A143395 Triangle read by rows: T(n,k) = number of forests of k labeled rooted trees of height at most 1, with n labels, where any root may contain >= 1 labels, n >= 0, 0 <= k <= n.
1, 0, 1, 0, 3, 1, 0, 7, 9, 1, 0, 15, 55, 18, 1, 0, 31, 285, 205, 30, 1, 0, 63, 1351, 1890, 545, 45, 1, 0, 127, 6069, 15421, 7770, 1190, 63, 1, 0, 255, 26335, 116298, 95781, 24150, 2282, 84, 1, 0, 511, 111645, 830845, 1071630, 416451, 62370, 3990, 108, 1
Offset: 0
Examples
T(3,2) = 9: {1}{2}<-3, {1}{3}<-2, {1}{2,3}, {2}{1}<-3, {2}{3}<-1, {2}{1,3}, {3}{1}<-2, {3}{2}<-1, {3}{1,2}.
Triangle begins:
1;
0, 1;
0, 3, 1;
0, 7, 9, 1;
0, 15, 55, 18, 1;
0, 31, 285, 205, 30, 1;
0, 63, 1351, 1890, 545, 45, 1;
0, 127, 6069, 15421, 7770, 1190, 63, 1;
...
From _Peter Bala_, Jan 07 2015: (Start)
T(4,2) = 55: There are 7 partitions of the set {1,2,3,4} into 2 blocks. For the 3 set partitions of the type {a,b}{c,d} we can choose a nonempty subset from each block in one of 3*3 ways giving 3*3*3 = 27 possibilities in all. The remaining 4 set partitions of {1,2,3,4} into 2 blocks are of the form {a,b,c}{d} and we can choose a nonempty subset from each block in 7*1 ways giving 4*7*1 = 28 possible choices. Thus in total T(4,2) = 27 + 28 = 55.
Recurrence equation example:
T(4,2) = sum {j = 1..3} (2^(4-j) - 1)*binomial(3,j)*T(j,1) = 7*3*1 + 3*3*3 + 1*1*7 = 55.
Connection constants:
Row 3 = [0, 7, 9, 1]. Hence x^3 = 7*x + 9*x*(x - 3) + x*(x - 4)*(x - 5); Row 4 = [0, 15, 55, 18, 1]. Hence x^4 = 15*x + 55*x*(x - 3) + 18*x*(x - 4)*(x - 5) + x*(x - 5)*(x - 6)*(x - 7).
With the array M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/ 1 \/1 \/1 \ / 1 \
| 3 1 ||0 1 ||0 1 | | 3 1 |
| 7 6 1 ||0 3 1 ||0 0 1 |... = | 7 9 1 |
|15 21 9 1 ||0 7 6 1 ||0 0 3 1 | |15 55 18 1 |
|... ||0 15 21 9 1||0 0 7 6 1| |... |
|... ||... ||... | | |
(End)
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- Eli Bagno, Riccardo Biagioli, and David Garber, Some identities involving second kind Stirling numbers of types B and D, arXiv:1901.07830 [math.CO], 2019.
- Peter Bala, A 3 parameter family of generalized Stirling numbers
- Takao Komatsu, Eli Bagno, and David Garber, A q,r-analogue of poly-Stirling numbers of second kind with combinatorial applications, arXiv:2209.06674 [math.CO], 2022.
- Index entries for sequences related to rooted trees
Crossrefs
Programs
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Magma
[[(&+[Binomial(n,j)*StirlingSecond(j,k)*k^(n-j): j in [k..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019
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Maple
T:= (n, k)-> add(binomial(n,t)*Stirling2(t,k)*k^(n-t), t=k..n): seq(seq(T(n, k), k=0..n), n=0..11);
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Mathematica
t[0, 0]=1; t[n_, k_]:= SeriesCoefficient[Exp[y*Exp[x]*(Exp[x]-1)], {x, 0, n}, {y, 0, k}]*n!; Table[t[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, Dec 05 2013, after Vladeta Jovovic *) Table[If[n==k==0, 1, If[k==0, 0, Sum[Binomial[n, j]*StirlingS2[j, k]* k^(n-j), {j,k,n}]]], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 07 2019 *) -
PARI
{T(n,k) = sum(j=k, n, binomial(n,j)*stirling(j,k,2)*k^(n-j))}; for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Mar 07 2019 -
Sage
# uses[bell_matrix from A264428] bell_matrix(lambda n: 2^(n+1)-1, 10) # Peter Luschny, Jan 18 2016
Formula
G.f. for column k: x^k/Product_{t=k..2*k} (1-t*x).
T(n,k) = Sum_{t=k..n} C(n,t) * Stirling2(t,k) * k^(n-t).
E.g.f.: exp(y*exp(x)*(exp(x)-1)). - Vladeta Jovovic, Dec 08 2008
T(n,k) = Sum_{m=0..k} Stirling2(n,k+m)*(k+m)!/(m!*(k-m)!). - Vladimir Kruchinin, Apr 06 2011
Let P be Pascal's triangle A007318. The first column of the array exp(t*(P^2-P)) gives the row generating polynomials of this triangle.
The row polynomials R(n,t) satisfy the recurrence R(n+1,t) = t*(Sum_{k = 0..n} (2^(k+1)-1)*C(n,k)*R(n-k,t)) with R(0,t) = 1. For example, the row 4 polynomial R(4,t) = 15*t + 55*t^2 + 18*t^3 + t^4 = t*((7*t + 9*t^2 + t^3) + 3*3*(3*t+t^2) + 7*3*t + 15*1). - Peter Bala, Oct 12 2011
From Peter Bala, Jan 07 2015: (Start)
T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(j + k)^n.
Recurrence equation: T(n+1,k+1) = Sum_{j = k..n} (2^(n-j+1) - 1)*binomial(n,j)*T(j,k) with T(0,0) = 1 and T(n,0) = 0 for n >= 1. This leads to the matrix factorization noted in the Comments section.
The inverse array is a signed version of A038455. (End)
Comments