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 1-3 of 3 results.

A303290 G.f. A(x) satisfies: 2 = Sum_{n>=0} (1/2^n) * (1+x)^(n^2) / A(x)^n.

Original entry on oeis.org

1, 3, 15, 225, 6003, 223029, 10403175, 577700889, 37009173207, 2679339499305, 216031850406327, 19187294118006057, 1861057604220294591, 195742656849628038465, 22192660352433291780159, 2698458809215198981964481, 350326879575505922875480047, 48370384900519379918253881361, 7078145146554395463373624118319, 1094300840117324691452685873392145
Offset: 0

Views

Author

Paul D. Hanna, Apr 20 2018

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 225*x^3 + 6003*x^4 + 223029*x^5 + 10403175*x^6 + 577700889*x^7 + 37009173207*x^8 + 2679339499305*x^9 + 216031850406327*x^10 + ...
such that A = A(x) satisfies:
2 = 1 + (1+x)/(2*A) + (1+x)^4/(2*A)^2 + (1+x)^9/(2*A)^3 + (1+x)^16/(2*A)^4 + (1+x)^25/(2*A)^5 + (1+x)^36/(2*A)^6 + (1+x)^49/(2*A)^7 + (1+x)^64/(2*A)^8 + ...

Links

Crossrefs

Cf. A303291, A303292, A303058.
Cf. A325286.

Programs

  • PARI
    /* Find A(x) that satisfies the continued fraction: */
{a(n) = my(A=[1],q=1+x,CF=1); for(i=1,n, A=concat(A,0); m=#A; for(k=0, m, CF = 1/(1 - q^(4*m-4*k+1)/(2*Ser(A) - q^(2*m-2*k+1)*(q^(2*m-2*k+2) - 1)*CF)) ); A[#A] = Vec(CF)[#A]/2 );A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

G.f.: 2 = 1/(1 - q/(2*A(x) - q*(q^2-1)/(1 - q^5/(2*A(x) - q^3*(q^4-1)/(1 - q^9/(2*A(x) - q^5*(q^6-1)/(1 - q^13/(2*A(x) - q^7*(q^8-1)/(1 - ...))))))))), where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
G.f.: 2 = Sum_{n>=0} (1+x)^n/(2^n*A(x)^n) * Product_{k=1..n} (2*A(x) - (1+x)^(4*k-3)) / (2*A(x) - (1+x)^(4*k-1)), due to a q-series identity.
a(n) ~ c * 2^(2*n) * n^n / (exp(n) * log(2)^(2*n)), where c = 0.339650521725496... - Vaclav Kotesovec, Oct 06 2020

A303291 G.f. A(x) satisfies: 3 = Sum_{n>=0} (2/3)^n * (1 + x)^(n^2) / A(x)^n.

Original entry on oeis.org

1, 5, 70, 3170, 252160, 27705800, 3806286820, 621124623740, 116766042046000, 24783363325335440, 5854493683431121840, 1522701357625214096240, 432347094526718807347480, 133078785461406479045306360, 44145742694332046133435657280, 15702781293109570148744738306240, 5962874290966165187708554294296880, 2407878412120285331813837276575565360
Offset: 0

Views

Author

Paul D. Hanna, Apr 22 2018

Keywords

Examples

			G.f.: A(x) = 1 + 5*x + 70*x^2 + 3170*x^3 + 252160*x^4 + 27705800*x^5 + 3806286820*x^6 + 621124623740*x^7 + 116766042046000*x^8 + ...
such that A = A(x) satisfies:
3 = 1 + (1+x)/(3*A/2) + (1+x)^4/(3*A/2)^2 + (1+x)^9/(3*A/2)^3 + (1+x)^16/(3*A/2)^4 + (1+x)^25/(3*A/2)^5 + (1+x)^36/(3*A/2)^6 + (1+x)^49/(3*A/2)^7 + ...

Links

Crossrefs

Cf. A303290, A303292.

Programs

  • PARI
    /* Find A(x) that satisfies the continued fraction: */
{a(n) = my(A=[1], q=1+x, CF=1); for(i=1, n, A=concat(A, 0); m=#A; for(k=0, m, CF = 1/(1 - q^(4*m-4*k+1)/(3/2*Ser(A) - q^(2*m-2*k+1)*(q^(2*m-2*k+2) - 1)*CF)) ); A[#A] = Vec(CF)[#A]/6 ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: 3 = 1/(1 - q/(3/2*A(x) - q*(q^2-1)/(1 - q^5/(3/2*A(x) - q^3*(q^4-1)/(1 - q^9/(3/2*A(x) - q^5*(q^6-1)/(1 - q^13/(3/2*A(x) - q^7*(q^8-1)/(1 - ...))))))))), where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
G.f.: 3 = Sum_{n>=0} (2/3)^n * (1+x)^n / A(x)^n * Product_{k=1..n} (3*A(x) - 2*(1+x)^(4*k-3)) / (3*A(x) - 2*(1+x)^(4*k-1)), due to a q-series identity.
a(n) ~ 2^(2*n + 2 + log(3/2)/8) * n^n / (3^(log(3/2)/8 + 7/2) * exp(n) * log(3/2)^(2*n + 1)). - Vaclav Kotesovec, Oct 14 2020

A303289 E.g.f. A(x) satisfies: e = Sum_{n>=0} (1/n!) * (1+x)^(n^2) / A(x)^n.

Original entry on oeis.org

1, 2, 5, 31, 390, 7926, 229448, 8769552, 421254088, 24578690456, 1699003652752, 136526757080176, 12565047627623648, 1308650039442105504, 152723805589647826368, 19806995417441865105472, 2834647872410303847945600, 444947841160313990957842304, 76198407065481146373641422336, 14170329519388795065500512696832
Offset: 0

Views

Author

Paul D. Hanna, Apr 23 2018

Keywords

Examples

			E.g.f.: A(x) = 1 + 2*x + 5*x^2/2! + 31*x^3/3! + 390*x^4/4! + 7926*x^5/5! + 229448*x^6/6! + 8769552*x^7/7! + 421254088*x^8/8! + 24578690456*x^9/9! + 1699003652752*x^10/10! + ...
such that A = A(x) satisfies:
e = 1 + (1+x)/A + (1+x)^4/(2!*A^2) + (1+x)^9/(3!*A^3) + (1+x)^16/(4!*A^4) + (1+x)^25/(5!*A^5) + (1+x)^36/(6!*A^6) + (1+x)^49/(7!*A^7) + ...

Links

Crossrefs

Cf. A303290, A303291, A303292.

Programs

  • PARI
    \p100; N=20;
A=[1]; for(i=1,N, A=concat(A,0); A[#A] = Vec( round( sum(n=0,200 + 2*#A, (1+x +x*O(x^#A))^(n^2)/Ser(A)^n/n!*1. )/exp(1)*(#A-1)! ) )[#A]/(#A-1)! ); Vec(serlaplace(Ser(A)))
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.