OFFSET
0,3
COMMENTS
If row elements are divided by row sums, one obtains a probability distribution that approaches a Poisson distribution with expected value 1 as n approaches infinity.
FORMULA
T(n,k) equals 2^k times the corresponding element of the triangle of A168422.
T(n,k) = 2^k * Sum_{j=k..n} (-1)^(j-k) * C(2*n-j,n) * C(n,j) * C(j,k) * (n-j)!.
Recurrence: T(n,k) = (1/k!) * Sum_{j=0..k} T(n-j,0) * (-1)^j * C(k,j) * Sum_{t=0..min(j,k-j)} (-1)^(j-t) * C(j,t) * (k-j)! / (k-j-t)!
= (1/k!) * Sum_{j=0..k} T(n-j,0) * (-1)^j * C(k,j) * R(k,j) where R(k,j) is an element of the triangle of A253667.
T(n, k) = 2^k*binomial(n, k)*KummerU(k-n, k-2*n, -1). - Peter Luschny, Mar 18 2026
EXAMPLE
Triangle begins:
1
1 2
7 8 4
71 78 36 8
1001 1072 504 128 16
18089 19090 9080 2480 400 32
398959 417048 199980 56960 10320 1152 64
10391023 10789982 5204556 1523480 295120 38304 3136 128
MAPLE
T := (n, k) -> 2^k*binomial(n, k)*KummerU(k-n, k-2*n, -1):
seq(print(seq(simplify(T(n, k)), k = 0..n)), n = 0..9); # Peter Luschny, Mar 18 2026
PROG
(SageMath)
def T(n, k):
return(2^k*sum((-1)^(j-k)*binomial(2*n-j, n)*binomial(n, j)
*binomial(j, k)*factorial(n-j) for j in range(k, n+1)))
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
William P. Orrick, Mar 08 2023
STATUS
approved
