A061691 Triangle of generalized Stirling numbers.
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
Offset: 1
Examples
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 4*4 = 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.
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- M. Aguiar and R. C. Orellana, The Hopf algebra of uniform block permutations, 17th International Conference on Formal Power Series and Algebraic Combinatorics, Taormina, July 2005.
- D. Aldous and P. Diaconis, Longest increasing subsequences: from patience sorting to the Baik-Deift-Johansson theorem, Bull. Amer. Math. Soc. 36 (1999), 413-432.
- D. G. Fitzgerald, A presentation for the monoid of uniform block permutations, Bull. Austral. Math. Soc. 68 (2003), 317-324.
- A. T. Irish, F. Quitin, U. Madhow, and M. Rodwell, Achieving multiple degrees of freedom in long-range mm-wave MIMO channels using randomly distributed relays, 2014.
- I. P. Lankham, Patience Sorting and Its Generalizations, arXiv:0705.4524 [math.CO], 2007.
- J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, Extended Bell and Stirling Numbers From Hypergeometric Exponentiation, J. Integer Seqs. Vol. 4 (2001), #01.1.4.
Crossrefs
Programs
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Maple
#A061691 #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!): seq(T(n), n=1..12); # Alois P. Heinz, Sep 10 2019
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Mathematica
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 *)
Formula
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
From Peter Bala, Jul 14 2011: (Start)
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:
T(n,k) = 1/k!*A192722(n,k).
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
Extensions
More terms from Vladeta Jovovic, Apr 23 2003
Comments