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 1-7 of 7 results.

A336575 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} 3^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.

Original entry on oeis.org

1, 1, 3, 1, 3, 3, 1, 3, 12, 3, 1, 3, 21, 57, 3, 1, 3, 30, 192, 300, 3, 1, 3, 39, 408, 2001, 1686, 3, 1, 3, 48, 705, 6402, 22539, 9912, 3, 1, 3, 57, 1083, 14799, 109137, 267276, 60213, 3, 1, 3, 66, 1542, 28488, 338430, 1964010, 3287496, 374988, 3, 1, 3, 75, 2082, 48765, 817743, 8181597, 36718680, 41556585, 2381322, 3
Offset: 0

Views

Author

Seiichi Manyama, Jul 26 2020

Keywords

Examples

			Square array begins:
  1,    1,     1,      1,      1,      1, ...
  3,    3,     3,      3,      3,      3, ...
  3,   12,    21,     30,     39,     48, ...
  3,   57,   192,    408,    705,   1083, ...
  3,  300,  2001,   6402,  14799,  28488, ...
  3, 1686, 22539, 109137, 338430, 817743, ...

Links

Crossrefs

Columns k=0-4 give: A122553, A047891, A219535, A336538, A336540.
Main diagonal gives A336578.
Cf. A336534, A336573, A336574.

Programs

  • Mathematica
    T[0, k_] := 1; T[n_, k_] := Sum[3^j * Binomial[n, j] * Binomial[k*n, j - 1], {j, 1, n}]/n; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 27 2020 *)
  • PARI
    T(n, k) = if(n==0, 1, sum(j=1, n, 3^j*binomial(n, j)*binomial(k*n, j-1))/n);
  • PARI
    T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k*(2+A)); polcoeff(A, n);
  • PARI
    T(n, k) = sum(j=0, n, 2^(n-j)*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));
  • PARI
    T(n, k) = sum(j=0, n, 2^j*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 * (2 + A_k(x)).
T(n,k) = Sum_{j=0..n} 2^(n-j) * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} 2^j * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = (1/n) * Sum_{j=0..n-1} (-2)^j * 3^(n-j) * binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0. - Seiichi Manyama, Aug 10 2023
T(n,k) = 3*hypergeom([1-n, -k*n], [2], 3) for n > 0. - Stefano Spezia, Aug 09 2025

A336573 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = (-1)^n * Sum_{j=0..n} (-2)^j * binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 11, 8, 1, 1, 5, 21, 45, 16, 1, 1, 6, 34, 126, 197, 32, 1, 1, 7, 50, 267, 818, 903, 64, 1, 1, 8, 69, 484, 2279, 5594, 4279, 128, 1, 1, 9, 91, 793, 5105, 20540, 39693, 20793, 256, 1, 1, 10, 116, 1210, 9946, 56928, 192350, 289510, 103049, 512
Offset: 0

Views

Author

Seiichi Manyama, Jul 26 2020

Keywords

Comments

T(n,k) is the number of Sylvester classes of k-packed words of degree n.

Examples

			Square array begins:
   1,   1,   1,    1,    1,    1, ...
   1,   1,   1,    1,    1,    1, ...
   2,   3,   4,    5,    6,    7, ...
   4,  11,  21,   34,   50,   69, ...
   8,  45, 126,  267,  484,  793, ...
  16, 197, 818, 2279, 5105, 9946, ...

Links

Crossrefs

Columns k = 0-5 are: A011782, A001003, A003168, A243659, A243667, A243668.
Main diagonal is A336495.
Cf. A336534, A336574, A336575.

Programs

  • Maple
    T := (n,k) -> `if`(k=0, `if`(n=0, 1, 2^(n-1)), (-1)^n*(binomial(k*n+1, n)* hypergeom([-n, k*n+1], [(k-1)*n+2], 2)) / (k*n+1)):
seq(lprint(seq(simplify(T(n, k)), k=0..9)), n=0..6); # Peter Luschny, Jul 26 2020
  • Mathematica
    T[n_, k_] := (-1)^n * Sum[(-2)^j * 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) = (-1)^n*sum(j=0, n, (-2)^j*binomial(n, j)*binomial(k*n+j+1, n)/(k*n+j+1));
  • PARI
    T(n, k) = my(A=1+x*O(x^n)); for(i=0, n, A=1-x*A^k*(1-2*A)); polcoeff(A, n);
  • PARI
    T(n, k) = (-1)^n*sum(j=0, n, (-2)^(n-j)*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 - 2 * A_k(x)).
T(n,k) = ( (-1)^n / (k*n+1) ) * Sum_{j=0..n} (-2)^(n-j) * binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = (-1)^n*binomial(k*n+1, n)*hypergeom([-n, k*n+1], [(k-1)*n+2], 2)/(k*n+1) for k >= 1. - Peter Luschny, Jul 26 2020
T(n,k) = (1/n) * Sum_{j=0..n-1} binomial(n,j) * binomial((k+1)*n-j,n-1-j) for n > 0. - Seiichi Manyama, Aug 08 2023

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

A219536 G.f. satisfies A(x) = 1 + x*(A(x)^2 + 2*A(x)^3).

