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A351583
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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-1 Y's and 2*n-1 Z's in which the X lies to the right of at least m Y's and at least n Z's.
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2
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2, 7, 7, 15, 52, 15, 26, 192, 192, 26, 40, 510, 1086, 510, 40, 57, 1115, 4098, 4098, 1115, 57, 77, 2142, 12075, 20840, 12075, 2142, 77, 100, 3752, 30072, 79600, 79600, 30072, 3752, 100, 126, 6132, 66276, 249408, 382510, 249408, 66276, 6132, 126
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OFFSET
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2,1
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COMMENTS
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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 A351584 and A351585.
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LINKS
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FORMULA
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T(n+1,1) = A(1,n) = 1/2*n*(3*n+1) = A005449(n), the n-th second pentagonal number.
T(n,k) = 1/(2*Beta(2*k, 2*n - 2*k)) - binomial(n, k)/(2*Beta(k, n - k)), where Beta(x,y) = Gamma(x)*Gamma(y)/Gamma(x+y) is the Beta-function (see A003506). [Connor and Fewster]
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EXAMPLE
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Triangle starts:
2;
7, 7;
15, 52, 15;
26, 192, 192, 26;
40, 510, 1086, 510, 40;
...
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MAPLE
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T:=(n, k) -> 1/(2*Beta(2*k, 2*n - 2*k)) - binomial(n, k)/(2*Beta(k, n - k)); [seq(seq(T(n, k), k=1..n-1), n=2..10)];
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MATHEMATICA
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t[n_, k_]:=1/(2*Beta[2*k, 2*n-2*k])-Binomial[n, k]/(2*Beta[k, n-k]); Table[t[n, k], {n, 2, 10}, {k, 1, n-1}]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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