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.

A121630 Finite sum involving signless Stirling numbers of the first kind and the Bell numbers. Appears in the process of normal ordering of n-th power of (a)^3*(a+*a), where a+ and a are boson creation and annihilation operators, respectively.

Original entry on oeis.org

1, 4, 29, 302, 4089, 68056, 1342949, 30635074, 792915057, 22952573484, 734630159341, 25757268041814, 981687991859689, 40407710444419072, 1786311057929722549, 84404172618241446506, 4244839086310722228449
Offset: 0

Views

Author

Karol A. Penson, Aug 12 2006

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[E^(((1-3*x)^(-1/3))-1)/(1-3*x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 14 2014 *)

Formula

a(n)=sum(abs(stirling1(n+1,p))*3^(n-p+1)*bell(p-1),p=1..n+1), n=0,1....
E.g.f.: exp(((1-3*x)^(-1/3))-1)/(1-3*x). - Vladeta Jovovic, Aug 13 2006
Recurrence: a(n) = 3*(4*n - 5)*a(n-1) - (54*n^2 - 189*n + 173)*a(n-2) + (108*n^3 - 729*n^2 + 1659*n - 1271)*a(n-3) - 9*(n-3)^2*(3*n - 8)*(3*n - 7)*a(n-4). - Vaclav Kotesovec, Mar 14 2014
a(n) ~ 1/2 * 3^(n+7/8) * exp(4*n^(1/4)/3^(3/4) - n - 1) * n^(n+3/8). - Vaclav Kotesovec, Mar 14 2014

A121631 Finite sum involving signless Stirling numbers of the first kind and the Bell numbers. Appears in the process of normal ordering of n-th power of (a)^4*(a+*a), where a+ and a are boson creation and annihilation operators, respectively.

Original entry on oeis.org

1, 5, 46, 613, 10679, 229576, 5868715, 173833661, 5853205468, 220767370219, 9219128625851, 422221005543250, 21041188313139901, 1133454896301865073, 65627299232007207934, 4064319309355535125201, 268077821490093243979235
Offset: 0

Views

Author

Karol A. Penson, Aug 12 2006

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[E^(((1-4*x)^(-1/4))-1)/(1-4*x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Mar 14 2014 *)

Formula

a(n)=sum(abs(stirling1(n+1,p))*4^(n-p+1)*bell(p-1),p=1..n+1), n=0,1....
E.g.f.: exp(((1-4*x)^(-1/4))-1)/(1-4*x). - Vladeta Jovovic, Aug 13 2006
Recurrence: a(n) = 2*(10*n - 17)*a(n-1) - (160*n^2 - 704*n + 811)*a(n-2) + 2*(320*n^3 - 2592*n^2 + 7138*n - 6675)*a(n-3) - (1280*n^4 - 16384*n^3 + 79120*n^2 - 170816*n + 139079)*a(n-4) + 32*(n-4)^2*(2*n - 7)*(4*n - 15)*(4*n - 13)*a(n-5). - Vaclav Kotesovec, Mar 14 2014
a(n) ~ 1/sqrt(5) * 2^(2*n+9/5) * exp(5*n^(1/5)/2^(8/5)-n-1) * n^(n+2/5). - Vaclav Kotesovec, Mar 14 2014

A239301 E.g.f.: exp((1-5*x)^(-1/5)-1)/(1-5*x).

Original entry on oeis.org

1, 6, 67, 1090, 23265, 614302, 19323163, 705288522, 29296813825, 1364468928022, 70414831288275, 3987980655931570, 245910243177940897, 16399345182278307822, 1176033825828643912747, 90242683036826223141370, 7377887848681408224106497, 640225878087732419052020134
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 14 2014

Keywords

Comments

Generally, for e.g.f.: exp((1-p*x)^(-1/p)-1)/(1-p*x), and p>1, we have a(n) ~ 1/sqrt(p+1) * p^(n+(2*p+1)/(2*p+2)) * exp((p+1)*p^(-p/(p+1)) *n^(1/(p+1))-n-1) * n^(n+p/(2*p+2)).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[E^((1-5*x)^(-1/5)-1)/(1-5*x),{x,0,20}],x]*Range[0,20]!

Formula

a(n) = 5*(6*n - 13)*a(n-1) - 5*(75*n^2 - 400*n + 557)*a(n-2) + 50*(50*n^3 - 475*n^2 + 1539*n - 1698)*a(n-3) - (9375*n^4 - 137500*n^3 + 764625*n^2 - 1910000*n + 1807524)*a(n-4) + (18750*n^5 - 390625*n^4 + 3267500*n^3 - 13716875*n^2 + 28896490*n - 24436079)*a(n-5) - 25*(n-5)^2*(5*n - 24)*(5*n - 23)*(5*n - 22)*(5*n - 21)*a(n-6).
a(n) ~ 1/sqrt(6) * 5^(n+11/12) * exp(6*5^(-5/6)*n^(1/6)-n-1) * n^(n+5/12).

A123332 a(n) = 2^n*(Gamma(n+1/2)/Gamma(1/2) + (n-1)!).

Original entry on oeis.org

3, 7, 31, 201, 1713, 18075, 227295, 3317265, 55103265, 1026318195, 21181092975, 479733356025, 11829834687825, 315481555464075, 9046941599670975, 277598531343758625, 9075051786962786625, 314884420627497595875, 11557482238066613223375, 447385119579169194047625
Offset: 1

Views

Author

Karol A. Penson, Sep 26 2006

Keywords

Comments

The EXP-transform of a(n) is equal to A121629(n).

Crossrefs

Cf. A121629.

Programs

  • Mathematica
    Table[2^n*(Gamma[n + 1/2]/Gamma[1/2] + (n - 1)!), {n, 0, 50}] (* G. C. Greubel, Oct 04 2017 *)
  • PARI
    a(n) = round(2^n*(gamma(n+1/2)/gamma(1/2) + (n-1)!)); \\ Michel Marcus, Oct 05 2017

Formula

E.g.f.: (1-2*x)^(-1/2) - 1 - log(1-2*x).
Showing 1-4 of 4 results.