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A351584
Triangle read by rows: T(n,k) = A(k,n-k), 1 <= k < n, 2 <= n, where A(m,n) is the number of distinct strings consisting of one X, 2*m Y's and 2*n Z's in which the X lies to the right of at least m Y's and at least n Z's.
2
16, 53, 53, 124, 306, 124, 240, 1103, 1103, 240, 412, 3043, 5664, 3043, 412, 651, 7056, 21095, 21095, 7056, 651, 968, 14476, 63480, 101950, 63480, 14476, 968, 1374, 27114, 163986, 386249, 386249, 163986, 27114, 1374, 1880, 47331, 377616, 1226540, 1798776, 1226540, 377616, 47331, 1880
OFFSET
2,1
COMMENTS
The general string enumeration problem of counting strings with k+k'-1 X's, m+m' Y's and n+n' Z's in which the k'th X is placed after at least m of the Y's and n of the Z's may be expressed in terms of an integral of incomplete Beta functions and evaluated in terms of Kampe de Feriet functions (see Connor & Fewster, 2022). Other special cases include A351583 and A351585.
LINKS
Stephen B. Connor and Christopher J. Fewster, Integrals of incomplete beta functions, with applications to order statistics, random walks and string enumeration, Brazilian Journal of Probability and Statistics 2022, Vol. 36, No. 1, 185-198; arXiv version, arXiv:2104.12216 [math.CA], 201.
FORMULA
T(n,k) = (4*(n - k)*k + 3*n + 2)*binomial(2*n + 2, 2*k + 1)/(4*n + 4) - n*binomial(n, k)^2/2. See Connor & Fewster (2022).
EXAMPLE
Triangle starts:
16;
53, 53;
124, 306, 124;
240, 1103, 1103, 240;
412, 3043, 5664, 3043, 412;
...
MAPLE
T:=(n, k)->(4*(n - k)*k + 3*n + 2)*binomial(2*n + 2, 2*k + 1)/(4*n + 4) - n*binomial(n, k)^2/2; [seq(seq(T(n, k), k=1..n-1), n=2..10)];
MATHEMATICA
t[n_, k_]:=(4*k*(n-k)+3*n+2)/(4*n+4)*Binomial[2*n+2, 2*k+1]- (n/2)*Binomial[n, k]^2; Table[t[n, k], {n, 2, 10}, {k, 1, n-1}]
CROSSREFS
Sequence in context: A211586 A232509 A087973 * A332591 A117273 A137515
KEYWORD
nonn,tabl,easy
AUTHOR
STATUS
approved