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.

A301311 G.f.: Sum_{n>=0} 2^n * (1-x)^(-n^2) / 3^(n+1).

Original entry on oeis.org

1, 10, 370, 22570, 1924270, 210821290, 28223418010, 4464779024650, 814901395935550, 168556843188104050, 38965275697707264970, 9955529477371346769010, 2785811940289987110605590, 847316256984037311888049090, 278329013908504193489288029090, 98197864581209379156337136722690, 37034491818759647215732974465421990, 14868275488492647637389364332301206490
Offset: 0

Views

Author

Paul D. Hanna, Mar 18 2018

Keywords

Comments

Is there a finite expression for the terms of this sequence?
a(n) is divisible by 10 for n>0 (conjecture).

Examples

			G.f.: A(x) = 1 + 10*x + 370*x^2 + 22570*x^3 + 1924270*x^4 + 210821290*x^5 + 28223418010*x^6 + 4464779024650*x^7 + 814901395935550*x^8 + ...
such that
A(x) = 1/3 + 2/(1-x)/3^2 + 2^2/(1-x)^4/3^3 + 2^3/(1-x)^9/3^4 + 2^4/(1-x)^16/3^5 + 2^5/(1-x)^25/3^6 + 2^6/(1-x)^36/3^7 + 2^7/(1-x)^49/3^8  + 2^8/(1-x)^64/3^9 + ...

Links

Crossrefs

Cf. A104209, A301310.

Programs

  • PARI
    /* Continued fraction expression: */
{a(n) = my(CF=1, q = 1/(1-x +x*O(x^n))); for(k=0, n, CF = 1/(3 - 2*q^(4*n-4*k+1)/(1 - 2*q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: 1/(3 - 2*q/(1 - 2*q*(q^2-1)/(3 - 2*q^5/(1 - 2*q^3*(q^4-1)/(3 - 2*q^9/(1 - 2*q^5*(q^6-1)/(3 - 2*q^13/(1 - 2*q^7*(q^8-1)/(3 - ...))))))))) where q = 1/(1-x), a continued fraction due to a partial elliptic theta function identity.
G.f.: Sum_{n>=0} 2^n/3^(n+1) * (1-x)^n * Product_{k=1..n} (3*(1-x)^(4*k-3) - 2) / (3*(1-x)^(4*k-1) - 2), due to a q-series identity.
a(n) = Sum_{k>=0} 2^k * binomial(k^2 + n-1, n) / 3^(k+1).
a(n) ~ 2^(2*n + 1/2 - log(3/2)/8) * 3^(log(3/2)/8 - 1) * n^n / (exp(n) * (log(3/2))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

A301312 G.f.: Sum_{n>=0} ( (1+x)^n + (1+2*x)^n )^n / 3^(n+1).

Original entry on oeis.org

1, 15, 818, 75237, 9704172, 1610219061, 326647152627, 78322576680405, 21671112063131181, 6796070676619188507, 2382079473846244973676, 922852837565535061715211, 391585126460697011112411576, 180608810765895380940822677853, 89966485333525032479641295540142, 48135006154108946423894447901622257, 27530139433795469892705229664410072563, 16761420605156187498505881406969283279001
Offset: 0

Views

Author

Paul D. Hanna, Mar 18 2018

Keywords

Examples

			G.f.: A(x) = 1 + 15*x + 818*x^2 + 75237*x^3 + 9704172*x^4 + 1610219061*x^5 + 326647152627*x^6 + 78322576680405*x^7 + 21671112063131181*x^8 + ...
such that
A(x) = 1/3  +  ((1+x) + (1+2*x))/3^2  +  ((1+x)^2 + (1+2*x)^2)^2/3^3  +  ((1+x)^3 + (1+2*x)^3)^3/3^4  +  ((1+x)^4 + (1+2*x)^4)^4/3^5  +  ((1+x)^5 + (1+2*x)^5)^5/3^6  + ... + ((1+x)^n + (1+2*x)^n)^n / 3^(n+1) + ...
Equivalently,
A(x) = 1/3 + (2 + 3*x)/3^2 + (2 + 6*x + 5*x^2)^2/3^3 + (2 + 9*x + 15*x^2 + 9*x^3)^3/3^4 + (2 + 12*x + 30*x^2 + 36*x^3 + 17*x^4)^4/3^5 + (2 + 15*x + 50*x^2 + 90*x^3 + 85*x^4 + 33*x^5)^5/3^6 + ...

Links

Crossrefs

Cf. A301307, A301310.

Formula

G.f.: Sum_{n>=0} Sum_{k=0..n} binomial(n,k) * (1+x)^(n*(n-k)) * (1+2*x)^(n*k) / 3^(n+1).
G.f.: Sum_{n>=0} [ Sum_{k=0..n} binomial(n,k) * (1+2^k)*x^k ]^n / 3^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = 36.9010493125536756798917509741716959... and c = 0.4484222753815457836094869794957853521... - Vaclav Kotesovec, Oct 10 2020

A301466 a(n) = Sum_{k>=0} binomial(k^3, n)/2^(k+1).

