A062196 -id:A062196 - cp's OEIS frontend

A107394 C(n+2,2)*C(n+4,2).

6, 30, 90, 210, 420, 756, 1260, 1980, 2970, 4290

Offset: 0

Zerinvary Lajos, May 25 2005

dead

Duplicates A033487.

			If n=0 then C(0+2,2)*C(0+4,2)= C(2,2)*C(4,2)=1*6=6
If n=4 then C(4+2,2)*C(4+4,2)= C(6,2)*C(8,2)=15*28=420

Cf. A062196.

a(n)=A033487(n+1). [From R. J. Mathar, Aug 24 2008]

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

Original entry on oeis.org

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

F. Chapoton, Sep 30 2021

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

			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
...

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.

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

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))

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)

cp's OEIS Frontend

A107394 C(n+2,2)*C(n+4,2).

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