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.

A336578 a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} 3^k * binomial(n,k) * binomial(n^2,k-1) for n > 0.

Original entry on oeis.org

1, 3, 21, 408, 14799, 817743, 61621806, 5921141502, 694008501627, 96176405390961, 15400332946269903, 2799678523675400832, 569877183695866859625, 128436925725088289658534, 31756620986815666396814796, 8548059658831271609064999978, 2488568825786280454788465874035, 779186768737628124697943895022101
Offset: 0

Views

Author

Seiichi Manyama, Jul 26 2020

Keywords

Links

Crossrefs

Main diagonal of A336575.
Cf. A336495, A336537, A336577.

Programs

  • Mathematica
    a[0] := 1; a[n_] := Sum[3^k * Binomial[n, k] * Binomial[n^2, k - 1], {k, 1, n}]/n;  Array[a, 18, 0] (* Amiram Eldar, Jul 27 2020 *)
  • PARI
    a(n) = if(n==0, 1, sum(k=1, n, 3^k*binomial(n, k)*binomial(n^2, k-1))/n);
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n^2+k+1, n)/(n^2+k+1)); \\ Seiichi Manyama, Jul 27 2020
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n^2+1, k)*binomial((n+1)*n-k, n-k))/(n^2+1); \\ Seiichi Manyama, Jul 27 2020

Formula

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(n^2+k+1,n)/(n^2+k+1).
a(n) = (1/(n^2+1)) * Sum_{k=0..n} 2^k * binomial(n^2+1,k) * binomial((n+1)*n-k,n-k).
a(n) ~ 3^n * exp(n - 1/6) * n^(n - 5/2) / sqrt(2*Pi). - Vaclav Kotesovec, Jul 31 2021
a(n) = (1/n) * Sum_{k=0..n-1} (-2)^k * 3^(n-k) * binomial(n,k) * binomial((n+1)*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 10 2023
a(n) = 3*hypergeom([1-n, -n^2], [2], 3) for n > 0. - Stefano Spezia, Aug 09 2025

A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 2, 6, 2, 1, 2, 10, 22, 2, 1, 2, 14, 66, 90, 2, 1, 2, 18, 134, 498, 394, 2, 1, 2, 22, 226, 1482, 4066, 1806, 2, 1, 2, 26, 342, 3298, 17818, 34970, 8558, 2, 1, 2, 30, 482, 6202, 52450, 226214, 312066, 41586, 2, 1, 2, 34, 646, 10450, 122762, 881970, 2984206, 2862562, 206098, 2
Offset: 0

Views

Author

Seiichi Manyama, Jul 25 2020

Keywords

Examples

			Square array begins:
  1,   1,    1,     1,     1,      1, ...
  2,   2,    2,     2,     2,      2, ...
  2,   6,   10,    14,    18,     22, ...
  2,  22,   66,   134,   226,    342, ...
  2,  90,  498,  1482,  3298,   6202, ...
  2, 394, 4066, 17818, 52450, 122762, ...

Links

Crossrefs

Columns k=0-3 give A040000, A006318, A027307, A144097.
If Michael D. Weiner's conjecture on A260332 is correct, column 4 is A260332 for n > 0.
Main diagonal gives A336537.
Cf. A336521, A336574. A336575.

Programs

  • Mathematica
    T[n_, k_] := Sum[Binomial[n, j] * Binomial[k*n+j+1, n]/(k*n+j+1), {j, 0, n}]; Table[T[k, n-k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 01 2021 *)
  • PARI
    T(n, k) = sum(j=0, n, binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1);

Formula

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k * (1 + A_k(x)).
T(n,k) = (1/n) * Sum_{j=1..n} 2^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = binomial(1+k*n, n)*hypergeom([-n, 1+k*n], [2+(k-1)*n], -1)/(1 + k*n) for k > 0. - Stefano Spezia, Aug 09 2025

A336577 a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(n^2+k+1,n)/(n^2+k+1).

Original entry on oeis.org

1, 3, 24, 498, 18708, 1055838, 80682414, 7829287392, 924359573112, 128815914107370, 20717986773639696, 3779867347688995698, 771666206195918154156, 174345811623642373266360, 43198501381068549879753648, 11648965476456962547182140512, 3396661425137920919866033312752
Offset: 0

Views

Author

Seiichi Manyama, Jul 26 2020

Keywords

Links

Crossrefs

Main diagonal of A336574.
Cf. A336495, A336537, A336578.

Programs

  • Mathematica
    a[n_] := Sum[2^k * Binomial[n, k] * Binomial[n^2 + k + 1, n]/(n^2 + k + 1), {k, 0, n}];  Array[a, 17, 0] (* Amiram Eldar, Jul 27 2020 *)
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n^2+k+1, n)/(n^2+k+1));
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(n^2+1, k)*binomial((n+1)*n-k, n-k))/(n^2+1);

Formula

a(n) = (1/(n^2+1)) * Sum_{k=0..n} 2^(n-k) * binomial(n^2+1,k) * binomial((n+1)*n-k,n-k).
a(n) ~ 3^n * exp(n + 1/6) * n^(n - 5/2) / sqrt(2*Pi). - Vaclav Kotesovec, Jul 31 2021
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 3^(n-k) * binomial(n,k) * binomial((n+1)*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * 2^(n-k) * binomial(n,k) * binomial(n^2,k-1) for n > 0. (End)
a(n) = binomial(1+n^2, n)*hypergeom([-n, 1+n^2], [2-n+n^2], -2)/(1 + n^2). - Stefano Spezia, Aug 09 2025

A336522 a(n) is the coefficient of x^(n^2) in expansion of ( (1 + x)/(1 - x) )^n.

Original entry on oeis.org

1, 2, 16, 326, 11008, 525002, 32497680, 2478629134, 224921989120, 23681262354194, 2838826197080080, 381825269929428822, 56949892477659339520, 9329658433405643973850, 1665421971238565711337488, 321771059958076157377283102, 66901218825369170336327860224, 14894388013750938445628478094370
Offset: 0

Views

Author

Seiichi Manyama, Jul 24 2020

Keywords

Links

Crossrefs

Main diagonal of A336521.
Cf. A336537.

Programs

  • Mathematica
    a[n_] := Sum[Binomial[n, k] * Binomial[n^2 + k - 1, n - 1], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Jul 24 2020 *)
  • PARI
    {a(n) = if(n==0, 1, sum(k=0, n, binomial(n^2, n-k) * binomial(n^2+k-1, k))/n)}
  • PARI
    {a(n) = if(n==0, 1, sum(k=1, n, 2^k*binomial(n, k) * binomial(n^2-1, k-1)))}

Formula

a(n) = (1/n) * [x^n] ( (1 + x)/(1 - x) )^(n^2) for n > 0.
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n^2+k-1,n-1).
a(n) = (1/n) * Sum_{k=0..n} binomial(n^2,n-k) * binomial(n^2+k-1,k) for n > 0.
a(n) = Sum_{k=1..n} 2^k * binomial(n,k) * binomial(n^2-1,k-1) for n > 0.
a(n) ~ 2^(n - 1/2) * exp(n) * n^(n - 3/2) / sqrt(Pi). - Vaclav Kotesovec, Jul 31 2021
a(n) = binomial(n^2-1, n-1)*hypergeom([-n, n^2], [1-n+n^2], -1). - Stefano Spezia, Aug 09 2025
Showing 1-4 of 4 results.

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