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.

A107267 A square array of Motzkin related transforms, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 3, 1, 0, 9, 20, 12, 4, 1, 0, 21, 72, 54, 20, 5, 1, 0, 51, 272, 261, 112, 30, 6, 1, 0, 127, 1064, 1323, 672, 200, 42, 7, 1, 0, 323, 4272, 6939, 4224, 1425, 324, 56, 8, 1, 0, 835, 17504, 37341, 27456, 10625, 2664, 490, 72, 9, 1
Offset: 0

Views

Author

Paul Barry, May 15 2005

Keywords

Comments

Rows are transforms of k^n, k>=0, under the matrix A107131. As a number triangle, with T(n,k)=if(k<=n,sum{j=0..n-k, (1/(j+1))C(j+1,n-k-j+1)C(n-k,j)k^j},0), row sums are A107268 and diagonal sums are A107269. Rows are series reversions of x/(1+kx+kx^2), k>=0. Conjecture: rows count weighted Motzkin paths.
Row k counts colored Motzkin paths, where H(1,0) and U(1,1) each have k colors and D(1,-1) one color. - Paul Barry, May 16 2005

Examples

			Array begins
  1, 0,  0,   0,    0,     0,      0, ...
  1, 1,  2,   4,    9,    21,     51, ...
  1, 2,  6,  20,   72,   272,   1064, ...
  1, 3, 12,  54,  261,  1323,   6939, ...
  1, 4, 20, 112,  672,  4224,  27456, ...
  1, 5, 30, 200, 1425, 10625,  81875, ...
  1, 6, 42, 324, 2664, 22896, 203256, ...

Links

Crossrefs

Rows n=0..6 give A000007, A001006, A071356, A107264, A003645, A107265, A107266.
Main diagonal gives A292716.
Cf. A000108.

Formula

Number array T(n,k) = Sum_{j=0..k} n^j * binomial(k,j) * binomial(j+1,k-j+1)/(j+1).
G.f. of row k: 1/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
From Seiichi Manyama, May 05 2019: (Start)
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * binomial(2*j,j)/(j+1) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * A000108(j).
(k+2) * T(n,k) = n * (2*k+1) * T(n,k-1) - n * (n-4) * (k-1) * T(n,k-2). (End)

A247496 a(n) = n!*[x^n](exp(n*x)*BesselI_{1}(2*x)/x), n>=0, main diagonal of A247495.

Original entry on oeis.org

1, 1, 5, 36, 354, 4425, 67181, 1200745, 24699662, 574795035, 14930563042, 428235433978, 13442267711940, 458373150076335, 16872717817840509, 666835739823870900, 28163028244810505622, 1265837029802096365275, 60330098878933736719190, 3039079334694016053006276
Offset: 0

Views

Author

Peter Luschny, Dec 12 2014

Keywords

Comments

Also coefficient of x^n in the expansion of 1/(n+1) * (1 + n*x + x^2)^(n+1). - Seiichi Manyama, May 06 2019

Links

Crossrefs

Cf. A001006, A186925, A247495, A292716.

Programs

  • Mathematica
    Flatten[{1,Table[n^n*HypergeometricPFQ[{1/2-n/2, -n/2}, {2}, 4/n^2],{n,1,20}]}] (* Vaclav Kotesovec, Dec 12 2014 *)
  • PARI
    {a(n) = sum(k=0, n\2, n^(n-2*k)*binomial(n, 2*k)*binomial(2*k, k)/(k+1))} \\ Seiichi Manyama, May 05 2019
  • PARI
    {a(n) = polcoef((1+n*x+x^2)^(n+1)/(n+1), n)} \\ Seiichi Manyama, May 06 2019
  • Sage
    a = lambda n: 1 if n==0 else n^n*hypergeometric([1/2-n/2, -n/2], [2], 4/n^2).simplify()
[a(n) for n in range(20)]

Formula

a(n) = Sum_{j=0..floor(n/2)} ((j+1)*n^(n-2*j)*n!)/((j+1)!^2*(n-2*j)!).
a(n) ~ BesselI(1,2) * n^n. - Vaclav Kotesovec, Dec 12 2014
From Ilya Gutkovskiy, Sep 21 2017: (Start)
a(n) = [x^n] (1 - n*x - sqrt(1 - 2*n*x + (n^2 - 4)*x^2))/(2*x^2).
a(n) = [x^n] 1/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - n*x - x^2/(1 - ...))))), a continued fraction. (End)

A307906 Coefficient of x^n in 1/(n+1) * (1 + x + n*x^2)^(n+1).

Original entry on oeis.org

1, 1, 3, 10, 57, 301, 2251, 15583, 138209, 1153603, 11592451, 111381348, 1235739385, 13276480803, 159935056555, 1884023828326, 24356065951617, 310189106485419, 4266048524240323, 58124516559463590, 844705360693479801, 12213285476055278959, 186543178982826381387
Offset: 0

Views

Author

Seiichi Manyama, May 05 2019

Keywords

Comments

Also coefficient of x^n in the expansion of 2/(1 - x + sqrt(1 - 2*x + (1 - 4*n)*x^2)).

Links

Crossrefs

Main diagonal of A306684.
Cf. A000108, A001006, A187018, A247496, A292716.

