A345013 - OEIS

%I #70 Jun 25 2023 09:39:26

%S 1,4,3,15,20,6,56,105,60,10,210,504,420,140,15,792,2310,2520,1260,280,

%T 21,3003,10296,13860,9240,3150,504,28,11440,45045,72072,60060,27720,

%U 6930,840,36

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

%C 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.

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

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

%C 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

%C 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

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

%F From _Peter Bala_, Jun 24 2023: (Start)

%F As conjectured above by Chapoton we have

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

%F 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)

%e Triangle begins:

%e [1] 1

%e [2] 4,    3

%e [3] 15,   20,    6

%e [4] 56,   105,   60,    10

%e [5] 210,  504,   420,   140,  15

%e [6] 792,  2310,  2520,  1260, 280,  21

%e [7] 3003, 10296, 13860, 9240, 3150, 504, 28

%e ...

%o (Sage)

%o def T_row(n):

%o     return [(n-k)*binomial(n,k)*binomial(2*n-k,n-1)//n for k in range(n)]

%o for n in range(1, 8): print(T_row(n))

%o (PARI) row(n) = vector(n, k, k--; (n-k)*binomial(n,k)*binomial(2*n-k, n-1)/n); \\ _Michel Marcus_, Sep 30 2021

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

%Y Cf. A089849, A062196, A063007, A253283.

%K tabl,nonn

%O 1,2

%A _F. Chapoton_, Sep 30 2021