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 1-3 of 3 results.

A378317 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n,r) * binomial(2*r+k,n)/(2*r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 4, 0, 1, 6, 12, 12, 0, 1, 8, 24, 40, 40, 0, 1, 10, 40, 92, 144, 144, 0, 1, 12, 60, 176, 360, 544, 544, 0, 1, 14, 84, 300, 752, 1440, 2128, 2128, 0, 1, 16, 112, 472, 1400, 3200, 5872, 8544, 8544, 0, 1, 18, 144, 700, 2400, 6352, 13664, 24336, 35008, 35008, 0
Offset: 0

Views

Author

Seiichi Manyama, Nov 23 2024

Keywords

Examples

			Square array begins:
  1,   1,    1,    1,     1,     1,     1, ...
  0,   2,    4,    6,     8,    10,    12, ...
  0,   4,   12,   24,    40,    60,    84, ...
  0,  12,   40,   92,   176,   300,   472, ...
  0,  40,  144,  360,   752,  1400,  2400, ...
  0, 144,  544, 1440,  3200,  6352, 11616, ...
  0, 544, 2128, 5872, 13664, 28480, 54768, ...

Crossrefs

Columns k=0..1 give A000007, A025227(n+1).
Main diagonal gives A333473.
Cf. A266213, A378318.

Programs

  • PARI
    T(n, k, t=0, u=2) = if(k==0, 0^n, k*sum(r=0, n, binomial(n, r)*binomial(t*n+u*r+k, n)/(t*n+u*r+k)));
matrix(7, 7, n, k, T(n-1, k-1))

Formula

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x + x * A_k(x)^(2/k) )^k for k > 0.
G.f. of column k: (B(x)/x)^k where B(x) is the g.f. of A025227.
B(x)^k = B(x)^(k-1) + x * B(x)^(k-1) + x * B(x)^(k+1). So T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k+1) for n > 0.

A378378 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+3*k-1,3*k).

Original entry on oeis.org

1, 2, 16, 170, 1920, 22402, 266800, 3222634, 39328768, 483752258, 5987236816, 74474238698, 930212870784, 11659157743170, 146567181170160, 1847198697449770, 23332153206562816, 295286370825453442, 3743540075432798608, 47532529217041519658, 604366048841146280320
Offset: 0

Views

Author

Seiichi Manyama, Nov 24 2024

Keywords

Crossrefs

Main diagonal of A378318.
Cf. A123164, A333473.

Programs

  • Mathematica
    a[n_]:=HypergeometricPFQ[{(1+n)/3,(2+n)/3,-n,n/3},{1/3,2/3,1},-1]; Array[a,21,0] (* Stefano Spezia, Nov 24 2024 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(n+3*k-1, 3*k));

Formula

a(n) = hypergeom([(1+n)/3, (2+n)/3, -n, n/3], [1/3, 2/3, 1], -1). - Stefano Spezia, Nov 24 2024

A333472 a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 1, 3, 13, 59, 276, 1317, 6371, 31131, 153292, 759428, 3780888, 18900389, 94805959, 476945913, 2405454213, 12158471195, 61574325840, 312365992620, 1587052145492, 8074474510884, 41131551386120, 209760563456920, 1070822078321520, 5471643738383781, 27982867986637151
Offset: 0

Views

Author

Peter Bala, Mar 23 2020

Keywords

Comments

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence u(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for any prime p and positive integers n and k. For certain power series F(x), stronger congruences may hold. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
Here we take F(x) = c(x/(1 + x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + ... (cf. A001006 and A086246) and conjecture that the associated sequence a(n) = [x^n] ( c(x/(1 + x)) )^n satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333473.
More generally, we conjecture that for any positive integer r and any integer s the sequence a(r,s;n) := [x^(r*n)] ( c(x/(1 + x)) )^(s*n) also satisfies the above congruences.
Note that the sequence b(n) := [x^n] c(x)^n = A025174(n) satisfies the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*c(x/(1 + x)) )^n = A333093(n) appears to satisfy the same congruences.

Examples

			Examples of congruences:
a(11) - a(1) = 3780888 - 1 = (11^2)*31247 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 41131551386120 - 13 = (7^2)*13*23671*2727841 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 27982867986637151 - 276 = (5^4)*13*74687*46113049 == 0 ( mod 5^4 ).

Crossrefs

Cf. A000108, A001006, A086246, A106228, A219537, A333093, A333473.

Programs

  • Maple
    Cat := x -> (1/2)*(1-sqrt(1-4*x))/x:
G := x -> Cat(x/(1+x)):
H := (x,n) -> series(G(x)^n, x, 51):
seq(coeff(H(x, n), x, n), n = 0..25);
  • Mathematica
    Table[SeriesCoefficient[((1 + x - Sqrt[1 - 2*x - 3*x^2]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 29 2020 *)

Formula

a(n) = [x^n] ( (1 + x - sqrt(1 - 2*x - 3*x^2)) / (2*x) )^n.
a(n) ~ sqrt(((9386 + 1026*sqrt(57))^(1/3) + (9386 - 1026*sqrt(57))^(1/3) - 19)/228) * (((1261 + 57*sqrt(57))^(1/3) + (1261 - 57*sqrt(57))^(1/3) + 10)/6)^n / sqrt(Pi*n). - Vaclav Kotesovec, Mar 29 2020
Showing 1-3 of 3 results.

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