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%I #11 Oct 18 2017 17:58:09
%S 1,1,1,7,103,2461,85236,4017959,247296043,19273646845,1856844859198,
%T 216816668773915,30183137125497500,4940046471894678145,
%U 939290181452023884094,205341788464993702617408
%N a(n) = [x^n] A(x)^n/(n+1) for n>1 where g.f. A(x) = Sum_{n>=0} a(n)*x^n with A(0)=A'(0)=1.
%H Vaclav Kotesovec, <a href="/A177752/b177752.txt">Table of n, a(n) for n = 0..250</a>
%F Given g.f. A(x), if D(x) satisfies:
%F . D(x) = A(x*D(x)), so that
%F . [x^n] D(x) = [x^n] A(x)^(n+1)/(n+1),
%F then in this case:
%F . D(x) = exp( Sum_{n>=1} (n+1)*a(n)*x^n/n - x).
%F a(n) ~ c * (n!)^2 / sqrt(n), where c = 0.500612869985729164508780668394780439... - _Vaclav Kotesovec_, Oct 18 2017
%e G.f.: A(x) = 1 + x + x^2 + 7*x^3 + 103*x^4 + 2461*x^5 + 85236*x^6 +...
%e Coefficients in the initial powers of A(x) begin:
%e [1, 1, 1, 7, 103, 2461, 85236, 4017959, 247296043,...];
%e [1, 2,(3), 16, 221, 5142, 175649, 8212754, 502843539,...];
%e [1, 3, 6,(28), 357, 8067, 271591, 12592554, 766917744,...];
%e [1, 4, 10, 44,(515), 11264, 373448, 17166028, 1039805373,...];
%e [1, 5, 15, 65, 700,(14766), 481645, 21942390, 1321805265,...];
%e [1, 6, 21, 92, 918, 18612,(596652), 26931450, 1613229105,...];
%e [1, 7, 28, 126, 1176, 22848, 718991,(32143672), 1914402210,...];
%e [1, 8, 36, 168, 1482, 27528, 849244, 37590240,(2225664387),...]; ...
%e where the above coefficients in parenthesis illustrate the property
%e that the coefficient of x^n in A(x)^n = (n+1)*a(n) for n>1:
%e 3 = 3*1; 28 = 4*7; 515 = 5*103; 14766 = 6*2461; 596652 = 7*85236; 32143672 = 8*4017959; ...
%e Therefore the logarithmic derivative of D(x) = A(x*D(x)) equals:
%e D'(x)/D(x) = 1 + 3*x + 28*x^2 + 515*x^3 + 14766*x^4 + 596652*x^5 +...
%e where [x^n] D(x) = [x^n] A(x)^(n+1)/(n+1) yields
%e D(x) = 1 + x + 2*x^2 + 11*x^3 + 140*x^4 + 3102*x^5 + 102713*x^6 +...
%t a[n_] := a[n] = SeriesCoefficient[Sum[a[k]*x^k, {k, 0, n-1}]^n, {x, 0, n}]; a[0] = 1; a[1] = 1; Table[a[n], {n, 0, 20}] (* _Vaclav Kotesovec_, Oct 18 2017 *)
%o (PARI) {a(n)=local(F=1+sum(m=1,n-1,a(m)*x^m)+x*O(x^n)); if(n<2,1,polcoeff(F^n,n))}
%Y Cf. A177753, A293864.
%K nonn
%O 0,4
%A _Paul D. Hanna_, May 13 2010
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