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A008957
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Triangle of central factorial numbers T(2*n,2*n-2*k), k >= 0, n >= 1 (in Riordan's notation).
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6
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1, 1, 1, 1, 5, 1, 1, 14, 21, 1, 1, 30, 147, 85, 1, 1, 55, 627, 1408, 341, 1, 1, 91, 2002, 11440, 13013, 1365, 1, 1, 140, 5278, 61490, 196053, 118482, 5461, 1, 1, 204, 12138, 251498, 1733303, 3255330, 1071799, 21845, 1, 1, 285, 25194, 846260, 10787231
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OFFSET
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1,5
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COMMENTS
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D. E. Knuth [1992] on page 10 gives the formula Sum n^(2m-1) = Sum_{k=1..m} (2k-1)! T(2m,2k) binomial(n+k, 2k) where T(m, k) is the central factorial number of the second kind. - Michael Somos, May 08 2018
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REFERENCES
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J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217, Table 6.2(a).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.
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LINKS
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FORMULA
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T(n, k) = T(n-1, k-1) + k^2 * T(n-1, k), where T(n, n) = T(n, 1) = 1.
E.g.f.: x^2 * cosh(sinh(y*x/2) / (x/2)) - 1) = (1*x^2)*y^2/2! + (1*x^2 + 1*x^4)*y^4/4! +(1*x^2 + 5*x^4 + x^6)*y^6/6! + (1*x^2 + 14*x^4 + 21*x^6 + 1*x^8)*y^8/8! + ... (End)
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EXAMPLE
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The triangle starts:
1;
1, 1;
1, 5, 1;
1, 14, 21, 1;
1, 30, 147, 85, 1;
1, 55, 627, 1408, 341, 1;
1, 91, 2002, 11440, 13013, 1365, 1;
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MAPLE
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A036969 := proc(n, k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!), j=1..k); end; # Gives rows of triangle in reversed order
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MATHEMATICA
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t[n_, n_] = t[n_, 1] = 1;
t[n_, k_] := t[n-1, k-1] + k^2 t[n-1, k];
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PROG
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(Haskell)
a008957 n k = a008957_tabl !! (n-1) (k-1)
a008957_row n = a008957_tabl !! (n-1)
a008957_tabl = map reverse a036969_tabl
(PARI) {T(n, k) = if( n<1 || k>n, 0, n==k || k==1, 1, T(n-1, k-1) + k^2 * T(n-1, k))}; \\ Michael Somos, May 08 2018
(Sage)
m = n - k
return 2*sum((-1)^(j+m)*(j+1)^(2*n)/(factorial(j+m+2)*factorial(m-j)) for j in (0..m))
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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