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.

A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) / (2 * (1-2*(k+4)*x+((k-4)*x)^2))).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 19, 20, 1, 5, 34, 141, 70, 1, 6, 51, 328, 1107, 252, 1, 7, 70, 587, 3334, 8953, 924, 1, 8, 91, 924, 7123, 34904, 73789, 3432, 1, 9, 114, 1345, 12870, 89055, 372436, 616227, 12870, 1, 10, 139, 1856, 20995, 184756, 1135005, 4027216, 5196627, 48620
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Examples

			Square array begins:
    1,    1,     1,     1,      1,      1, ...
    2,    3,     4,     5,      6,      7, ...
    6,   19,    34,    51,     70,     91, ...
   20,  141,   328,   587,    924,   1345, ...
   70, 1107,  3334,  7123,  12870,  20995, ...
  252, 8953, 34904, 89055, 184756, 337877, ...

Links

Crossrefs

Columns k=0..5 give A000984, A082758, A337390, A245926, A001448, A243946.
Main diagonal gives A337388.
Cf. A337369, A337419.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    {T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).
T(0,k) = 1, T(1,k) = k+2 and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (4*(k+4)*n^2-6*(k+4)*n+k+6) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 28 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 1/2) / sqrt(8*Pi*n). - Vaclav Kotesovec, Aug 31 2020
Conjecture: the k-th column entries, k >= 0, are given by [x^n] ( (1 + (k-2)*x + x^2)*(1 + x)^2/(1 - x)^2 )^n. This is true for k = 0 and k = 4. - Peter Bala, May 03 2022

A337464 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1+2*(k-4)*x+((k+4)*x)^2) * (1-(k+4)*x+sqrt(1+2*(k-4)*x+((k+4)*x)^2)) )).

Original entry on oeis.org

1, 1, 6, 1, 5, 30, 1, 4, 11, 140, 1, 3, -6, -29, 630, 1, 2, -21, -120, -365, 2772, 1, 1, -34, -139, -266, -1409, 12012, 1, 0, -45, -92, 531, 2520, -155, 51480, 1, -1, -54, 15, 1654, 6489, 17380, 29485, 218790, 1, -2, -61, 176, 2755, 4828, -9723, -13104, 170035, 923780
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2020

Keywords

Examples

			Square array begins:
     1,     1,    1,    1,    1,     1, ...
     6,     5,    4,    3,    2,     1, ...
    30,    11,   -6,  -21,  -34,   -45, ...
   140,   -29, -120, -139,  -92,    15, ...
   630,  -365, -266,  531, 1654,  2755, ...
  2772, -1409, 2520, 6489, 4828, -5853, ...

Links

Crossrefs

Columns k=0..4 give A002457, A337394, A337466, A337467, A337397.
Main diagonal gives A337465.
Cf. A337369, A337419.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == n-j == 0, 1, (-k)^(n-j)] * Binomial[2*j, j] * Binomial[2*n+1, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 29 2021 *)
  • PARI
    {T(n, k) = sum(j=0, n, (-k)^(n-j)*binomial(2*j, j)*binomial(2*n+1, 2*j))}

Formula

T(n,k) = Sum_{j=0..n} (-k)^(n-j) * binomial(2*j,j) * binomial(2*n+1,2*j).
T(0,k) = 1, T(1,k) = 6-k and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (-4*(k-4)*n^2+2*(k-4)*n+k-2) * T(n-1,k) - (k+4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 29 2020

A337370 Expansion of sqrt(2 / ( (1-12*x+4*x^2) * (1-2*x+sqrt(1-12*x+4*x^2)) )).

Original entry on oeis.org

1, 8, 74, 736, 7606, 80464, 864772, 9400192, 103061158, 1137528688, 12623082284, 140697113792, 1574005263676, 17663830073504, 198760191043784, 2241743315230208, 25335473017856774, 286850379192127664, 3252960763923781276, 36942512756224955456, 420084161646913792724
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Links

Crossrefs

Column k=2 of A337369.
Cf. A337390.

Programs

  • Maple
    Rec:= 8*(2*n - 3)*(n - 2)*a(n - 3) - 4*(10*n^2 - 35*n + 27)*a(n - 2) - 2*(10*n^2 + 5*n - 3)*a(n - 1) + (2*n + 1)*n*a(n) = 0:
f:= gfun:-rectoproc({Rec,a(0)=1,a(1)=8,a(2)=74},a(n),remember):
map(f, [$0..30]); # Robert Israel, Aug 27 2020
  • Mathematica
    a[n_] := Sum[2^(n - k) * Binomial[2*k, k] * Binomial[2*n + 1, 2*k], {k, 0, n}]; Array[a, 21, 0] (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    N=40; x='x+O('x^N); Vec(sqrt(2/((1-12*x+4*x^2)*(1-2*x+sqrt(1-12*x+4*x^2)))))
  • PARI
    {a(n) = sum(k=0, n, 2^(n-k)*binomial(2*k, k)*binomial(2*n+1, 2*k))}

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*k,k) * binomial(2*n+1,2*k).
8*(2*n - 3)*(n - 2)*a(n - 3) - 4*(10*n^2 - 35*n + 27)*a(n - 2) - 2*(10*n^2 + 5*n - 3)*a(n - 1) + (2*n + 1)*n*a(n) = 0. - Robert Israel, Aug 27 2020
a(0) = 1, a(1) = 8 and n * (2*n+1) * (4*n-3) * a(n) = (4*n-1) * (24*n^2-12*n-4) * a(n-1) - 4 * (n-1) * (2*n-1) * (4*n+1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 29 2020
a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(2*n + 3/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 31 2020

A337387 a(n) = Sum_{k=0..n} n^(n-k) * binomial(2*k,k) * binomial(2*n+1,2*k).

Original entry on oeis.org

1, 7, 74, 1175, 24310, 610897, 17920356, 598099077, 22305598630, 917158184525, 41148369048876, 1997720107411613, 104241356841544636, 5813083330109559415, 344783011379207286920, 21660231928192698604995, 1436143861200146476260102, 100179915387243084700279349
Offset: 0

Views

Author

Seiichi Manyama, Aug 25 2020

Keywords

Crossrefs

Main diagonal of A337369.
Cf. A337388.

Programs

  • Mathematica
    a[n_] := Sum[If[n == 0, Boole[n == k], n^(n - k)] * Binomial[2*k, k] * Binomial[2*n + 1, 2*k], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    {a(n) = sum(k=0, n, n^(n-k)*binomial(2*k, k)*binomial(2*n+1, 2*k))}

Formula

From Vaclav Kotesovec, Aug 31 2020: (Start)
a(n) ~ (2 + sqrt(n))^(2*n + 3/2) / (2*n*sqrt(2*Pi)).
a(n) ~ exp(4*sqrt(n) - 4) * n^(n - 1/4) / sqrt(8*Pi) * (1 + 25/(3*sqrt(n)) + 427/(18*n)). (End)
Showing 1-4 of 4 results.

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