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.

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

A325287 G.f. satisfies 1 = Sum_{n>=0} ((1+x)^(n*(n-1)/2) / A(x)^n) * (2^n/3^(n+1)).

Original entry on oeis.org

1, 2, 16, 380, 15280, 842672, 57985144, 4735508672, 445364211760, 47281191656960, 5586025249211056, 726588091176753152, 103169269785836042656, 15880361395424986644320, 2634307488850605478606240, 468569833279898692863674720, 88975116507316444085923086400, 17966290253142630862386608565440, 3844488506759131598435757854078080, 869080066111317591084733034309229440, 206969312517505574682143594517889278400
Offset: 0

Views

Author

Paul D. Hanna, Apr 23 2019

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 16*x^2 + 380*x^3 + 15280*x^4 + 842672*x^5 + 57985144*x^6 + 4735508672*x^7 + 445364211760*x^8 + 47281191656960*x^9 + 5586025249211056*x^10 + ...

Links

Crossrefs

Cf. A325286.

Programs

  • Mathematica
    a[n_] := Module[{A}, A = {1}; Do[AppendTo[A, 0]; A[[-1]] = Round[ Coefficient[ Sum[(1+x + x*O[x]^Length[A])^(m*(m-1)/2)/(A.x^Range[0, Length[A] - 1])^m*2^m/3^(m + 1), {m, 0, 30 Length[A] + 200}]/2, x, Length[A] - 1]], {i, 1, n}]; Print[A[[n + 1]]]; A[[n + 1]]];
a /@ Range[0, 25] (* Jean-François Alcover, Oct 02 2019 *)
  • PARI
    /* Requires adequate precision */
\p500
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0); A[#A] = round( polcoeff( sum(m=0, 30*#A+200, (1+x+x*O(x^#A))^(m*(m-1)/2)/Ser(A)^m*2^m/3^(m+1)*1.)/2, #A-1))); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

Extensions

Added missing parentheses to definition. - N. J. A. Sloane, Aug 01 2019
Showing 1-2 of 2 results.

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

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