Original entry on oeis.org

1, 3, 24, 255, 3102, 40854, 566934, 8164263, 120864390, 1827982362, 28122626760, 438720097638, 6923868098820, 110346550539780, 1773394661610258, 28707809007278775, 467677404522668742, 7661583171651546786, 126137791939032756960, 2085923447593966281378
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2012

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...
A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...

Links

Crossrefs

Column k=2 of A336574.
Cf. A003168, A103210, A219534, A219535.

Programs

  • Mathematica
    CoefficientList[1/x*InverseSeries[Series[4*x^2/(1-x-Sqrt[1-10*x+x^2]), {x, 0, 20}], x],x] (* Vaclav Kotesovec, Dec 28 2013 *)
  • PARI
    /* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */
{a(n)=my(A=1);for(i=1,n,A=1+x*(A^2+2*A^3) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))
  • PARI
    /* Formula using Series Reversion: */
{a(n)=my(A=1,G=(1-x-sqrt(1-10*x+x^2+x^3*O(x^n)))/(4*x));A=(1/x)*serreverse(x/G);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 26 2020
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 26 2020

Formula

Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where G(x) is the g.f. of A103210.
Recurrence: 4*n*(2*n+1)*(19*n-26)*a(n) = (2717*n^3 - 6435*n^2 + 4342*n - 840)*a(n-1) + 2*(n-2)*(2*n-3)*(19*n-7)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ (3/19)^(1/4) * (5+sqrt(57)) * ((143 + 19*sqrt(57))/16)^n / (16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 26 2020: (Start)
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
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(3*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(2*n,k-1) for n > 0. (End)
a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - Peter Bala, Sep 08 2024

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

A336539 G.f. A(x) satisfies A(x) = 1 + x * A(x)^3 * (1 + 2 * A(x)).

Original entry on oeis.org

1, 3, 33, 498, 8691, 164937, 3305868, 68855862, 1475636055, 32327521077, 720713175441, 16298128820568, 372946723698516, 8619565476744156, 200920644131737992, 4718057697038124750, 111505342455507462207, 2650261296098965752669, 63308992564445668959795
Offset: 0

Views

Author

Seiichi Manyama, Jul 25 2020

Keywords

Links

Crossrefs

Column k=3 of A336574.
Cf. A243659, A336538.

Programs

  • Mathematica
    a[n_] := Sum[2^k * Binomial[n, k] * Binomial[3*n + k + 1, n]/(3*n + k + 1), {k, 0, n}];  Array[a, 19, 0] (* Amiram Eldar, Jul 27 2020 *)
  • PARI
    a(n) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^3*(1+2*A)); polcoeff(A, n);
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(3*n+k+1, n)/(3*n+k+1));
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(3*n+1, k)*binomial(4*n-k, n-k))/(3*n+1); \\ Seiichi Manyama, Jul 26 2020

Formula

a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(3*n+k+1,n)/(3*n+k+1).
a(n) = (1/(3*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(3*n+1,k) * binomial(4*n-k,n-k).
a(n) ~ sqrt(168 + 97*sqrt(3)) * (26 + 15*sqrt(3))^(n - 1/2) / (3*sqrt(Pi) * n^(3/2) * 2^(n + 3/2)). - 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(4*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(3*n,k-1) for n > 0. (End)
a(n) = binomial(1+3*n, n)*hypergeom([-n, 1+3*n], [2+2*n], -2)/(1 + 3*n). - Stefano Spezia, Aug 09 2025

A336572 G.f. A(x) satisfies A(x) = 1 + x * A(x)^4 * (1 + 2 * A(x)).

Original entry on oeis.org

1, 3, 42, 822, 18708, 464115, 12175368, 332156784, 9328004700, 267870927324, 7829893576878, 232189300430454, 6968123350684692, 211232335919261178, 6458598626291716128, 198949096401788859636, 6168233789851179030684, 192334850789654814053700, 6027727888877572168027368
Offset: 0

Views

Author

Seiichi Manyama, Jul 25 2020

Keywords

Links

Crossrefs

Column k=4 of A336574.
Cf. A243667, A336540.

Programs

  • Mathematica
    a[n_] := Sum[2^k * Binomial[n, k] * Binomial[4*n + k + 1, n]/(4*n + k + 1), {k, 0, n}];  Array[a, 19, 0] (* Amiram Eldar, Jul 27 2020 *)
  • PARI
    a(n) = my(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^4*(1+2*A)); polcoeff(A, n);
  • PARI
    a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(4*n+k+1, n)/(4*n+k+1));
  • PARI
    a(n) = sum(k=0, n, 2^(n-k)*binomial(4*n+1, k)*binomial(5*n-k, n-k))/(4*n+1); \\ Seiichi Manyama, Jul 26 2020

Formula

a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(4*n+k+1,n)/(4*n+k+1).
a(n) = (1/(4*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(4*n+1,k) * binomial(5*n-k,n-k).
a(n) ~ sqrt(95781603 + 7199237*sqrt(177))*(69845 + 5251*sqrt(177))^(n - 1/2) / (sqrt(59*Pi) * n^(3/2) * 2^(12*n + 9/2)). - 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(5*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(4*n,k-1) for n > 0. (End)
a(n) = binomial(1+4*n, n)*hypergeom([-n, 1+4*n], [2+3*n], -2)/(1 + 4*n). - Stefano Spezia, Aug 09 2025
Showing 1-7 of 7 results.

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