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 1-10 of 19 results. Next

A187536 Partial sums of the central Lah numbers (A187535).

Original entry on oeis.org

1, 3, 39, 1239, 60039, 3870279, 311229639, 29993362119, 3369233266119, 432276047602119, 62366420037720519, 9994350965362162119, 1761334292457572030919, 338557476887113316030919, 70488382605888266852030919, 15802755831536546966525630919
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Crossrefs

Cf. A187535, A008297, A111596, A187538 - A187540, A187542 - A187548.

Programs

  • Maple
    A187536 := proc(n) add(A187535(i),i=0..n) ; end proc:
seq(A187536(n),n=0..10) ; # R. J. Mathar, Mar 20 2011
  • Mathematica
    Table[1 + Sum[Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist(1+sum(binomial(2*k-1,k-1)*(2*k)!/k!,k,1,n),n,0,12);

Formula

a(n) = 1 + Sum_{k=0..n} binomial(2k-1,k-1)*(2k)!/k!.
(n+2)*a(n+2) - (16n^2 + 49n +3 8)*a(n+1) + 4 *(2n+3)^2*a(n) = 0.
Asymptotically a(n) ~ 2^(4n)n^n exp(-n)/sqrt(2n*pi).

A187538 Alternating partial sums of the central Lah numbers (A187535).

Original entry on oeis.org

1, 1, 35, 1165, 57635, 3752605, 303606755, 29378525725, 3309861378275, 425596952957725, 61508547037160675, 9870475998287280925, 1741469465493922587875, 335054673129161821412125, 69814770455871991714587875, 15662452678474786707959012125, 3764014801927115965888623387875
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187539, A187540, A187542 - A187548.

Programs

  • Maple
    A187538 := proc(n) add( (-1)^(n+k)*A187535(k),k=0..n) ; end proc:
seq(A187538(n),n=0..10) ; # R. J. Mathar, Mar 21 2011
  • Mathematica
    Table[(-1)^n + Sum[(-1)^(n-k)Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist((-1)^n+sum((-1)^(n-k)*binomial(2*k-1,k-1)*(2*k)!/k!,k,1,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*A187535(k).
(n+2)*a(n+2) - (16*n^2 + 47*n + 34)*a(n+1) - 4*(2*n+3)^2*a(n) = 0.
a(n) ~ 2^(4*n - 1/2) * n^(n - 1/2) / (sqrt(Pi) * exp(n)). - Vaclav Kotesovec, Mar 30 2018

A187542 Convolutions of the central Lah numbers (A187535).

Original entry on oeis.org

1, 4, 76, 2544, 123696, 7942080, 635633280, 61009159680, 6831940227840, 874493448514560, 125946241018214400, 20156433977646489600, 3548609812373223628800, 681555522002874494976000, 141810253720479017017344000
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187543, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    a := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(sum(a(k)*a(n-k), k=0..n),n=0..12);
  • Mathematica
    a[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[a[k]a[n - k], {k, 0, n}], {n, 0, 20}]
  • Maxima
    a(n) := if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(a(k)*a(n-k),k,0,n),n,0,12);

Formula

a(n) = sum(L(k)L(n-k),k=0..n), where L(n) is a central Lah number.
a(n) ~ n! * 16^n / (Pi*n). - Vaclav Kotesovec, Oct 06 2019

A187539 Alternated binomial partial sums of central Lah numbers (A187535).

Original entry on oeis.org

1, 1, 33, 1097, 54209, 3527889, 285356449, 27608615257, 3110179582593, 399896866564001, 57791843384031521, 9273757516482276201, 1636151050649025202753, 314786007405793614831217, 65590496972310741712688289, 14714600180590751334321307769
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187540, A187542, A187543, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    seq((-1)^n+add((-1)^(n-k)*binomial(n,k)*binomial(2*k-1,k-1)*(2*k)!/k!, k=1..n), n=0..20);
  • Mathematica
    Table[(-1)^n + Sum[(-1)^(n-k)Binomial[n,k]Binomial[2k-1,k-1](2k)!/k!, {k, 1, n}], {n, 0, 20}]
  • Maxima
    makelist((-1)^n+sum((-1)^(n-k)*binomial(n,k)*binomial(2*k-1,k-1) *(2*k)!/k!, k,1,n), n,0,12);

Formula

a(n) = 1+sum((-1)^(n-k)*C(n,k)*C(2k-1,k-1)*(2k)!/k!, k=0..n).
Recurrence: n>=3, a(n) = (2*(-1)^n + (32 - 48*n + 16*n^2)*a(n-3) + (33 - 65*n + 32*n^2)*a(n-2) + (5 - 18*n + 16*n^2)*a(n-1))/n
E.g.f.: exp(-x) (1/2 + 1/pi K(16x) ), where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ 16^n*n^(n-1/2)/(sqrt(2*Pi)*exp(n+1/16)). - Vaclav Kotesovec, Aug 10 2013

A187543 Binomial convolutions of the central Lah numbers (A187535).

