789472093044, 1442162566738036, 8016343531922084, 44697615509640615, 249596790724248848
Offset: 1

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Author

Paul D. Hanna, Feb 03 2024

Keywords

Comments

Note: theta_4(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n^2) - see A002448.
Congruences:
(C.1) a(2*n) == 0 (mod 4) for n >= 1.
(C.2) a(n) == A369672(n) (mod 4) for n >= 1.
(C.3) a(2*n)/4 == -A369672(2*n)/4 (mod 4) for n >= 1.

Examples

			G.f.: A(x) = x + 4*x^2 + 15*x^3 + 52*x^4 + 177*x^5 + 664*x^6 + 3038*x^7 + 16268*x^8 + 90660*x^9 + 490456*x^10 + 2541387*x^11 + 12819184*x^12 + ...
where Sum_{n=-oo..+oo} (x^n - 4*A(x))^n = theta_4(x), and
theta_4(x) = 1 - 2*x + 2*x^4 - 2*x^9 + 2*x^16 + ... + (-1)^n*2*x^(n^2) + ...
RELATED SERIES.
When we break up the doubly infinite sum into the following parts
P = Sum_{n>=0} (x^n - 4*A(x))^n = 1 - 3*x - 4*x^3 - 15*x^4 - 76*x^5 - 336*x^6 - 1516*x^7 - 7040*x^8 - 34403*x^9 - 175616*x^10 - 918968*x^11 - 4847040*x^12 + ...
N = Sum_{n>=1} x^(n^2) / (1 - 4*x^n*A(x))^n = x + 4*x^3 + 17*x^4 + 76*x^5 + 336*x^6 + 1516*x^7 + 7040*x^8 + 34401*x^9 + 175616*x^10 + 918968*x^11 + 4847040*x^12 + ...
we see that the sum equals P + N = theta_4(x).
SPECIAL VALUES.
(V.1) A(exp(-Pi)) = 0.05210763699884104351595933706426840151754418802521727110...
where Sum_{n=-oo..+oo} (exp(-n*Pi) - 4*A(exp(-Pi)))^n = (Pi/2)^(1/4)/gamma(3/4) = 0.91357913815611682140724...
(V.2) A(exp(-2*Pi)) = 0.001881490423764068063219673469053308038171175452456126483...
where Sum_{n=-oo..+oo} (exp(-2*n*Pi) - 4*A(exp(-2*Pi)))^n = (Pi/2)^(1/4)/gamma(3/4) * 2^(1/8) = 0.99626511456090713578995...
(V.3) A(exp(-4*Pi)) = 0.000003487391003072013497532566545785034046098962165471423...
where Sum_{n=-oo..+oo} (exp(-4*n*Pi) - 4*A(exp(-4*Pi)))^n = Pi^(1/4)/gamma(3/4) * (sqrt(2) + 1)^(1/4)/2^(7/16) = 0.99999302531528758200931...
(V.4) A(exp(-10*Pi)) = 0.000000000000022711010683243001546817769702787327972263611...
where Sum_{n=-oo..+oo} (exp(-10*n*Pi) - 4*A(exp(-10*Pi)))^n = Pi^(1/4)/gamma(3/4) * 2^(7/8)/((5^(1/4) - 1)*sqrt(5*sqrt(5) + 5)) = 0.99999999999995457797863...

Links

Crossrefs

Cf. A369672 (dual), A002448 (theta_4), A355868.

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0); M=sqrtint(#A+4);
A[#A] = polcoeff( (-sum(n=-M,M, (-1)^n * x^(n^2)) + sum(n=-#A,#A, (x^n - 4*x*Ser(A))^n) )/4, #A); ); A[n]}
for(n=1,30,print1(a(n),", "))
  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0); M=sqrtint(#A+4);
A[#A] = polcoeff( (-sum(n=-M,M, (-1)^n * x^(n^2)) + sum(n=-#A,#A, x^(n^2)/(1 - 4*x^(n+1)*Ser(A))^n) )/4, #A); ); A[n]}
for(n=1,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) Sum_{n=-oo..+oo} (x^n - 4*A(x))^n = Sum_{n=-oo..+oo} (-1)^n * x^(n^2).
(2) Sum_{n=-oo..+oo} x^n * (x^n + 4*A(x))^(n-1) = Sum_{n=-oo..+oo} (-1)^n * x^(n^2).
(3) Sum_{n=-oo..+oo} (-1)^n * x^n * (x^n - 4*A(x))^n = 0.
(4) Sum_{n=-oo..+oo} x^(n^2) / (1 - 4*x^n*A(x))^n = Sum_{n=-oo..+oo} (-1)^n * x^(n^2).
(5) Sum_{n=-oo..+oo} x^(n^2) / (1 + 4*x^n*A(x))^(n+1) = Sum_{n=-oo..+oo} (-1)^n * x^(n^2).
(6) Sum_{n=-oo..+oo} (-1)^n * x^(n*(n-1)) / (1 - 4*x^n*A(x))^n = 0.
a(n) ~ c * d^n / n^(3/2), where d = 5.9085050558... and c = 0.2952711268... - Vaclav Kotesovec, Feb 03 2024

A369683 Expansion of g.f. A(x) satisfying Sum_{n>=0} (-1)^n * x^n * Product_{k=0..n} (x^(2*k+1) + A(x)) = theta_3(x).

