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 1-3 of 3 results.

A121629 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)^2*(a+*a), where a+ and a are boson creation and annihilation operators, respectively.

Original entry on oeis.org

1, 3, 16, 121, 1179, 14026, 196783, 3177861, 58019356, 1181098459, 26515026561, 650572403218, 17316566815441, 496889918749251, 15288155067806104, 502024850361876481, 17522822345606176083, 647790109599863145106, 25283238154309049107231
Offset: 0

Views

Author

Karol A. Penson, Aug 12 2006

Keywords

Crossrefs

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(((1-2*x)^(-1/2))-1)/(1-2*x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, May 17 2018
  • Mathematica
    CoefficientList[Series[E^(((1-2*x)^(-1/2))-1)/(1-2*x), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 29 2013 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace(exp(((1-2*x)^(-1/2))-1)/(1-2*x))) \\ G. C. Greubel, May 17 2018
    

Formula

a(n) = Sum_{p=1..n+1} abs(stirling1(n+1,p))*2^(n-p+1)*bell(p-1), n=0,1...
E.g.f.: exp(((1-2*x)^(-1/2))-1)/(1-2*x). - Vladeta Jovovic, Aug 13 2006
Recurrence: a(n) = (6*n-5)*a(n-1) - (2*n-3)*(6*n-7)*a(n-2) + 4*(2*n-3)*(n-2)^2*a(n-3). - Vaclav Kotesovec, Jun 29 2013
a(n) ~ 2^(n+5/6)*exp(3/2*(2*n)^(1/3)-1-n)*n^(n+1/3)/sqrt(3). - Vaclav Kotesovec, Jun 29 2013

Extensions

Terms a(17) onward added by G. C. Greubel, May 17 2018

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
Showing 1-3 of 3 results.