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.

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A320951 G.f.: A(x) satisfies: A(x) = Sum_{n>=0} x^n * (1+x)^(n*(n+1)) / A(x)^n.

Original entry on oeis.org

1, 1, 2, 3, 9, 28, 110, 485, 2358, 12486, 70726, 425747, 2702837, 18004835, 125337381, 908737863, 6843536374, 53407750147, 431075414218, 3592384229312, 30862831600689, 272976843937138, 2482698463801148, 23192576636266041, 222310388884578760, 2184486850658804107, 21985733344615744620, 226455749821063728474, 2385331864619907236147, 25676170688883138634306, 282253492062060457638824
Offset: 0

Views

Author

Paul D. Hanna, Nov 20 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 3*x^3 + 9*x^4 + 28*x^5 + 110*x^6 + 485*x^7 + 2358*x^8 + 12486*x^9 + 70726*x^10 + 425747*x^11 + 2702837*x^12 + ...
such that
A(x) = 1 + x*(1+x)^2/A(x) + x^2*(1+x)^6/A(x)^2 + x^3*(1+x)^12/A(x)^3 + x^4*(1+x)^20/A(x)^4 + x^5*(1+x)^30/A(x)^5 + ...
Also
1 + x = 1 + x/A(x) + x^2*(1+x)^2/A(x)^2 + x^3*(1+x)^6/A(x)^3 + x^4*(1+x)^12/A(x)^4 + x^5*(1+x)^20/A(x)^5 + x^6*(1+x)^30/A(x)^6 + ...
RELATED SERIES.
Sum_{n>=0} x^n * (1+x)^(n^2) / A(x)^n = 1 + x + x^2 + x^3 + 3*x^4 + 8*x^5 + 32*x^6 + 135*x^7 + 649*x^8 + 3381*x^9 + 18894*x^10 + 112382*x^11 + 705174*x^12 + ...
A(A(x)-1) = 1 + x + 4*x^2 + 14*x^3 + 56*x^4 + 251*x^5 + 1239*x^6 + 6627*x^7 + 38112*x^8 + 233692*x^9 + 1517788*x^10 + 10384824*x^11 + ...
where A(A(x)-1) = Sum_{n>=0} (A(x)-1)^n * A(x)^(n*(n+1)) / A(A(x)-1)^n.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0); A[#A] = Vec(sum(n=0, #A, ((1+x)^n +x*O(x^#A))^(n+1) * x^n/Ser(A)^n ) )[#A] ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = Sum_{n>=0} x^n * (1+x)^(n*(n+1)) / A(x)^n.
(2) 1 + x = Sum_{n>=0} x^n * (1+x)^(n*(n-1)) / A(x)^n.
(3) A(x) = Sum_{n>=0} (A(x)-1)^n * A(x)^(n*(n-1)) / A(A(x)-1)^n.

A325155 G.f. A(x) satisfies: x = Sum_{n>=1} x^n * (1+x)^(n^2/2) / A(x)^(n/2).

Original entry on oeis.org

1, 3, 5, 7, 11, 21, 49, 133, 408, 1376, 5020, 19564, 80741, 350551, 1593066, 7547792, 37163568, 189662934, 1001046684, 5453972462, 30622950955, 176942133603, 1050773432990, 6405898358012, 40048848677954, 256521565555908, 1681897617101795, 11278819380424173, 77301464920178158, 541084956406886214, 3865540113371340736, 28167799470180443028, 209238063076396838375, 1583562040116769584091, 12204247180832799551059, 95731651337427271893873
Offset: 0

Views

Author

Paul D. Hanna, Apr 06 2019

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 5*x^2 + 7*x^3 + 11*x^4 + 21*x^5 + 49*x^6 + 133*x^7 + 408*x^8 + 1376*x^9 + 5020*x^10 + 19564*x^11 + 80741*x^12 + ...
such that
A(x) = 1 + x*(1+x)^(1/2)/A(x)^(1/2) + x^2*(1+x)^2/A(x) + x^3*(1+x)^(9/2)/A(x)^(3/2) + x^4*(1+x)^8/A(x)^2 + x^5*(1+x)^(25/2)/A(x)^(5/2) + x^6*(1+x)^18/A(x)^3 + x^7*(1+x)^(49/2)/A(x)^(7/2) + x^8*(1+x)^32/A(x)^4 + ...
Note that
sqrt(A(x))*sqrt(1+x) = 1 + x + x^2 + x^3 + 2*x^4 + 4*x^5 + 11*x^6 + 32*x^7 + 106*x^8 + 376*x^9 + 1433*x^10 + 5782*x^11 + ... + A318644(n)*x^n + ...
		

Crossrefs

Programs

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

Formula

G.f. A(x) satisfies the following identities.
(1) 1 + x = Sum_{n>=0} x^n * (1+x)^(n^2/2) / A(x)^(n/2).
(2) 1 + x = 1/(1 - q*x/(sqrt(A(x)) - q*(q^2-1)*x/(1 - q^5*x/(sqrt(A(x)) - q^3*(q^4-1)*x/(1 - q^9*x/(sqrt(A(x)) - q^5*(q^6-1)*x/(1 - q^13*x/(sqrt(A(x)) - q^7*(q^8-1)*x/(1 - ...))))))))), where q = sqrt(1+x), a continued fraction due to a partial elliptic theta function identity.
(3) 1 + x = Sum_{n>=0} x^n * (1+x)^(n/2) / A(x)^(n/2) * Product_{k=1..n} (sqrt(A(x)) - x*sqrt(1+x)^(4*k-3)) / (sqrt(A(x)) - x*sqrt(1+x)^(4*k-1)), due to a q-series identity.
(4) A(x) = (1+x)*G(x)^2 where G(x) is the g.f. of A318644.
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