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.

A243660 Triangle read by rows: the x = 1+q Narayana triangle at m=2.

Original entry on oeis.org

1, 3, 2, 12, 16, 5, 55, 110, 70, 14, 273, 728, 702, 288, 42, 1428, 4760, 6160, 3850, 1155, 132, 7752, 31008, 50388, 42432, 19448, 4576, 429, 43263, 201894, 395010, 418950, 259350, 93366, 18018, 1430, 246675, 1315600, 3010700, 3853696, 3010700, 1466080, 433160, 70720, 4862
Offset: 1

Views

Author

N. J. A. Sloane, Jun 13 2014

Keywords

Comments

See Novelli-Thibon (2014) for precise definition.
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)*...*(x+2n+1) / (n! * (2n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022
The Maple code T(n,k) := binomial(3*n+1-k,n-k)*binomial(2*n,k-1)/n: with(sumtools): sumrecursion( (-1)^(k+1)*T(n,k)*binomial(x+3*n-k+1, 3*n-k+1), k, s(n) ); returns the recurrence 2*(2*n+1)*n^2*s(n) = (x+n)*(x+2*n)*(x+2*n+1)*s(n-1). The above observation follows from this. - Peter Bala, Oct 30 2022

Examples

			Triangle begins:
     1;
     3,    2;
    12,   16,    5;
    55,  110,   70,   14;
   273,  728,  702,  288,   42;
  1428, 4760, 6160, 3850, 1155,  132;
  ...
		

Crossrefs

Row sums give A034015(n-1).
The case m=1 is A126216 or A033282 (its mirror image).
The case m=3 is A243661.
The right diagonal is A000108.
The left column is A001764.
Same table as A383453, transposed. - Dan Eilers, May 06 2025

Programs

  • Mathematica
    polrecip[P_, x_] := P /. x -> 1/x // Together // Numerator;
    P[n_, m_] := Sum[Binomial[m n + 1, k] Binomial[(m+1) n - k, n - k] (1-x)^k x^(n-k), {k, 0, n}]/(m n + 1);
    T[m_] := Reap[For[i=1, i <= 20, i++, z = polrecip[P[i, m], x] /. x -> 1+q; Sow[CoefficientList[z, q]]]][[2, 1]];
    T[2] // Flatten (* Jean-François Alcover, Oct 08 2018, from PARI *)
  • PARI
    N(n,m)=sum(k=0,n,binomial(m*n+1,k)*binomial((m+1)*n-k,n-k)*(1-x)^k*x^(n-k))/(m*n+1);
    T(m)=for(i=1,20,z=subst(polrecip(N(i,m)),x,1+q);print(Vecrev(z)));
    T(2)  /* Lars Blomberg, Jul 17 2017 */
    
  • PARI
    T(n,k) = binomial(3*n+1-k,n-k) * binomial(2*n,k-1) / n; \\ Andrew Howroyd, Nov 23 2018

Formula

From Werner Schulte, Nov 23 2018: (Start)
T(n,k) = binomial(3*n+1-k,n-k) * binomial(2*n,k-1) / n.
More generally: T(n,k) = binomial((m+1)*n+1-k,n-k) * binomial(m*n,k-1) / n, where m = 2.
Sum_{k=1..n} (-1)^k * T(n,k) = -1. (End)

Extensions

Corrected example and a(22)-a(43) from Lars Blomberg, Jul 12 2017
a(44)-a(45) from Werner Schulte, Nov 23 2018