A243660 - OEIS

%I #73 Nov 01 2022 16:37:32

%S 1,3,2,12,16,5,55,110,70,14,273,728,702,288,42,1428,4760,6160,3850,

%T 1155,132,7752,31008,50388,42432,19448,4576,429,43263,201894,395010,

%U 418950,259350,93366,18018,1430,246675,1315600,3010700,3853696,3010700,1466080,433160,70720,4862

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

%C See Novelli-Thibon (2014) for precise definition.

%C 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

%C 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

%H Andrew Howroyd, <a href="/A243660/b243660.txt">Table of n, a(n) for n = 1..1275</a> (first 50 rows)

%H J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014. See Fig. 8.

%F From _Werner Schulte_, Nov 23 2018: (Start)

%F T(n,k) = binomial(3*n+1-k,n-k) * binomial(2*n,k-1) / n.

%F More generally: T(n,k) = binomial((m+1)*n+1-k,n-k) * binomial(m*n,k-1) / n, where m = 2.

%F Sum_{k=1..n} (-1)^k * T(n,k) = -1. (End)

%e Triangle begins:

%e      1;

%e      3,    2;

%e     12,   16,    5;

%e     55,  110,   70,   14;

%e    273,  728,  702,  288,   42;

%e   1428, 4760, 6160, 3850, 1155,  132;

%e   ...

%t polrecip[P_, x_] := P /. x -> 1/x // Together // Numerator;

%t 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 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 T[2] // Flatten (* _Jean-François Alcover_, Oct 08 2018, from PARI *)

%o (PARI)

%o 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);

%o T(m)=for(i=1,20,z=subst(polrecip(N(i,m)),x,1+q);print(Vecrev(z)));

%o T(2)  /* _Lars Blomberg_, Jul 17 2017 */

%o (PARI) T(n,k) = binomial(3*n+1-k,n-k) * binomial(2*n,k-1) / n; \\ _Andrew Howroyd_, Nov 23 2018

%Y Row sums give A034015(n-1).

%Y The case m=1 is A126216 or A033282 (its mirror image).

%Y The case m=3 is A243661.

%Y The right diagonal is A000108.

%Y The left column is A001764.

%K nonn,tabl

%O 1,2

%A _N. J. A. Sloane_, Jun 13 2014

%E Corrected example and a(22)-a(43) from _Lars Blomberg_, Jul 12 2017

%E a(44)-a(45) from _Werner Schulte_, Nov 23 2018