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.

A193281 E.g.f.: A(x) = 1/[(1 - x^2)^(1 + 1/x)].

Original entry on oeis.org

1, 1, 3, 10, 49, 271, 1861, 14218, 125945, 1222525, 13274551, 155993586, 2006410033, 27654327715, 411146102641, 6498852124066, 109654601938801, 1954603251773273, 36899871955766155, 732088187170903162, 15288727416236948241, 334089096072180624631
Offset: 0

Views

Author

Paul D. Hanna, Jul 21 2011

Keywords

Comments

More generally, we have the identity:
Sum_{n>=0} (x^n/n!)*Product_{k=1..n} (1+k*y) = 1/(1 - x*y)^(1 + 1/y); here y=x.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 10*x^3/3! + 49*x^4/4! + 271*x^5/5! +...
where:
A(x)^(x/(1+x)) = 1 + x^2 + x^4 + x^6 + x^8 + x^10 + x^12 +...
Also,
A(x) = 1 + x*(1+x) + x^2*(1+x)*(1+2*x)/2! + x^3*(1+x)*(1+2*x)*(1+3*x)/3! + x^4*(1+x)*(1+2*x)*(1+3*x)*(1+4*x)/4! +...
The logarithm begins:
log(A(x)) = x + x^2 + x^3/2 + x^4/2 + x^5/3 + x^6/3 + x^7/4 + x^8/4 +...

Links

Crossrefs

Cf. A110654, A193287, A193288, A193289, A193290, A246689.

Programs

  • Mathematica
    CoefficientList[Series[1/((1-x^2)^(1+1/x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 08 2013 *)
  • PARI
    {a(n)=n!*polcoeff(1/(1 - x^2 +x^2*O(x^n))^((1+x)/x),n)}
  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n,x^m/m!*prod(k=1,m,1+k*x+x*O(x^n))),n)}
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, j/((j+1)\2)*v[i-j+1]/(i-j)!)); v; \\ Seiichi Manyama, Apr 30 2022

Formula

E.g.f.: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (1 + k*x).
E.g.f.: A(x)=1/[ (1 - x^2)*(1 - x^2)^(1/x)] = 1 + x*(1+x)/(G(0) - x*(1+x)) ; G(k) = 1 + k + x*(k*x+x+1) - x*(k+1)*(k*x+2*x+1)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011
a(n) ~ n! * (n + log(n) + 2 + gamma - log(2))/4, where gamma is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 08 2013
From Peter Bala, Aug 31 2014: (Start)
E.g.f.: exp( sum{n = 1..inf} x^n/A110654(n) ).
It appears that a(n) = 1 ( mod n*(n - 1) ) for n even, while a(n) = 1 ( mod n*(n - 1)/2 ) for n odd [checked up to a(100)]. (End)
a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} k/A110654(k) * a(n-k)/(n-k)!. - Seiichi Manyama, Apr 30 2022

A193287 E.g.f.: A(x) = 1/(1 - 2*x^2)^(1 + 1/(2*x)).

Original entry on oeis.org

1, 1, 5, 19, 145, 981, 10141, 98575, 1289569, 16314121, 258568021, 4023553931, 74961787825, 1383475135069, 29636315118957, 632414472704071, 15316605861040321, 370875832116841105, 10021723060544059429, 271409166367070755843
Offset: 0

Views

Author

Paul D. Hanna, Jul 21 2011

Keywords

Comments

More generally, we have the identity:
Sum_{n>=0} (x^n/n!)*Product_{k=1..n} (1+k*y) = 1/(1 - x*y)^(1 + 1/y); here y=2*x.

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 19*x^3/3! + 145*x^4/4! + 981*x^5/5! +...
where A(x) satisfies:
A(x)^(2*x/(1+2*x)) = 1 + 2*x^2 + 4*x^4 + 8*x^6 + 16*x^8 + 32*x^10 +...
Also,
A(x) = 1 + x*(1+2*x) + x^2*(1+2*x)*(1+4*x)/2! + x^3*(1+2*x)*(1+4*x)*(1+6*x)/3! + x^4*(1+2*x)*(1+4*x)*(1+6*x)*(1+8*x)/4! +...
The logarithm begins:
log(A(x)) = x + 2*x^2 + 2*x^3/2 + 4*x^4/2 + 4*x^5/3 + 8*x^6/3 + 8*x^7/4 +...
a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} k * 2^floor(k/2)/floor((k+1)/2) * a(n-k)/(n-k)!. - _Seiichi Manyama_, Apr 30 2022

Links

Crossrefs

Cf. A193281, A193288, A193289, A193290.

Programs

  • Mathematica
    CoefficientList[Series[1/(1-2*x^2)^(1+1/(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 25 2013 *)
  • PARI
    {a(n)=n!*polcoeff(1/(1 - 2*x^2 +x^2*O(x^n))^((1+2*x)/(2*x)),n)}
  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n,x^m/m!*prod(k=1,m,1+2*k*x+x*O(x^n))),n)}
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, j*2^(j\2)/((j+1)\2)*v[i-j+1]/(i-j)!)); v; \\ Seiichi Manyama, Apr 30 2022

Formula

E.g.f.: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (1 + 2*k*x).
a(n) ~ n! * 2^(n/2-1/2-1/sqrt(2))*n^(1/sqrt(2))/Gamma(1/sqrt(2)). - Vaclav Kotesovec, Jun 25 2013

A193288 E.g.f.: A(x) = 1/(1 - 3*x^2)^(1 + 1/(3*x)).

