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%I #64 Jul 18 2022 19:46:28
%S 1,1,4,22,148,1156,10192,99688,1069168,12468208,157071424,2126386912,
%T 30797423680,475378906432,7793485765888,135284756985472,
%U 2479535560687360,47860569736036096,970606394944476160,20635652201785613824,459015456156148876288,10662527360021306782720
%N Row 2 of table A111528.
%H Robert Israel, <a href="/A111529/b111529.txt">Table of n, a(n) for n = 0..410</a>
%H Paul Barry, <a href="https://arxiv.org/abs/1804.06801">A note on number triangles that are almost their own production matrix</a>, arXiv:1804.06801 [math.CO], 2018.
%H Richard J. Martin and Michael J. Kearney, <a href="http://dx.doi.org/10.1007/s00493-014-3183-3">Integral representation of certain combinatorial recurrences</a>, Combinatorica: 35:3 (2015), 309-315.
%H A. N. Stokes, <a href="https://doi.org/10.1017/S0004972700005219">Continued fraction solutions of the Riccati equation</a>, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
%F G.f.: (1/2)*log(Sum_{n >= 0} (n+1)!*x^n) = Sum_{n >= 1} a(n)*x^n/n.
%F G.f.: 1/(1+2*x - 3*x/(1+3*x - 4*x/(1+4*x - ... (continued fraction).
%F a(n) = Sum_{k = 0..n} 2^(n-k)*A089949(n,k). - _Philippe Deléham_, Oct 16 2006
%F G.f. 1/(2*x-G(0)) where G(k) = 2*x - 1 - k*x - x*(k+1)/G(k+1); G(0)=x (continued fraction, Euler's 1st kind, 1-step). - _Sergei N. Gladkovskii_, Aug 14 2012
%F G.f.: 1/(2*x) - 1/(G(0) - 1) where G(k) = 1 + x*(k+1)/(1 - 1/(1 + 1/G(k+1)));(continued fraction, 3-step). - _Sergei N. Gladkovskii_, Nov 20 2012
%F G.f.: 1 + x/(G(0)-2*x) where G(k) = 1 + (k+1)*x - x*(k+3)/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Dec 26 2012
%F G.f.: (1 + 1/Q(0))/2, where Q(k) = 1 + k*x - x*(k+2)/Q(k+1); (continued fraction). In general, the g.f. for row (r+2) is (r + 1 + 1/Q(0))/(r + 2). - _Sergei N. Gladkovskii_, May 04 2013
%F G.f.: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+3)/( x*(k+3) - 1/W(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Aug 26 2013
%F a(n) ~ n! * n^2/2 * (1 - 1/n - 2/n^2 - 8/n^3 - 52/n^4 - 436/n^5 - 4404/n^6 - 51572/n^7 - 683428/n^8 - 10080068/n^9 - 163471284/n^10), where the coefficients are given by (n+2)*(n+1)/n^2 * Sum_{k>=0} A260491(k)/(n+2)^k. - _Vaclav Kotesovec_, Jul 27 2015
%F a(n) = -A077607(n+2)/2. - _Vaclav Kotesovec_, Jul 29 2015
%F From _Peter Bala_, Jul 12 2022: (Start)
%F O.g.f: A(x) = ( Sum_{k >= 0} ((k+2)!/2!)*x^k )/( Sum_{k >= 0} (k+1)!*x^k ).
%F A(x)/(1 - 2*x*A(x)) = Sum_{k >= 0} ((k+2)!/2!)*x^k.
%F Riccati differential equation: x^2*A'(x) + 2*x*A^2(x) - (1 + x)*A(x) + 1 = 0.
%F Apply Stokes 1982 to find that A(x) = 1/(1 - x/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 3*x/(1 - 5*x/(1 - ... - n*x/(1 - (n+2)*x/(1 - ...))))))))), a continued fraction of Stieltjes type. (End)
%e (1/2)*log(1 + 2*x + 6*x^2 + ... + ((n+1)!/1!)*x^n + ...)
%e = x + (4/2)*x^2 + (22/3)*x^3 + (148/4)*x^4 + (1156/5)*x^5 + ...
%p N:= 30: # to get a(0) to a(N)
%p g:= 1/2*log(add((n+1)!*x^n,n=0..N+1)):
%p S:= series(g,x,N+1);
%p 1, seq(j*coeff(S,x,j),j=0..N); # _Robert Israel_, Jul 10 2015
%t T[n_, k_] := T[n, k] = Which[n<0 || k<0, 0, k==0 || k==1, 1, n==0, k!, True, (T[n-1, k+1]-T[n-1, k])/n - Sum[T[n, j] T[n-1, k-j], {j, 1, k-1}]];
%t a[n_] := T[2, n];
%t Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Aug 09 2018 *)
%o (PARI) {a(n)=if(n<0,0,if(n==0,1, (n/2)*polcoeff(log(sum(m=0,n,(m+1)!/1!*x^m)),n)))}
%Y Cf. A111528 (table), A003319 (row 1), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).
%Y Cf. A077607, A089949, A260491.
%K nonn,easy
%O 0,3
%A _Paul D. Hanna_, Aug 06 2005
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