Original entry on oeis.org

1, 2, 4, 8, 19, 40, 86, 181, 383, 811, 1709, 3598, 7554, 15839, 33158, 69319, 144731, 301813, 628727, 1308487, 2720908, 5653743, 11740260, 24365703, 50544832, 104810967, 217270721, 450287996, 933043086, 1933125817, 4004865745, 8296690701, 17188106646, 35609996584
Offset: 0

Views

Author

Paul D. Hanna, Feb 04 2024

Keywords

Comments

Note: theta_3(x) = Sum_{n=-oo..+oo} x^(n^2) - see A000122.
a(n+1)/a(n) tends to 2.07474... - Vaclav Kotesovec, Feb 05 2024

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^2 + 8*x^3 + 19*x^4 + 40*x^5 + 86*x^6 + 181*x^7 + 383*x^8 + 811*x^9 + 1709*x^10 + 3598*x^11 + 7554*x^12 + ...
By definition, A = A(x) satisfies the sum of products
theta_3(x) = (x + A) - x*(x + A)*(x^3 + A) + x^2*(x + A)*(x^3 + A)*(x^5 + A) - x^3*(x + A)*(x^3 + A)*(x^5 + A)*(x^7 + A) + x^4*(x + A)*(x^3 + A)*(x^5 + A)*(x^7 + A)*(x^9 + A) -+ ...
also, A = A(x) satisfies another sum of products
1 - x*theta_3(x) = 1/(1 + x*A) - x^2/((1 + x*A)*(1 + x^3*A)) + x^6/((1 + x*A)*(1 + x^3*A)*(1 + x^5*A)) - x^12/((1 + x*A)*(1 + x^3*A)*(1 + x^5*A)*(1 + x^7*A)) + x^20/((1 + x*A)*(1 + x^3*A)*(1 + x^5*A)*(1 + x^7*A)*(1 + x^9*A)) + ...
Further, A = A(x) satisfies the continued fraction given by
theta_3(x) = (x + A)/(1 + x*(x^3 + A)/(1 - x*(x^3 + A) + x*(x^5 + A)/(1 - x*(x^5 + A) + x*(x^7 + A)/(1 - x*(x^7 + A) + x*(x^9 + A)/(1 - x*(x^9 + A) + x*(x^11 + A)/(1 - ...))))))
where theta_3(x) = 1 + 2*x + 2*x^4 + 2*x^9 + 2*x^16 + 2*x^25 + 2*x^36 + ... + 2*x^(n^2) + ...

Links

Crossrefs

Cf. A369684, A369682, A369672, A000122 (theta_3).

Programs

  • PARI
    {a(n) = my(A=[1], M = sqrtint(n)+1); for(i=1,n, A = concat(A,0);
A[#A] = polcoeff( sum(n=-M,M, x^(n^2) ) - sum(n=0,#A, (-1)^n * x^n * prod(k=0,n, x^(2*k+1) + Ser(A)) ), #A-1) ); H=A; A[n+1]}
for(n=0,40, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) Sum_{n>=0} (-1)^n * x^n * Product_{k=0..n} (x^(2*k+1) + A(x)) = Sum_{n=-oo..+oo} x^(n^2).
(2) Sum_{n>=0} (-1)^n * x^(n*(n+1)) / Product_{k=0..n} (1 + x^(2*k+1)*A(x)) = 1 - x * Sum_{n=-oo..+oo} (-1)^n * x^(n^2).
(3) theta_3(x) = (x + A(x))/(1 + F(1)), where F(n) = x*(x^(2*n+1) + A(x))/(1 - x*(x^(2*n+1) + A(x)) + F(n+1)), a continued fraction.
(4) 1 - x*theta_3(x) = 1/((1 + x*A(x))*(1 + F(1))), where F(n) = x^(2*n) / (1 - x^(2*n) + x^(2*n+1)*A + (1 + x^(2*n+1)*A)*F(n+1)), a continued fraction.
Showing 1-2 of 2 results.

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