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.

A135148 A binomial recursion: a(n) = q(n) (see formula).

Original entry on oeis.org

0, 1, 6, 45, 400, 4115, 48146, 631729, 9189972, 146829039, 2556200086, 48167698733, 976792093784, 21211601837803, 491112582793626, 12077021182230057, 314362864408454236, 8635229233659916007, 249631741661080132766, 7575921686807827601701, 240827454421807200901728
Offset: 1

Views

Author

Benoit Cloitre, Nov 20 2007

Keywords

Links

Crossrefs

Cf. A135147, A135149, A135150, A135074, A135075, A142980.

Programs

  • Mathematica
    z[1] := x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x, 0], {n, 1, 20}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(E^x - 2*x - 1)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)
  • PARI
    r=1; s=2; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Formula

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (2 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.177398899124179661610768...
a(n) ~ (2*log(2) - 1) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: (1 - exp(x)) * (exp(x) - 2*x - 1) / (2*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020
a(n+1) = Sum_{k = 1..n} Stirling2(n, k)*A142980(k). - Peter Bala, Dec 10 2024

A135147 A binomial recursion : a(n) = p(n) (see formula).

Original entry on oeis.org

1, 4, 25, 188, 1671, 17190, 201125, 2638984, 38390179, 613363466, 10678267425, 201215691660, 4080450217247, 88609322165902, 2051573162708125, 50450534991347216, 1313219083705400475, 36072797094375866898, 1042811362801447763225, 31647646914322017237652, 1006032342980535954429463
Offset: 1

Views

Author

Benoit Cloitre, Nov 20 2007

Keywords

Links

Crossrefs

Cf. A135148, A135149, A135150, A135074, A135075.

Programs

  • Mathematica
    z[1]:= x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x], {n, 1, 20}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(-1 - E^x + 2*x)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)
  • PARI
    r=1; s=2; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Formula

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (2 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.17739889912417966161076...
a(n) ~ (3 - 2*log(2)) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: (1 - exp(x)) * (2*x - 1 - exp(x)) / (2*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020

Extensions

More terms from Amiram Eldar, Nov 25 2020

A135150 A binomial recursion: a(n) = q(n) (see formula).

Original entry on oeis.org

0, 1, 7, 59, 577, 6435, 80731, 1126321, 17306899, 290514275, 5290386805, 103892269503, 2188786203451, 49246871008285, 1178620260610039, 29898497436003155, 801364442718809233, 22629823094599476315, 671564575318740405283, 20894818098241648524577, 680161672262047334987995
Offset: 1

Views

Author

Benoit Cloitre, Nov 20 2007

Keywords

References

  • Benoit Cloitre, Binomial recursions, Pi and log2, in preparation 2007.

Links

Crossrefs

Cf. A135147, A135148, A135149, A135074, A135075.

Programs

  • Mathematica
    z[1] := x; z[n_] := 1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[
Coefficient[z[n], x, 0], {n, 1, 10}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[((-24 + 44*E^x - 46*E^(2*x))/(2 - E^x)^2 - 15*x + E^(5*x/2)*(52 + 15*Pi - 60*ArcSin[E^(x/2)/Sqrt[2]])/(2*(2 - E^x)^(5/2)))/75, {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)
  • PARI
    r=1; s=3; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Formula

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (3 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (15*Pi - 22)/(52 - 15*Pi) = 5.1524450418835554775446337...
a(n) ~ 2 * (52 - 15*Pi) * n^(3/2) * n! / (225 * sqrt(Pi) * log(2)^(n + 5/2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: ((-24 + 44*exp(x) - 46*exp(2*x))/(2 - exp(x))^2 - 15*x + exp(5*x/2)*(52 + 15*Pi - 60*arcsin(exp(x/2)/sqrt(2))) /(2*(2 - exp(x))^(5/2)))/75. - Vaclav Kotesovec, Nov 25 2020

Extensions

More terms from Vaclav Kotesovec, Nov 25 2020

A132436 A binomial recursion: a(n) = p(n) (see comment).

Original entry on oeis.org

1, 1, 4, 20, 129, 1020, 9542, 103063, 1262134, 17279744, 261531315, 4335950346, 78146040374, 1521220672933, 31808447321848, 711019048106744, 16919695824732249, 427046133330613512, 11394750238551713066, 320486422239301377007, 9476411014096567341034
Offset: 1

Views

Author

Benoit Cloitre, Nov 20 2007

Keywords

Comments

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (-1 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).

Links

Crossrefs

Cf. A135147, A135148, A135149, A135150, A135074, A135075.

Programs

  • Mathematica
    z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(-1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[Simplify[CoefficientList[Series[1 + x + E^(x/2)*(2*ArcSin[E^(x/2)/Sqrt[2]] - 1 - Pi/2)/Sqrt[2 - E^x], {x, 0, nmax}], x] * Range[0, nmax]!]] (* Vaclav Kotesovec, Nov 25 2020 *)
  • PARI
    r=1; s=-1; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Formula

Limit_{n->oo} p(n)/q(n) = (Pi-2)/(4-Pi) = 1.329896183162743847239353...
From Vaclav Kotesovec, Nov 25 2020: (Start)
a(n) ~ (Pi - 2) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).
a(n) ~ (Pi - 2) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)).
E.g.f.: 1 + x + exp(x/2)*(2*arcsin(exp(x/2)/sqrt(2)) - 1 - Pi/2) / sqrt(2 - exp(x)).
(End)

A132437 A binomial recursion: a(n) = q(n) (see comment).

Original entry on oeis.org

0, 1, 3, 15, 97, 767, 7175, 77497, 949047, 12993303, 196655437, 3260367539, 58761008087, 1143864229549, 23917992791139, 534642521054391, 12722568903456817, 321112383611040455, 8568150193087139231, 240986045600284560553, 7125677277725450247087
Offset: 1

Views

Author

Benoit Cloitre, Nov 20 2007

Keywords

Comments

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (-1 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).

Links

Crossrefs

Cf. A135147, A135148, A135149, A135150, A135074, A135075.

Programs

  • Mathematica
    z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(-1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
Rest[CoefficientList[Series[-2 - x + E^(x/2)*((4 + Pi)/2 - 2*ArcSin[E^(x/2) / Sqrt[2]]) / Sqrt[2 - E^x], {x, 0, 20}], x] * Range[0, 20]!] (* Vaclav Kotesovec, Nov 25 2020 *)
  • PARI
    r=1; s=-1; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Formula

Limit_{n->oo} p(n)/q(n) = (Pi-2)/(4-Pi) = 1.329896183162743847239353...
From Vaclav Kotesovec, Nov 25 2020: (Start)
E.g.f.: -2-x + exp(x/2)*((4+Pi)/2 - 2*arcsin(exp(x/2)/sqrt(2))) / sqrt(2-exp(x)).
a(n) ~ (4 - Pi) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).
a(n) ~ (4 - Pi) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)). (End)
Showing 1-5 of 5 results.

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