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%I #33 Jul 26 2018 03:35:29
%S 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,
%T 110,136,117,92,70,43,32,17,12,5,3,1,1,1,21,140,385,616,784,694,687,
%U 478,411,255,222,127,91,50,39,17,12,5,3,1,1,1,28,266,1106,2471,4032,4887,5189
%N 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.
%C Row sums equal the factorials: F(n,1) = n!.
%C Limit of reversed rows equals A126471. Largest term in rows equal A126472.
%H Paul D. Hanna, <a href="/A126470/b126470.txt">Rows n = 0..40, flattened.</a>
%H Miguel A. Mendez, <a href="https://arxiv.org/abs/1610.03602">Combinatorial differential operators in: Faà di Bruno formula, enumeration of ballot paths, enriched rooted trees and increasing rooted trees</a>, arXiv:1610.03602 [math.CO], 2016.
%F From _Paul D. Hanna_, Oct 04 2008: (Start)
%F E.g.f. satisfies: A(x,q) = exp( Integral A(q*x,q) dx ); further,
%F 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)
%F 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
%e Number of terms in row n is: n*(n-1)/2 + 1.
%e Row functions B(n,q) begin:
%e F(0,q) = F(1,q) = 1;
%e F(2,q) = 1 + q;
%e F(3,q) = 1 + 3*q + q^2 + q^3;
%e F(4,q) = 1 + 6*q + 7*q^2 + 5*q^3 + 3*q^4 + q^5 + q^6.
%e Triangle begins:
%e 1;
%e 1;
%e 1, 1;
%e 1, 3, 1, 1;
%e 1, 6, 7, 5, 3, 1, 1;
%e 1, 10, 25, 25, 26, 11, 12, 5, 3, 1, 1;
%e 1, 15, 65, 110, 136, 117, 92, 70, 43, 32, 17, 12, 5, 3, 1, 1;
%e 1, 21, 140, 385, 616, 784, 694, 687, 478, 411, 255, 222, 127, 91, 50, 39, 17, 12, 5, 3, 1, 1;
%e 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;
%e 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;
%e 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 ...
%e E.g.f.: A(x,q) = 1 + x + x^2*(1+q)/2! + x^3*(1+3*q+q^2+q^3)/3! +...
%e where A(x,q) = exp( Integral A(q*x,q) dx ),
%e A(q*x,q) = exp( q * Integral A(q^2*x,q) dx ),
%e A(q^2*x,q) = exp( q^2 * Integral A(q^3*x,q) dx ), ...
%e A(q^n*x,q) = exp( q^n * Integral A(q^(n+1)*x,q) dx ) for n>=0.
%e Here the Integral is always in the limits 0..x.
%t 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}];
%t row[n_] := CoefficientList[F[n, q], q];
%t Table[row[n], {n, 0, 8}] // Flatten (* _Jean-François Alcover_, Jul 26 2018 *)
%o (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))
%o {T(n,k)=Vec(F(n,q)+O(q^(n*(n-1)/2+1)))[k+1]}
%o for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))
%o (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
%o for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))
%o (PARI) /* Faster to use: A(x,q) = 1 + Integral A(x,q)*A(qx,q) dx */
%o {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)}
%o for(n=0,10,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print()) \\ _Paul D. Hanna_, Oct 04 2008
%Y Cf. A126471, A126472; Bell number variant: A126347.
%K nonn,tabl
%O 0,6
%A _Paul D. Hanna_, Dec 31 2006
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