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A136630
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Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.
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9
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1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0
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OFFSET
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0,13
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COMMENTS
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For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
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REFERENCES
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L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.
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LINKS
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Table of n, a(n) for n=0..83.
Ch. A. Charalambides, Central factorial numbers and related expansions
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FORMULA
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G.f. for column k: x^k/Product_{j=0..[k/2]} (1 - (2*j + k-2*[k/2])^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
Contribution from Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*sum {j = 0..k}(-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = sum {n = 0..inf} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) +x*R'(n,x) +x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
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EXAMPLE
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Triangle begins:
1;
0, 1;
0, 0, 1;
0, 1, 0, 1;
0, 0, 4, 0, 1;
0, 1, 0, 10, 0, 1;
0, 0, 16, 0, 20, 0, 1;
0, 1, 0, 91, 0, 35, 0, 1;
0, 0, 64, 0, 336, 0, 56, 0, 1;
0, 1, 0, 820, 0, 966, 0, 84, 0, 1;
0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1;
0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
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PROG
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(PARI) {T(n, k)=polcoeff(x^k/prod(j=0, k\2, 1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)), n)}
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CROSSREFS
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Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.
Sequence in context: A036859 A036861 A120324 * A111728 A143784 A147311
Adjacent sequences: A136627 A136628 A136629 * A136631 A136632 A136633
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna, Jan 14 2008
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STATUS
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approved
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