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A179495
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E.g.f. satisfies: A'(x) = [A(x)^2 + A(x)^3]/(x^2 + x^3).
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2
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0, 1, 2, 12, 84, 820, 9540, 132888, 2129232, 38760048, 788500800, 17740459440, 437238410400, 11716457100192, 339129808346784, 10544636706428160, 350515939418507520, 12404398847785793280, 465618362609300313600
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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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).
...
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.
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
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EXAMPLE
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E.g.f. A(x) = x + 2*x^2/2! + 12*x^3/3! + 84*x^4/4! + 820*x^5/5! +...
Related expansions:
A(x)^2 + A(x)^3 = 2*x^2/2! + 18*x^3/3! + 192*x^4/4! + 2400*x^5/5! +...
A'(x) = 1 + 2*x + 12*x^2/2! + 84*x^3/3! + 820*x^4/4! + 9540*x^5/5! +...
A(x)/x = 1 + x + 4*x^2/2! + 21*x^3/3! + 164*x^4/4! + 1590*x^5/5! +...
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Define a triangular matrix where the e.g.f. of column k equals A(x)^k:
1;
1, 1;
4/2!, 2, 1;
21/3!, 10/2!, 3, 1;
164/4!, 66/3!, 18/2!, 4, 1;
1590/5!, 592/4!, 141/3!, 28/2!, 5, 1;
18984/6!, 6500/5!, 1428/4!, 252/3!, 40/2!, 6, 1;
266154/7!, 85548/6!, 17430/5!, 2840/4!, 405/3!, 54/2!, 7, 1;
...
then the matrix log of the above matrix equals:
0;
1, 0;
1, 2, 0;
0, 2, 3, 0;
0, 0, 3, 4, 0;
0, 0, 0, 4, 5, 0;
0, 0, 0, 0, 5, 6, 0; ...
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MATHEMATICA
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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 *)
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PROG
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(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)}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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