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.

A303288 G.f. A(x) satisfies: 1 = Sum_{n>=0} ( (1+x)^n - 5*x*A(x) )^n * 2^n / 3^(n+1).

Original entry on oeis.org

1, 14, 688, 56738, 6347176, 881241656, 144796770004, 27351977086556, 5826096152426212, 1380051673281134312, 359720002818554238352, 102317793242070983628176, 31540355035889303797419616, 10475792506313141986771902704, 3730248479020018845292570520560, 1417811189172027111629537752756520, 572992474515466430293335350543824096
Offset: 0

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Author

Paul D. Hanna, Apr 23 2018

Keywords

Examples

			G.f.: A(x) = 1 + 14*x + 688*x^2 + 56738*x^3 + 6347176*x^4 + 881241656*x^5 + 144796770004*x^6 + 27351977086556*x^7 + 5826096152426212*x^8 + ...
such that
1 = 1/3 + 2*((1+x) - 5*x*A(x))/3^2 + 2^2*((1+x)^2 - 5*x*A(x))^2/3^3 + 2^3*((1+x)^3 - 5*x*A(x))^3/3^4 + 2^4*((1+x)^4 - 5*x*A(x))^4/3^5 + 2^5*((1+x)^5 - 5*x*A(x))^5/3^6 + ...
also,
1 = 1/(3 + 10*x*A(x)) + 2*(1+x)/(3 + 10*x*A(x))^2 + 2^2*(1+x)^4/(3 + 10*x*A(x))^3 + 2^3*(1+x)^9/(3 + 10*x*A(x))^4 + 2^4*(1+x)^16/(3 + 10*x*A(x))^5 + 2^5*(1+x)^25/(3 + 10*x*A(x))^6 + ...
		

Crossrefs

Formula

G.f. A(x) satisfies:
(1) 1 = Sum_{n>=0} 2^n * ( (1+x)^n - 5*x*A(x) )^n / 3^(n+1).
(2) 1 = Sum_{n>=0} 2^n * (1+x)^(n^2) / (3 + 10*x*A(x)*(1+x)^n)^(n+1). - Paul D. Hanna, Jan 10 2019
(3) 1 = Sum_{k>=0} (-5*x)^k * A(x)^k * Sum_{n>=0} C(n+k,k) * (1+x)^(n*(n+k)) * 2^(n+k) / 3^(n+k+1).
(4) 1 = Sum_{n>=0} Sum_{k=0..n} C(n,k) * (1+x)^(n*(n-k)) * 2^n / 3^(n+1) * (-5*x)^k * A(x)^k.

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

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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 + ...
		

Crossrefs

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

A303292 G.f. A(x) satisfies: 4 = Sum_{n>=0} (3/4)^n * (1 + x)^(n^2) / A(x)^n.

Original entry on oeis.org

1, 7, 189, 17283, 2755053, 604260531, 165416203197, 53736069429315, 20098682471065149, 8484270818691168963, 3985069388942026022589, 2060504358592580623699011, 1162904612283296975554475517, 711422819982429170172765550083, 469007739834268780510389856367613, 331521891387779056571085490125831171, 250157485456407234540581483486760865533
Offset: 0

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Author

Paul D. Hanna, Apr 22 2018

Keywords

Examples

			G.f.: A(x) = 1 + 7*x + 189*x^2 + 17283*x^3 + 2755053*x^4 + 604260531*x^5 + 165416203197*x^6 + 53736069429315*x^7 + 20098682471065149*x^8 + ...
such that A = A(x) satisfies:
4 = 1 + (1+x)/(4*A/3) + (1+x)^4/(4*A/3)^2 + (1+x)^9/(4*A/3)^3 + (1+x)^16/(4*A/3)^4 + (1+x)^25/(4*A/3)^5 + (1+x)^36/(4*A/3)^6 + (1+x)^49/(4*A/3)^7 + ...
		

Crossrefs

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)/(4/3*Ser(A) - q^(2*m-2*k+1)*(q^(2*m-2*k+2) - 1)*CF)) ); A[#A] = Vec(CF)[#A]/12 ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: 4 = 1/(1 - q/(4/3*A(x) - q*(q^2-1)/(1 - q^5/(4/3*A(x) - q^3*(q^4-1)/(1 - q^9/(4/3*A(x) - q^5*(q^6-1)/(1 - q^13/(4/3*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.: 4 = Sum_{n>=0} (3/4)^n * (1+x)^n / A(x)^n * Product_{k=1..n} (4*A(x) - 3*(1+x)^(4*k-3)) / (4*A(x) - 3*(1+x)^(4*k-1)), due to a q-series identity.
a(n) ~ 3^(5/2) * 2^(2*n - 17/2) * n^n / (exp(n + log(4/3)^2 / 8) * log(4/3)^(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) + ...
		

Crossrefs

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 1-4 of 4 results.