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 A133399 Triangle T(n,k)=number of forests of labeled rooted trees with n nodes, containing exactly k trees of height one, all others having height zero (n>=0, 0<=k<=floor(n/2)). 5

%I

%S 1,1,1,2,1,9,1,28,12,1,75,120,1,186,750,120,1,441,3780,2100,1,1016,

%T 16856,21840,1680,1,2295,69552,176400,45360,1,5110,272250,1224720,

%U 705600,30240,1,11253,1026300,7692300,8316000,1164240,1,24564,3762132,45018600

%N Triangle T(n,k)=number of forests of labeled rooted trees with n nodes, containing exactly k trees of height one, all others having height zero (n>=0, 0<=k<=floor(n/2)).

%H Alois P. Heinz, <a href="/A133399/b133399.txt">Rows n = 0..200, flattened</a>

%H A. P. Heinz, <a href="https://doi.org/10.1007/3-540-53504-7_68">Finding Two-Tree-Factor Elements of Tableau-Defined Monoids in Time O(n^3)</a>, Ed. S. G. Akl, F. Fiala, W. W. Koczkodaj: Advances in Computing and Information, ICCI90 Niagara Falls, LNCS 468, Springer-Verlag (1990), pp. 120-128.

%F T(n,k) = C(n,k) * k! * stirling2(n-k+1,k+1).

%F E.g.f.: exp(y*x*(exp(x)-1))*exp(x). - _Geoffrey Critzer_, Feb 09 2013

%F Sum_{k=1..floor(n/2)} T(n,k) = A235596(n+1). - _Alois P. Heinz_, Jun 21 2019

%e Triangle begins:

%e 1;

%e 1;

%e 1, 2;

%e 1, 9;

%e 1, 28, 12;

%e 1, 75, 120;

%e 1, 186, 750, 120;

%e 1, 441, 3780, 2100;

%e 1, 1016, 16856, 21840, 1680;

%e 1, 2295, 69552, 176400, 45360;

%e 1, 5110, 272250, 1224720, 705600, 30240;

%e ...

%p T:= (n,k)-> binomial(n,k)*k!*Stirling2(n-k+1,k+1): for n from 0 to 10 do lprint(seq(T(n, k), k=0..floor(n/2))) od;

%t nn=12;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[ Series[Exp[y x (Exp[x]-1)] Exp[x],{x,0,nn}],{x,y}]]//Grid (* _Geoffrey Critzer_, Feb 09 2013 *)

%t t[n_, k_] := Binomial[n, k]*k!*StirlingS2[n-k+1, k+1]; Table[t[n, k], {n, 0, 12}, {k, 0, n/2}] // Flatten (* _Jean-François Alcover_, Dec 19 2013 *)

%o (MAGMA) /* As triangle */ [[Binomial(n,k)*Factorial(k)*StirlingSecond(n-k+1,k+1): k in [0..Floor(n/2)]]: n in [0.. 15]]; // _Vincenzo Librandi_, Jun 06 2019

%Y Columns k=1,2 give: A058877, A133386.

%Y Row sums give: A000248.

%Y T(2n,n) = A001813(n), T(2n+1,n) = A002691(n).

%Y Reading the table by diagonals gives triangle A198204. - _Peter Bala_, Jul 31 2012

%Y Cf. A235596.

%K nonn,tabf

%O 0,4

%A _Alois P. Heinz_, Nov 24 2007

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Last modified May 28 12:31 EDT 2020. Contains 334681 sequences. (Running on oeis4.)