Original entry on oeis.org

1, 13, 2335, 1178873, 1168712311, 1916687692685, 4697337224419543, 16082097033630615185, 73313708225823014181097, 429319086610079876821621425, 3140585308524019620784003889263, 28066697522114849327295724261347841, 300886927215791917153044786581553617063
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 21 2018

Keywords

Links

Crossrefs

Cf. A173217, A301310, A301432, A301468.

Programs

  • Mathematica
    Table[Sum[Binomial[k^3, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 15}]
Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -3*j, 0]/2, {j, 0, n}] / n!, {n, 0, 15}]

Formula

a(n) ~ 3^(3*n + 1/2) * n^(2*n) / (2 * exp(2*n) * (log(2))^(3*n + 1)).
G.f.: Sum_{n>=0} (1 + x)^(n^3) / 2^(n+1).

A301432 G.f.: Sum_{n>=0} 3^n * (1+x)^(n^2) / 4^(n+1).

Original entry on oeis.org

1, 21, 1512, 182448, 30845052, 6706403424, 1782361433664, 559861341721920, 202922346528231120, 83358099246202940880, 38271708686845732234752, 19421327571536329073316864, 10794249397953336851774993664, 6521104275997643157262604783616, 4254768377324045826054766465227264, 2981719456871640091643441908508931072
Offset: 0

Views

Author

Paul D. Hanna, Mar 21 2018

Keywords

Comments

a(n) is divisible by 21 for n>0 (conjecture).
In general, for s > 1, Sum_{k>=0} binomial(k^2, n) / s^k is asymptotic to 2^(2*n + 1/2) * n^n / (s^(log(s)/8) * exp(n) * (log(s))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

Examples

			G.f.: A(x) = 1 + 21*x + 1512*x^2 + 182448*x^3 + 30845052*x^4 + 6706403424*x^5 + 1782361433664*x^6 + 559861341721920*x^7 + ...
such that
A(x) = 1/4 + 3*(1+x)/4^2 + 3^2*(1+x)^4/4^3 + 3^3*(1+x)^9/4^4 + 3^4*(1+x)^16/4^5 + 3^5*(1+x)^25/4^6 + 3^6*(1+x)^36/4^7 + 3^7*(1+x)^49/4^8  + 3^8*(1+x)^64/4^9 + ...

Links

Crossrefs

Cf. A301310, A173217, A301311.

Programs

  • Mathematica
    Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[3/4, -2*j, 0]/4, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *)
  • PARI
    /* Continued fraction expression: */
{a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(4 - 3*q^(4*n-4*k+1)/(1 - 3*q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f.: 1/(4 - 3*q/(1 - 3*q*(q^2-1)/(4 - 3*q^5/(1 - 3*q^3*(q^4-1)/(4 - 3*q^9/(1 - 3*q^5*(q^6-1)/(4 - 3*q^13/(1 - 3*q^7*(q^8-1)/(4 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
G.f.: Sum_{n>=0} 3^n/4^(n+1) * (1+x)^n * Product_{k=1..n} (4 - 3*(1+x)^(4*k-3)) / (4 - 3*(1+x)^(4*k-1)), due to a q-series identity.
a(n) = Sum_{k>=0} 3^k * binomial(k^2, n) / 4^(k+1).
a(n) ~ 2^(2*n - 3/2) * n^n / ((4/3)^(log(4/3)/8) * exp(n) * (log(4/3))^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018

A301468 a(n) = Sum_{k>=0} binomial(k^4, n)/2^(k+1).

Original entry on oeis.org

1, 75, 272880, 4681655040, 221478589107480, 22313622005672849712, 4108665216956980742226192, 1249503956658157724969373808320, 583952821303314451291898006535866460, 397372225886096887788939487944785734626120, 377577476850495509525002042506806447493291890064
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 21 2018

Keywords

Comments

In general, for m > 2, Sum_{k>=0} binomial(k^m, n) / 2^(k+1) is asymptotic to m^(m*n + 1/2) * n^((m-1)*n) / (2*exp((m-1)*n) * (log(2))^(m*n + 1)).

Crossrefs

Cf. A173217 (m=2), A301466 (m=3), A301310.

Programs

  • Mathematica
    Table[Sum[Binomial[k^4, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 12}]
Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -4*j, 0]/2, {j, 0, n}] / n!, {n, 0, 12}]

Formula

a(n) ~ 2^(8*n) * n^(3*n) / (exp(3*n) * (log(2))^(4*n+1)).
Showing 1-5 of 5 results.

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