OFFSET
1,2
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
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)
EXAMPLE
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
...
PROG
(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))
(PARI) row(n) = vector(n, k, k--; (n-k)*binomial(n, k)*binomial(2*n-k, n-1)/n); \\ Michel Marcus, Sep 30 2021
CROSSREFS
KEYWORD
tabl,nonn
AUTHOR
F. Chapoton, Sep 30 2021
STATUS
approved