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 1-5 of 5 results.

A303058 G.f. A(x) satisfies: A(x) = Sum_{n>=0} (1+x)^(n^2) * x^n / A(x)^n.

Original entry on oeis.org

1, 1, 1, 2, 5, 16, 61, 259, 1228, 6284, 34564, 201978, 1246652, 8084728, 54862377, 388266809, 2857708840, 21822753453, 172550972216, 1410144139982, 11892084248959, 103343300813517, 924223611649636, 8496346816801059, 80201063980292729, 776585923239589681, 7706568335863727817, 78311132374535936605
Offset: 0

Views

Author

Paul D. Hanna, Apr 20 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 16*x^5 + 61*x^6 + 259*x^7 + 1228*x^8 + 6284*x^9 + 34564*x^10 + 201978*x^11 + 1246652*x^12 + ...
such that
A(x) = 1 + (1+x)*x/A(x) + (1+x)^4*x^2/A(x)^2 + (1+x)^9*x^3/A(x)^3 + (1+x)^16*x^4/A(x)^4 + (1+x)^25*x^5/A(x)^5 + (1+x)^36*x^6/A(x)^6 + ...

Links

Crossrefs

Cf. A303057, A303290, A301929, A321607, A321608.

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 * x^n/Ser(A)^n ) )[#A] );A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

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

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

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

Views

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 + ...

Links

Crossrefs

Cf. A303290, A303291.

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

A325286 G.f. satisfies: 1 = Sum_{n>=0} (1+x)^(n*(n-1)/2) / A(x)^n * 1/2^(n+1).

Original entry on oeis.org

1, 1, 3, 25, 343, 6441, 150975, 4203201, 134852079, 4886641681, 197154406591, 8760602600193, 425074860993439, 22363792326962881, 1268239233311498079, 77129745316500047745, 5008173999379887257151, 345838251972031108425345, 25309861534968595801377279, 1956926079593452273940279169, 159406563966400881627947865279, 13645204581985719926987977747329, 1224591755319676016226530026499583, 114980206425267526899287638805977857
Offset: 0

Views

Author

Paul D. Hanna, Apr 18 2019

Keywords

Comments

a(n) == 1 (mod 2) for n >= 0.
Conjecture: a(n) == 1 (mod 3) iff n is a number whose base-3 representation contains no 2 (cf. A005836), otherwise a(n) == 0 (mod 3).

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 25*x^3 + 343*x^4 + 6441*x^5 + 150975*x^6 + 4203201*x^7 + 134852079*x^8 + 4886641681*x^9 + 197154406591*x^10 + ...
such that
1 = 1/2 + 1/(2^2*A(x)) + (1+x)/(2^3*A(x)^2) + (1+x)^3/(2^4*A(x)^3) + (1+x)^6/(2^5*A(x)^4) + (1+x)^10/(2^6*A(x)^5) + (1+x)^15/(2^7*A(x)^6) + (1+x)^21/(2^8*A(x)^7) + ...
also,
1 = 1/(2*A(x)) + (1+x)/(2*A(x))^2 + (1+x)^3/(2*A(x))^3 + (1+x)^6/(2*A(x))^4 + (1+x)^10/(2*A(x))^5 + (1+x)^15/(2*A(x))^6 + (1+x)^21/(2*A(x))^7 + (1+x)^28/(2*A(x))^8 + ...

Links

Crossrefs

Cf. A303290.

Programs

  • PARI
    /* Requires adequate precision */
{a(n) = my(A=[1]); for(i=1,n, A = concat(A,0); A[#A] = round( polcoeff( sum(m=0,10*#A+100, (1+x+x*O(x^#A))^(m*(m-1)/2)/Ser(A)^m/2^(m+1)*1.),#A-1)));A[n+1]}
for(n=0,25,print1(a(n),", "))

Formula

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

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 1-5 of 5 results.

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