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A061691
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Triangle of generalized Stirling numbers.
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8
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1, 1, 2, 1, 9, 6, 1, 34, 72, 24, 1, 125, 650, 600, 120, 1, 461, 5400, 10500, 5400, 720, 1, 1715, 43757, 161700, 161700, 52920, 5040, 1, 6434, 353192, 2361016, 4116000, 2493120, 564480, 40320, 1, 24309, 2862330, 33731208, 96960024, 97161120, 39372480, 6531840, 362880
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OFFSET
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1,3
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COMMENTS
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The Eulerian-type number triangle associated with this triangle of generalized Stirling numbers is A192721. The table entry T(n,k) gives the number of uniform block permutations of the set {1,2,...,n} partitioned into k blocks. An example is given below. T(n,k) also gives the number of games of simple patience with n cards resulting in k piles (adapt Algorithm 1.1.22 of Lankham). [Peter Bala, Jul 14 2011]
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LINKS
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FORMULA
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T(n, k) = 1/k!*Sum multinomial(n, n_1, n_2, ..n_k)^2, where the sum extends over all compositions (n_1, n_2, .., n_k) of n into exactly k nonnegative parts. - Vladeta Jovovic, Apr 23 2003
The table entry T(n,k) may also be expressed as a sum over (unordered) partitions of n into k parts:
T(n,k) = sum {partitions m_1*1+...+m_n*n = n, m_1+...+m_n = k} 1/(m_1!*...*m_n!)*{n!/(1!^(m_1)*...*n!^(m_n))}^2.
Generating function:
Let J(z) = sum {n>=0} z^n/n!^2. Then
exp(x*(J(z)-1)) = 1 + x*z + (x + 2*x^2)*z^2/2!^2 + (x + 9*x^2 + 6*x^3)*z^3/3!^2 + ....
Relations with other sequences:
Row sums [1,3,16,131,...] = A023998. (End)
The row polynomials R(n,x) satisfy the recurrence equation R(n,x) = x*( sum {k = 0..n-1} binomial(n,k)*binomial(n-1,k)*R(k,x) ) with R(0,x) = 1. Also R(n,x + y) = sum {k = 0..n} binomial(n,k)^2*R(k,x)*R(n-k,y). - Peter Bala, Sep 17 2013
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EXAMPLE
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Triangle begins:
1;
1,2;
1,9,6;
1,34,72,24;
1,125,650,600,120;
...
T(4,2) = 34:
There are 7 partitions of the set {1,2,3,4} into 2 blocks. The four partitions {1,2,3}{4), {1,2,4}{3}, {1,3,4}{2} and {2,3,4}{1} give rise to 4x4 = 16 uniform block permutations while the remaining 3 partitions {1,2}{3,4}, {1,3}{2,4} and {1,4}{2,3} give 2!*3*3 = 18 uniform block permutations : thus in total there are 16+18 = 34 block permutations between the set partitions of {1,2,3,4} into 2 blocks.
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MAPLE
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#J = sum {n>=0} z^n/n!^2
J := BesselJ(0, 2*i*sqrt(z)):
G := exp(x*(J(z)-1)):
Gser := simplify(series(G, z = 0, 12)):
for n from 1 to 10 do
P[n] := n!^2*sort(coeff(Gser, z, n)) od:
for n from 1 to 10 do seq(coeff(P[n], x, k), k = 1..n) od;
# yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; expand(`if`(n=0, 1,
add(x*b(n-i)*binomial(n, i)/i!, i=1..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i)/i!, i=1..n))(b(n)*n!):
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MATHEMATICA
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max = 9; g := Exp[x*(BesselI[0, 2*Sqrt[z]] - 1)]; gser = Series[g, {z, 0, max}, {x, 0, max}]; t[n_, k_] := n!^2*SeriesCoefficient[ gser // Normal, {z, 0, n}, {x, 0, k}]; Flatten[ Table[ t[n, k], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Apr 04 2012, after Maple *)
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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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