A126470 Triangle, read by rows, where row n lists coefficients of q in F(n,q) that satisfies: F(n,q) = Sum_{k=0..n-1} C(n-1,k)*F(k,q)*F(n-k-1,q)*q^k for n>0, with F(0,q) = 1.
1, 1, 1, 1, 1, 3, 1, 1, 1, 6, 7, 5, 3, 1, 1, 1, 10, 25, 25, 26, 11, 12, 5, 3, 1, 1, 1, 15, 65, 110, 136, 117, 92, 70, 43, 32, 17, 12, 5, 3, 1, 1, 1, 21, 140, 385, 616, 784, 694, 687, 478, 411, 255, 222, 127, 91, 50, 39, 17, 12, 5, 3, 1, 1, 1, 28, 266, 1106, 2471, 4032, 4887, 5189
Offset: 0
Examples
Number of terms in row n is: n*(n-1)/2 + 1. Row functions B(n,q) begin: F(0,q) = F(1,q) = 1; F(2,q) = 1 + q; F(3,q) = 1 + 3*q + q^2 + q^3; F(4,q) = 1 + 6*q + 7*q^2 + 5*q^3 + 3*q^4 + q^5 + q^6. Triangle begins: 1; 1; 1, 1; 1, 3, 1, 1; 1, 6, 7, 5, 3, 1, 1; 1, 10, 25, 25, 26, 11, 12, 5, 3, 1, 1; 1, 15, 65, 110, 136, 117, 92, 70, 43, 32, 17, 12, 5, 3, 1, 1; 1, 21, 140, 385, 616, 784, 694, 687, 478, 411, 255, 222, 127, 91, 50, 39, 17, 12, 5, 3, 1, 1; 1, 28, 266, 1106, 2471, 4032, 4887, 5189, 4832, 4240, 3426, 2658, 2143, 1534, 1143, 790, 575, 351, 262, 151, 99, 58, 39, 17, 12, 5, 3, 1, 1; 1, 36, 462, 2730, 8589, 17892, 28519, 35613, 40639, 39200, 37934, 31508, 28076, 21570, 18288, 13451, 11009, 7747, 6120, 4089, 3106, 2056, 1530, 943, 683, 396, 289, 160, 108, 58, 39, 17, 12, 5, 3, 1, 1; 1, 45, 750, 6000, 25977, 70497, 141499, 220500, 291877, 336945, 357638, 347396, 323795, 288162, 247473, 207630, 170336, 139565, 109967, 87581, 66534, 51411, 37845, 28948, 20626, 15284, 10727, 7810, 5169, 3731, 2446, 1700, 1063, 733, 426, 299, 170, 108, 58, 39, 17, 12, 5, 3, 1, 1; ... E.g.f.: A(x,q) = 1 + x + x^2*(1+q)/2! + x^3*(1+3*q+q^2+q^3)/3! +... where A(x,q) = exp( Integral A(q*x,q) dx ), A(q*x,q) = exp( q * Integral A(q^2*x,q) dx ), A(q^2*x,q) = exp( q^2 * Integral A(q^3*x,q) dx ), ... A(q^n*x,q) = exp( q^n * Integral A(q^(n+1)*x,q) dx ) for n>=0. Here the Integral is always in the limits 0..x.
Links
- Paul D. Hanna, Rows n = 0..40, flattened.
- Miguel A. Mendez, Combinatorial differential operators in: Faà di Bruno formula, enumeration of ballot paths, enriched rooted trees and increasing rooted trees, arXiv:1610.03602 [math.CO], 2016.
Programs
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Mathematica
F[0, ] = 1; F[n, q_] := F[n, q] = Sum[Binomial[n-1, k] F[k, q] F[n-k-1, q] q^k, {k, 0, n-1}]; row[n_] := CoefficientList[F[n, q], q]; Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jul 26 2018 *)
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PARI
F(n,q)=if(n==0,1,sum(k=0,n-1,binomial(n-1,k)*F(k,q)*F(n-k-1,q)*q^k)) {T(n,k)=Vec(F(n,q)+O(q^(n*(n-1)/2+1)))[k+1]} for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print("")) -
PARI
T(n,k)=local(A=vector(n+2, j, 1+j*x)); for(i=0, n+1, for(j=0, n, m=n+1-j; A[m]=exp(q^(m-1)*intformal(A[m+1]+x*O(x^n))))); polcoeff(n!*polcoeff(A[1], n, x),k,q) \\ From Paul D. Hanna, Oct 04 2008 for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))
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PARI
/* Faster to use: A(x,q) = 1 + Integral A(x,q)*A(qx,q) dx */ {T(n,k)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A*subst(A,x,q*x))); polcoeff(n!*polcoeff(A,n,x),k,q)} for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Oct 04 2008
Formula
From Paul D. Hanna, Oct 04 2008: (Start)
E.g.f. satisfies: A(x,q) = exp( Integral A(q*x,q) dx ); further,
A(q^n*x,q) = exp( q^n * Integral A(q^(n+1)*x,q) dx ) for n>=0, where A(x,q) = Sum_{n>=0} x^n*[Sum_{k=0..n(n-1)/2} T(n,k)*q^k]/n!. (End)
E.g.f. satisfies: d/dx A(x,q) = A(x,q) * A(q*x,q) with A(0,q)=1; i.e., the logarithmic derivative of A(x,q) with respect to x equals A(q*x,q). - Paul D. Hanna, Oct 04 2008
Comments