Programs

  • Mathematica
    Table[Hypergeometric2F1[1/2 - n/2, -n/2, 2, 4*n], {n, 0, 20}] (* Vaclav Kotesovec, May 05 2019 *)
  • PARI
    {a(n) = polcoef((1+x+n*x^2)^(n+1)/(n+1), n)}
  • PARI
    {a(n) = sum(k=0, n\2, n^k*binomial(n, k)*binomial(n-k, k)/(k+1))}
  • PARI
    {a(n) = sum(k=0, n\2, n^k*binomial(n, 2*k)*binomial(2*k, k)/(k+1))}

Formula

a(n) = [x^n] (1 - x - sqrt(1 - 2*x + (1 - 4*n)*x^2))/(2*n*x^2).
a(n) = Sum_{k=0..floor(n/2)} n^k * binomial(n,k) * binomial(n-k,k)/(k+1) = Sum_{k=0..floor(n/2)} n^k * binomial(n,2*k) * A000108(k).
a(n) ~ exp(sqrt(n)/2 - 1/8) * 2^(n + 1/2) * n^((n-3)/2) / sqrt(Pi). - Vaclav Kotesovec, May 05 2019

A307946 Coefficient of x^n in 1/(n+1) * (1 - n*x - n*x^2)^(n+1).

Original entry on oeis.org

1, -1, 2, 0, -96, 1875, -32184, 554631, -9773056, 172718325, -2874200000, 35973317666, 218394869760, -46968959184459, 2890848443624064, -147665402789062500, 7121567693920010240, -337669517265832692843, 15985827659730523364352, -759295252512454596032456
Offset: 0

Views

Author

Seiichi Manyama, May 07 2019

Keywords

Comments

Also coefficient of x^n in the expansion of 2/(1 + n*x + sqrt(1 + 2*n*x + n*(n+4)*x^2)).

Links

Crossrefs

Cf. A000108, A292716, A307911.

Programs

  • Mathematica
    a[0] = 1; a[n_] := Sum[(-n)^(n-k) * Binomial[n, 2*k] * CatalanNumber[k], {k, 0, Floor[n/2]}]; Array[a, 20, 0] // Flatten (* Amiram Eldar, May 12 2021 *)
Join[{1}, Table[(-n)^n * Hypergeometric2F1[1/2 - n/2, -n/2, 2, -4/n], {n, 1, 20}]] (* Vaclav Kotesovec, May 12 2021 *)
  • PARI
    {a(n) = polcoef((1-n*x-n*x^2)^(n+1)/(n+1), n)}
  • PARI
    {a(n) = sum(k=0, n\2, (-n)^(n-k)*binomial(n, k)*binomial(n-k, k)/(k+1))}
  • PARI
    {a(n) = sum(k=0, n\2, (-n)^(n-k)*binomial(n, 2*k)*binomial(2*k, k)/(k+1))}

Formula

a(n) = Sum_{k=0..floor(n/2)} (-n)^(n-k) * binomial(n,k) * binomial(n-k,k)/(k+1) = Sum_{k=0..floor(n/2)} (-n)^(n-k) * binomial(n,2*k) * A000108(k).
For n>0, a(n) = (-n)^n * Hypergeometric2F1(1/2 - n/2, -n/2, 2, -4/n). - Vaclav Kotesovec, May 12 2021

A294642 a(n) = n! * [x^n] exp(n*x)*BesselI(1,2*sqrt(2)*x)/(sqrt(2)*x).

Original entry on oeis.org

1, 1, 6, 45, 456, 5825, 89896, 1627437, 33822944, 793783233, 20765009344, 599157626925, 18904594000128, 647524807918209, 23929038677825152, 948995910652193325, 40203601321988822528, 1812025020244371552897, 86577002960871477916672, 4371100278517527047687213
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 05 2017

Keywords

Links

Crossrefs

Cf. A000108, A001003, A025235, A068764, A071356, A151374, A247496, A292716.

Programs

  • Mathematica
    Simplify[Table[n! SeriesCoefficient[Exp[n x] BesselI[1, 2 Sqrt[2] x]/(Sqrt[2] x), {x, 0, n}], {n, 0, 19}]]
Table[SeriesCoefficient[(1 - n x - Sqrt[1 - 2 n x + (n^2 - 8) x^2])/(4 x^2), {x, 0, n}], {n, 0, 19}]
Table[SeriesCoefficient[1/(1 - n x + ContinuedFractionK[-2 x^2, 1 - n x, {i, 1, n}]), {x, 0, n}], {n, 0, 19}]
Join[{1}, Table[Sum[2^k n^(n - 2 k) Binomial[n, 2 k] CatalanNumber[k], {k, 0, Floor[n/2]}], {n, 1, 19}]]
Join[{1}, Table[n^n HypergeometricPFQ[{1/2 - n/2, -n/2}, {2}, 8/n^2], {n, 1, 19}]]

Formula

a(n) = [x^n] (1 - n*x - sqrt(1 - 2*n*x + (n^2 - 8)*x^2))/(4*x^2).
a(n) = [x^n] 1/(1 - n*x - 2*x^2/(1 - n*x - 2*x^2/(1 - n*x - 2*x^2/(1 - n*x - 2*x^2/(1 - ...))))), a continued fraction.
a(n) = Sum_{k=0..floor(n/2)} 2^k*n^(n-2*k)*binomial(n,2*k)*A000108(k).
a(n) = n^n*2F1(1/2-n/2,-n/2; 2; 8/n^2).
a(n) ~ c * n^n, where c = BesselI(1, 2*sqrt(2))/sqrt(2) = 2.3948330992734... - Vaclav Kotesovec, Nov 06 2017
Showing 1-5 of 5 results.

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