A345013 Triangle read by rows, related to clusters of type D.

1, 4, 3, 15, 20, 6, 56, 105, 60, 10, 210, 504, 420, 140, 15, 792, 2310, 2520, 1260, 280, 21, 3003, 10296, 13860, 9240, 3150, 504, 28, 11440, 45045, 72072, 60060, 27720, 6930, 840, 36

Offset: 1

Author

F. Chapoton, Sep 30 2021

Keywords

• tabl
• nonn

Comments

Let C_{n+1} be the cyclic quiver with n+1 vertices. Empirically, the n-th row is related to the green-mutation partial order on clusters for this quiver, restricted to clusters that do not meet the initial seed.

Apparently, value of the associated polynomials at -2 is A089849, up to sign.

By evaluating the associated polynomials at x-1, one apparently gets A062196.

The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)*(x+n+2) / (n! * (n+2)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 31 2022

Chapoton's observation above is correct: the precise expansion is (x+1)^2*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)*(x+n+2) / (n! * (n+2)!) = Sum_{k = 0..n} (-1)^k*T(n+1,k)*binomial(x+2*n+2-k, 2*n+2-k), as can be verified using the WZ algorithm. For example, n = 2 gives (x+1)^2*(x+2)^2*(x+3)*(x+4)/(2!*4!) = 15*binomial(x+6,6) - 20*binomial(x+5,5) + 6*binomial(x+4,4). - Peter Bala, Jun 24 2023

Examples

			Triangle begins:
[1] 1
[2] 4,    3
[3] 15,   20,    6
[4] 56,   105,   60,    10
[5] 210,  504,   420,   140,  15
[6] 792,  2310,  2520,  1260, 280,  21
[7] 3003, 10296, 13860, 9240, 3150, 504, 28
...
		

Crossrefs

Cf. A001791 (T(n,1)), A000217 (T(n,n)), A026002 (row sums), A000012 (alternating row sum), A051924 (number of clusters of type D_n).

Cf. A089849, A062196, A063007, A253283.

Programs

PARI

row(n) = vector(n, k, k--; (n-k)*binomial(n,k)*binomial(2*n-k, n-1)/n); \\ Michel Marcus, Sep 30 2021

Sage

def T_row(n):
    return [(n-k)*binomial(n,k)*binomial(2*n-k,n-1)//n for k in range(n)]
for n in range(1, 8): print(T_row(n))

Formula

T(n, k) = (n-k)*binomial(n,k)*binomial(2*n-k, n-1)/n, for n >= 1 and 0 <= k < n.

From Peter Bala, Jun 24 2023: (Start)

As conjectured above by Chapoton we have

Sum_{k = 0..n-1} T(n,k)*(x - 1)^k = Sum_{k = 0..n-1} A062196(n-1,k)*x^k and

Sum_{k = 0..n-1} T(n,k)*(-2)^k = (-1)^floor(n/2)*A089849(n) for n >= 1 (both easily verified using the WZ algorithm). (End)