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%I #23 Jun 04 2018 08:19:49
%S 1,1,1,2,1,6,3,6,1,12,24,28,24,12,24,1,20,90,140,245,120,240,140,120,
%T 60,120,1,30,240,660,1320,1626,1920,2100,1560,1830,1440,1440,840,720,
%U 360,720,1,42,525,2450,6195,12432,15127,23310,21000,26250,19320,26502,19320,19740,13440,17850,10080,10080,5880,5040,2520,5040,1,56,1008,7336,26320,68208,121072,190968,245616,304080,335440,332136,371616,346136,324240,310800,330960,235200,259056,194880,178080,147840,142800,80640,80640,47040,40320,20160,40320
%N Triangle of coefficients of q in e.g.f. that satisfies: A(x,q) = exp( q*x*A(q*x,q) ), read by rows of [n*(n-1)/2 + 1] terms in row n for n>=0.
%C Row sums equals A000272(n+1) = (n+1)^(n-1). Last term in rows are the factorials. Coefficients of q in {[x^n] A(x,q)} when read backward converge to the sequence: [1,1/2,1,7/6,2,2,85/24,11/3,65/12,19/3,357/40,19/2,111/8,123/8,81/4,...].
%H Alois P. Heinz, <a href="/A126265/b126265.txt">Rows n = 0..50, flattened</a> (first 19 rows from Paul D. Hanna)
%F G.f.: A(x,q) = Sum_{n>=0} (x^n/n!) * q^n * Sum_{k=0..n*(n-1)/2} T(n,k)*q^k.
%e G.f.: A(x,q) = 1 + q*x + (1 + 2*q)*q^2*x^2/2! +
%e (1 + 6*q + 3*q^2 + 6*q^3)*q^3*x^3/3! +
%e (1 + 12*q + 24*q^2 + 28*q^3 + 24*q^4 + 12*q^5 + 24*q^6)*q^4*x^4/4! +
%e (1 + 20*q + 90*q^2 + 140*q^3 + 245*q^4 + 120*q^5 + 240*q^6 + 140*q^7 + 120*q^8 + 60*q^9 + 120*q^10)*q^5*x^5/5! +
%e (1 + 30*q + 240*q^2 + 660*q^3 + 1320*q^4 + 1626*q^5 + 1920*q^6 + 2100*q^7 + 1560*q^8 + 1830*q^9 + 1440*q^10 + 1440*q^11 + 840*q^12 + 720*q^13 + 360*q^14 + 720*q^15)*q^6*x^6/6! +
%e (1 + 42*q + 525*q^2 + 2450*q^3 + 6195*q^4 + 12432*q^5 + 15127*q^6 + 23310*q^7 + 21000*q^8 + 26250*q^9 + 19320*q^10 + 26502*q^11 + 19320*q^12 + 19740*q^13 + 13440*q^14 + 17850*q^15 + 10080*q^16 + 10080*q^17 + 5880*q^18 + 5040*q^19 + 2520*q^20 + 5040*q^21)*q^7*x^7/7! +...
%e This irregular triangle begins:
%e 1;
%e 1;
%e 1, 2;
%e 1, 6, 3, 6;
%e 1, 12, 24, 28, 24, 12, 24;
%e 1, 20, 90, 140, 245, 120, 240, 140, 120, 60, 120;
%e 1, 30, 240, 660, 1320, 1626, 1920, 2100, 1560, 1830, 1440, 1440, 840, 720, 360, 720;
%e 1, 42, 525, 2450, 6195, 12432, 15127, 23310, 21000, 26250, 19320, 26502, 19320, 19740, 13440, 17850, 10080, 10080, 5880, 5040, 2520, 5040;
%e 1, 56, 1008, 7336, 26320, 68208, 121072, 190968, 245616, 304080, 335440, 332136, 371616, 346136, 324240, 310800, 330960, 235200, 259056, 194880, 178080, 147840, 142800, 80640, 80640, 47040, 40320, 20160, 40320;
%e 1, 72, 1764, 18648, 96894, 319536, 784980, 1420848, 2389977, 3140424, 4338432, 4618656, 5847408, 5619096, 6366528, 5687712, 6565104, 5579280, 6227424, 4985064, 5219424, 4445784, 4566240, 3522960, 3704400, 2540160, 2694384, 1935360, 1965600, 1330560, 1285200, 725760, 725760, 423360, 362880, 181440, 362880; ...
%p b:= proc(n) option remember; expand(`if`(n=0, 1,add(j*
%p binomial(n-1, j-1)*x^(j-1)*b(j-1)*b(n-j), j=1..n)))
%p end:
%p T:= n-> (p-> seq(coeff(p,x,i), i=0..degree(p)))(b(n)):
%p seq(T(n), n=0..10); # _Alois P. Heinz_, Aug 17 2017
%t b[n_] := b[n] = Expand[If[n == 0, 1, Sum[j*Binomial[n - 1, j - 1]*x^(j - 1)*b[j - 1]*b[n - j], {j, 1, n}]]];
%t T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n]];
%t Table[T[n], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Jun 04 2018, after _Alois P. Heinz_ *)
%o (PARI) T(n,k)=local(A=x);for(i=1,n,A=x*exp(subst(A,x,q*x+x*O(x^n)))); if(k>n*(n-1)/2 || k<0,0,Vec(Vec(A)[n+1]*n!+q*O(q^(n*(n+1)/2)))[k+1])
%o for(n=0,9,for(k=0,n*(n-1)/2,print1(T(n,k),", "));print(""))
%Y Cf. A000272 (row sums); diagonals: A126266, A126267.
%K nonn,tabf
%O 0,4
%A _Paul D. Hanna_, Dec 22 2006
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