Original entry on oeis.org

1, 1, 7, 28, 289, 2131, 29161, 316072, 5395993, 77326165, 1583326171, 28229026156, 674412621697, 14384156661343, 392879390385301, 9753823992141496, 299849358712509361, 8492478062686906057, 290226665437376352463, 9233909417529486840412
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2011

Keywords

Comments

More generally, we have the identity:
Sum_{n>=0} (x^n/n!)*Product_{k=1..n} (1+k*y) = 1/(1 - x*y)^(1 + 1/y); here y=3*x.

Examples

			E.g.f.: A(x) = 1 + x + 7*x^2/2! + 28*x^3/3! + 289*x^4/4! + 2131*x^5/5! +...
where A(x) satisfies:
A(x)^(3*x/(1+3*x)) = 1 + 3*x^2 + 9*x^4 + 27*x^6 + 81*x^8 + 243*x^10 +...
Also,
A(x) = 1 + x*(1+3*x) + x^2*(1+3*x)*(1+6*x)/2! + x^3*(1+3*x)*(1+6*x)*(1+9*x)/3! + x^4*(1+3*x)*(1+6*x)*(1+9*x)*(1+12*x)/4! +...
The logarithm begins:
log(A(x)) = x + 3*x^2 + 3*x^3/2 + 9*x^4/2 + 9*x^5/3 + 27*x^6/3 + 27*x^7/4 +...
a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} k * 3^floor(k/2)/floor((k+1)/2) * a(n-k)/(n-k)!. - _Seiichi Manyama_, Apr 30 2022

Links

Crossrefs

Cf. A193281, A193287, A193289, A193290.

Programs

  • Mathematica
    CoefficientList[Series[1/(1-3*x^2)^(1+1/(3*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 25 2013 *)
  • PARI
    {a(n)=n!*polcoeff(1/(1 - 3*x^2 +x^2*O(x^n))^((1+3*x)/(3*x)),n)}
  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n,x^m/m!*prod(k=1,m,1+3*k*x+x*O(x^n))),n)}
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, j*3^(j\2)/((j+1)\2)*v[i-j+1]/(i-j)!)); v; \\ Seiichi Manyama, Apr 30 2022

Formula

E.g.f.: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (1 + 3*k*x).
a(n) ~ n! * n^(1/sqrt(3))*3^(n/2+1/2)/(2^(1+1/sqrt(3))*Gamma(1/sqrt(3))). - Vaclav Kotesovec, Jun 25 2013

A193289 E.g.f.: A(x) = 1/(1 - 6*x^2)^(1 + 1/(3*x)).

Original entry on oeis.org

1, 2, 16, 116, 1456, 18272, 315424, 5592512, 123304192, 2814746624, 75639399424, 2108241486848, 66872341633024, 2198914617257984, 80437062279012352, 3046047243283570688, 126259635313097506816, 5408763597941368291328, 250569314672586154835968
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2011

Keywords

Comments

More generally, we have the identity:
Sum_{n>=0} (x^n/n!)*Product_{k=1..n} (1+k*y) = 1/(1 - x*y)^(1 + 1/y); here x=2*x, y=3*x.

Examples

			E.g.f.: A(x) = 1 + 2*x + 16*x^2/2! + 116*x^3/3! + 1456*x^4/4! + 18272*x^5/5! +...
where A(x) satisfies:
A(x)^(3*x/(1+3*x)) = 1 + 6*x^2 + 36*x^4 + 216*x^6 +...+ 6^n*x^(2*n) +...
Also,
A(x) = 1 + 2*x*(1+3*x) + 4*x^2*(1+3*x)*(1+6*x)/2! + 8*x^3*(1+3*x)*(1+6*x)*(1+9*x)/3! + 16*x^4*(1+3*x)*(1+6*x)*(1+9*x)*(1+12*x)/4! +...
The logarithm begins:
log(A(x)) = 2*x + 6*x^2 + 2*6*x^3/2 + 6^2*x^4/2 + 2*6^2*x^5/3 + 6^3*x^6/3 + 2*6^3*x^7/4 + 6^4*x^8/4 + 2*6^4*x^9/5 + 6^5*x^10/5 +...

Links

Crossrefs

Cf. A193281, A193287, A193288, A193290.

Programs

  • Mathematica
    CoefficientList[Series[1/(1-6*x^2)^(1+1/(3*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 22 2013 *)
  • PARI
    {a(n)=n!*polcoeff(1/(1 - 6*x^2 +x^2*O(x^n))^((1+3*x)/(3*x)),n)}
  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n,2^m*x^m/m!*prod(k=1,m,1+3*k*x+x*O(x^n))),n)}

Formula

E.g.f.: A(x) = Sum_{n>=0} 2^n*x^n/n! * Product_{k=1..n} (1 + 3*k*x).
a(n) ~ n! * 6^(n/2)*(n/2)^sqrt(2/3)/(2*Gamma(1+sqrt(2/3))). - Vaclav Kotesovec, Sep 22 2013
Showing 1-4 of 4 results.

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