A075856 Triangle formed from coefficients of the polynomials p(1)=x, p(n+1) = (n + x*(n+1))*p(n) + x*x*(d/dx)p(n).
1, 1, 3, 2, 10, 15, 6, 40, 105, 105, 24, 196, 700, 1260, 945, 120, 1148, 5068, 12600, 17325, 10395, 720, 7848, 40740, 126280, 242550, 270270, 135135, 5040, 61416, 363660, 1332100, 3213210, 5045040, 4729725, 2027025
Offset: 1
Examples
Triangle begins
1;
1, 3;
2, 10, 15;
6, 40, 105, 105;
24, 196, 700, 1260, 945;
120, 1148, 5068, 12600, 17325, 10395;
...
p(1) = x, p(2) = 3*x^2 + x, p(3) = 15*x^3 + 10*x^2 + 2*x, etc. - _Michael Somos_, Mar 17 2011
Links
- P. Bala, Diagonals of triangles with generating function exp(t*F(x)).
- J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013.
- Brian Drake, Ira M. Gessel, and Guoce Xin, Three Proofs and a Generalization of the Goulden-Litsyn-Shevelev Conjecture on a Sequence Arising in Algebraic Geometry, J. of Integer Sequences, Vol. 10 (2007), #07.3.7.
- Dominique Dumont and Armand Ramamonjisoa, Grammaire de Ramanujan et Arbres de Cayley, Electr. J. Combinatorics, Volume 3, Issue 2 (1996) R17 (see page 16).
- H. W. Gould, A Set of Polynomials Associated with the Higher Derivatives of y = x^x, Rocky Mountain J. Math. Volume 26, Number 2 (1996), 615-625.
- M. Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
- S. Ramanujan, Notebook entry
- P. W. Shor, Problem 78-6: A combinatorial identity, in Problems and Solutions column, SIAM Review; problem in 20, p. 394 (1978); solution in 21, pp. 258-260 (1979). [N. Sato, Feb 19 2010]
- P. W. Shor, A = B (but not quite); 3-d array with multiple recurrences, MathOverflow, Nov 2010-Nov 2011.
- J. Zeng, A Ramanujan sequence that refines the Cayley formula for trees, manuscript, 1996.
- J. Zeng, A Ramanujan sequence that refines the Cayley formula for trees, Ramanujan J., 3(1999) 1, 45-54.
Crossrefs
Programs
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Mathematica
p[1] = x; p[n_] := p[n] = (n - 1 + x*n)*p[n - 1] + x*x*D[p[n - 1], x]; Flatten[Rest[CoefficientList[#1, x]] & /@ Table[p[n], {n, 8}]] (* Jean-François Alcover, May 31 2011 *) -
PARI
{T(n, k) = if( k<1 || n
Formula
T(n, k) = (n-1) * T(n-1, k) + (n+k-1) * T(n-1, k-1). - Michael Somos, Mar 17 2011
G.f.: A(x, t) = Sum_{n>0} p[n] t^n / n! satisfies (dA / dt) * (x + t - 1) = x * (1 + A)^2 * (x * (1 + A) - 1). - Michael Somos, Mar 17 2011
T(n, 1) = (n-1)! = A000142(n-1). T(n, n) = A001147(n). Sum_{k>0} T(n, k) = n^n = A000312(n). Sum_{k>0} T(n, k) x^k = p[n].
From Peter Bala, Mar 14 2012: (Start)
This triangle is A185164 read by diagonals.
Let F(x) = x + (1-x)*log(1-x). The e.g.f. is given by the compositional inverse
(x - t*F(x))^(-1) = x + t*x^2/2! + (t + 3*t^2)x^3/3! + (2*t + 10*t^2 + 15*t^3)*x^4/4! + ....
Let f(x) = 1/log(1+x) and define inductively D^(n+1)(f(x)) = f(x)*(d/dx)(D^n(f(x))) with D^(0)f(x) = f(x). Then D^(n)f = (-1)^n*Sum_{k = 1..n} T(n,k)*f^(n-k)/((1+x)^n*f^(2n+1)).
(End)
Comments