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A112493
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Coefficient triangle of polynomials used for e.g.f.s of Stirling2 diagonals.
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11
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1, 1, 1, 1, 4, 3, 1, 11, 25, 15, 1, 26, 130, 210, 105, 1, 57, 546, 1750, 2205, 945, 1, 120, 2037, 11368, 26775, 27720, 10395, 1, 247, 7071, 63805, 247555, 460845, 405405, 135135, 1, 502, 23436, 325930, 1939630, 5735730, 8828820, 6756750, 2027025, 1
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| For the o.g.f.s of diagonal k of the Stirling2 triangle one has a similar result. See A008517 (second-order Eulerian triangle).
A(m,x), the o.g.f. for column m satisfies the recurrence A(m,x)= x*(x*diff(A(m-1,x),x) + m*A(m-1,x))/(1-(m+1)*x), for m>=1 and A(0,x)=1/(1-x).
The column sequences start with A000012 (powers of 1), A000295 (Eulerian numbers), A112495-A112497.
The e.g.f. for the sequence in column k+1, k>=0, of A008278, i.e. for the diagonal k>=0 of the Stirling2 triangle A048993, is exp(x)*sum(a(k,m)*(x^(m+k))/(m+k)!,m=0..k).
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LINKS
| F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48. D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA]
W. Lang, First ten rows.
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FORMULA
| a(k, m) = 0 if k<m, a(k, -1):=0, a(0, 0)=1, a(k, m)=(m+1)*a(k-1, m)+(k+m-1)*a(k-1, m-1) else.
From Peter Bala, Sep 30 2011: (Start)
E.g.f.: A(x,t) = -1-(1+t)/t*LambertW(-t/(1+t)*exp((x-t)/(1+t)) = x + (1+t)*x^2/2! + (1+4*t+3*t^2)*x^3/3! + .... A(x,t) is the inverse function of (1+t)*log(1+x)-t*x.
A(x,t) satisfies the partial differential equation (1-x*t)*dA/dx = 1 + A + t*(1+t)*dA/dt. It follows that the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) =(n*t+1)*R(n,t) + t*(1+t)*dR(n,t)/dt. Cf. A054589 and A075856. The polynomials t/(1+t)*R(n,t) are the row polynomials of A134991.
The generating function A(x,t) satisfies the autonomous differential equation dA/dx = (1+A)/(1-t*A). Applying [Bergeron et al, Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees on n+1 vertices where the non-leaf vertices of outdegree k come in t^(k-1)*(1+t) colors. An example is given below. Cf. A006351, which corresponds to the case t = 1. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x) = (1+x)/(1-x*t). Then R(n,t) = (f(x)*d/dx)^n(f(x)) evaluated at x = 0. (End)
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EXAMPLE
| [1];[1,1];[1,4,3];[1,11,25,15];[1,26,130,210,105];...
The e.g.f. of [0,0,1,7,25,65,...], the k=3 column of A008278, but with offset n=0, is exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!).
Third row [1,4,3]: There are three plane increasing trees on 3 vertices. The number of colors are shown to the right of a vertex.
...................................................
....1o.(1+t)...........1o.t*(1+t).....1o.t*(1+t)...
....|................. /.\............/.\..........
....|................ /...\........../...\.........
....2o.(1+t)........2o.....3o......3o....2o........
....|..............................................
....|..............................................
....3o.............................................
...................................................
The total number of trees is (1+t)^2 + t*(1+t) + t*(1+t) = 1+4*t+3*t^2 = R(2,t).
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CROSSREFS
| Row sums give A006351(k+1), k>=0. A054589, A075856, A134991.
Sequence in context: A172106 A128813 A109062 * A010305 A098234 A193795
Adjacent sequences: A112490 A112491 A112492 * A112494 A112495 A112496
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KEYWORD
| nonn,easy,tabl
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AUTHOR
| Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Oct 14 2005
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