Original entry on oeis.org

1, 4, 80, 2832, 144576, 9660480, 798468480, 78670609920, 9002061573120, 1173384611804160, 171641216823552000, 27843893955582566400, 4961007038613633638400, 963075987422089673932800, 202333751987206944654950400
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187542, A187544, A187545, A187546, A187547, A187548.

Programs

  • Maple
    a := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(add(binomial(n,k)*a(k)*a(n-k), k=0..n),n=0..12);
  • Mathematica
    a[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[Binomial[n, k]a[k]a[n - k], {k, 0, n}], {n, 0, 20}]
CoefficientList[Series[(1/2 + EllipticK[16*x]/Pi)^2, {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Oct 06 2019 *)
  • Maxima
    a(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(binomial(n,k)*a(k)*a(n-k),k,0,n),n,0,12);

Formula

a(n) = sum(binomial(n,k)*L(k)*L(n-k),k=0..n), where L(n) is a central Lah number.
E.g.f.: (1/2 + 1/Pi*K(16x))^2, where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
Recurrence: (n-1)*n^2*(4*n^2-15*n+13)*a(n) = 4*(n-1)*(48*n^5-292*n^4+672*n^3-747*n^2+399*n-76)*a(n-1) - 32*(96*n^7-1000*n^6+4408*n^5-10628*n^4+15034*n^3-12312*n^2+5265*n-854)*a(n-2) + 1024*(2*n-5)^2*(4*n^2-7*n+2)*(n-2)^4*a(n-3). - Vaclav Kotesovec, Aug 10 2013
a(n) ~ n! * log(n) * 2^(4*n-1) / (Pi^2 * n) * (1 + (gamma + Pi + 4*log(2)) / log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Oct 06 2019

A187544 Stirling transform (of the second kind) of the central Lah numbers (A187535).

Original entry on oeis.org

1, 2, 38, 1310, 66254, 4428782, 368444078, 36691056110, 4256199137774, 563672814445742, 83921091641375918, 13875375391723852910, 2522552600160248918894, 500141581330626431059502, 107400097037199576065830958
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187542, A187543, A187545, A187546, A187547, A187548.

Programs

  • Maple
    a := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(sum(combinat[stirling2](n,k)*a(k), k=0..n),n=0..12);
  • Mathematica
    a[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[StirlingS2[n, k]a[k], {k, 0, n}], {n, 0, 20}]
CoefficientList[Series[1/2 + EllipticK[16*(E^x - 1)]/Pi, {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Oct 06 2019 *)
  • Maxima
    a(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(stirling2(n,k)*a(k),k,0,n),n,0,12);

Formula

a(n) = sum(S(n,k)*L(k),k=0..n), where S(n,k) are the Stirling numbers of the second kind and L(n) are the central Lah numbers.
E.g.f.: 1/2 + 1/Pi*K(16(exp(x)-1)) where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ n! / (2*Pi*n * (log(17/16))^n). - Vaclav Kotesovec, Oct 06 2019

A187545 Stirling transform (of the first kind) of the central Lah numbers (A187535).

Original entry on oeis.org

1, 2, 38, 1312, 66408, 4442088, 369791064, 36848702784, 4277191653888, 566809715422464, 84441103242634176, 13970100487593468480, 2541362625439551554880, 504185908064687887996800, 108336183242510523080868480
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187542, A187543, A187544, A187546, A187547, A187548.

Programs

  • Maple
    lahc := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(add(abs(combinat[stirling1](n,k))*lahc(k), k=0..n), n=0..20);
  • Mathematica
    lahc[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[Abs[StirlingS1[n, k]]*lahc[k], {k, 0, n}], {n, 0, 20}]
  • Maxima
    lahc(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(abs(stirling1(n,k))*lahc(k),k,0,n),n,0,12);

Formula

a(n) = sum(s(n,k)*L(k), k=0..n), where s(n,k) are the (signless) Stirling numbers of the first kind and L(n) are the central Lah numbers.
E.g.f.: 1/2 + 1/Pi*K(-16*log(1-x)), where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ n! / (2*Pi*n * (1 - exp(-1/16))^n). - Vaclav Kotesovec, Apr 10 2018

A187546 Stirling transform (of the first kind, with signs) of the central Lah numbers (A187535).

