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%I #19 Oct 16 2019 11:31:46
%S 0,1,2,12,84,820,9540,132888,2129232,38760048,788500800,17740459440,
%T 437238410400,11716457100192,339129808346784,10544636706428160,
%U 350515939418507520,12404398847785793280,465618362609300313600
%N E.g.f. satisfies: A'(x) = [A(x)^2 + A(x)^3]/(x^2 + x^3).
%C a(n) = n * A179496(n-1). - _Vaclav Kotesovec_, Dec 25 2013
%H Vincenzo Librandi, <a href="/A179495/b179495.txt">Table of n, a(n) for n = 0..200</a>
%H Roland Bacher, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i3p7">Counting Packings of Generic Subsets in Finite Groups</a>, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
%F E.g.f. satisfies: d/dx A_n(x) = [A_n(x)^2 + A_n(x)^3]/(x^2 + x^3) where A_n(x) denotes the n-th iteration of e.g.f. A(x).
%F ...
%F Define a triangular matrix where the e.g.f. of column k equals (A(x)/x)^k, then the matrix log is the matrix L with L(n+1,n)=L(n+2,n)=n+1 and zeros elsewhere.
%F a(n) ~ sqrt(1+r) * n^n * r^(n-1) / exp(n), where r = -1-LambertW(-1, -exp(-2)) = 2.146193220620582585237... is the root of the equation log(1+r)=r-1. - _Vaclav Kotesovec_, Jan 04 2014
%e E.g.f. A(x) = x + 2*x^2/2! + 12*x^3/3! + 84*x^4/4! + 820*x^5/5! +...
%e Related expansions:
%e A(x)^2 + A(x)^3 = 2*x^2/2! + 18*x^3/3! + 192*x^4/4! + 2400*x^5/5! +...
%e A'(x) = 1 + 2*x + 12*x^2/2! + 84*x^3/3! + 820*x^4/4! + 9540*x^5/5! +...
%e A(x)/x = 1 + x + 4*x^2/2! + 21*x^3/3! + 164*x^4/4! + 1590*x^5/5! +...
%e ...
%e Define a triangular matrix where the e.g.f. of column k equals A(x)^k:
%e 1;
%e 1, 1;
%e 4/2!, 2, 1;
%e 21/3!, 10/2!, 3, 1;
%e 164/4!, 66/3!, 18/2!, 4, 1;
%e 1590/5!, 592/4!, 141/3!, 28/2!, 5, 1;
%e 18984/6!, 6500/5!, 1428/4!, 252/3!, 40/2!, 6, 1;
%e 266154/7!, 85548/6!, 17430/5!, 2840/4!, 405/3!, 54/2!, 7, 1;
%e ...
%e then the matrix log of the above matrix equals:
%e 0;
%e 1, 0;
%e 1, 2, 0;
%e 0, 2, 3, 0;
%e 0, 0, 3, 4, 0;
%e 0, 0, 0, 4, 5, 0;
%e 0, 0, 0, 0, 5, 6, 0; ...
%t nmax = 20; aa = ConstantArray[0, nmax]; aa[[1]] = 1; aa[[2]] = 2; Do[AGF = Sum[aa[[n]]*x^n/n!, {n, 1, j - 1}] + koef*x^j/j!; sol = Solve[Coefficient[D[AGF, x]*(x^2 + x^3) - (AGF^2 + AGF^3), x, j + 1] == 0, koef][[1]]; aa[[j]] = koef /. sol[[1]], {j, 3, nmax}]; Flatten[{0, aa}] (* _Vaclav Kotesovec_, Dec 25 2013 *)
%o (PARI) {a(n)=local(A=x+x^2+O(x^(n+1)),D=1);n!*polcoeff(1+sum(m=1, n+1, (D=A*deriv(x*D+O(x^(n+1))))/m!),n-1)}
%Y Cf. A179496.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 25 2010
%E Minor edits by _Vaclav Kotesovec_, Mar 31 2014
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