Original entry on oeis.org

1, 2, 34, 1096, 51984, 3262488, 254943384, 23853046656, 2600024557248, 323588157732096, 45276442446814656, 7035574740347812800, 1202158966644148296000, 224022356544364922931840, 45215509996613004825121920
Offset: 0

Views

Author

Emanuele Munarini, Mar 11 2011

Keywords

Links

Crossrefs

Cf. A187536, A008297, A111596, A187536, A187538, A187539, A187540, A187542, A187543, A187544, A187545, A187547, A187548.

Programs

  • Maple
    lahc := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(add(combinat[stirling1](n,k)*lahc(k), k=0..n), n=0..20);
  • Mathematica
    lahc[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[StirlingS1[n, k]*lahc[k], {k, 0, n}], {n, 0, 20}]
  • Maxima
    lahc(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(stirling1(n,k)*lahc(k),k,0,n),n,0,12);

Formula

a(n) = sum((-1)^(n-k)*s(n,k)*L(k), k=0..n), where s(n,k) are the (signless) Stirling numbers of the first kind and L(n) are the central Lah numbers.
E.g.f.: 1/2 + 1/Pi*K(16*log(1+x)), where K(z) is the elliptic integral of the first kind (defined as in Mathematica).
a(n) ~ n! / (2*Pi*n * (exp(1/16) - 1)^n). - Vaclav Kotesovec, Apr 10 2018

A187662 Convolution of the (signless) central Lah numbers (A187535) and the central Stirling numbers of the second kind (A007820).

Original entry on oeis.org

1, 3, 45, 1340, 62133, 3926607, 313159138, 30077004204, 3373855596485, 432604296358341, 62396125789568633, 9997677582465775336, 1761777732880595653932, 338625441643226149909356, 70500059235176885929427760
Offset: 0

Views

Author

Emanuele Munarini, Mar 12 2011

Keywords

Crossrefs

Cf. A187535, A007820.

Programs

  • Maple
    L := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(sum(L(k)*combinat[stirling2](2*(n-k),n-k),k=0..n),n=0..12);
  • Mathematica
    L[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[L[k]StirlingS2[2n - 2k, n - k], {k, 0, n}], {n, 0, 14}]
  • Maxima
    L(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(L(k)*stirling2(2*n-2*k,n-k),k,0,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} Lah(2*k,k)*S(2*n-2*k,n-k).
a(n) ~ 2^(4*n) * n^n / (exp(n) * sqrt(2*Pi*n)). - Vaclav Kotesovec, May 21 2014

A187664 Convolution of the (signless) central Lah numbers (A187535) and the (signless) central Stirling numbers of the first kind (A187646).

Original entry on oeis.org

1, 3, 49, 1483, 67615, 4173203, 326208269, 30880075203, 3430574739759, 437145190334383, 62803806114813801, 10038354053796477099, 1766255133182030548351, 339166069936077378326187, 70571377417819411767223541
Offset: 0

Views

Author

Emanuele Munarini, Mar 12 2011

Keywords

Crossrefs

Cf. A187535, A187646.

Programs

  • Maple
    L := n -> if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n! fi;
seq(sum(L(k)*abs(combinat[stirling1](2*(n-k),n-k)),k=0..n),n=0..12);
  • Mathematica
    L[n_] := If[n == 0, 1, Binomial[2n - 1, n - 1](2n)!/n!]
Table[Sum[L[k]Abs[StirlingS1[2n - 2k, n - k]], {k, 0, n}], {n, 0, 14}]
  • Maxima
    L(n):= if n=0 then 1 else binomial(2*n-1,n-1)*(2*n)!/n!;
makelist(sum(L(k)*abs(stirling1(2*n-2*k,n-k)),k,0,n),n,0,12);

Formula

a(n) = Sum_{k=0..n} Lah(2*k,k)*s(2*n-2*k,n-k).
a(n) ~ 2^(4*n - 1/2) * n^(n - 1/2) / (sqrt(Pi) * exp(n)). - Vaclav Kotesovec, May 30 2025
Showing 1-10 of 19 results. Next

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