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%I #15 Jan 07 2024 07:16:56
%S 1,2,7,43,431,6466,135787,3802037,136873333,6159299986,338761499231,
%T 22358258949247,1743944198041267,158698922021755298,
%U 16663386812284306291,1999606417474116754921,271946472776479878669257,41607810334801421436396322
%N Recurrence a(n) = (1/2)*n*(n+1)*a(n-1) + 1 with a(0) = 1.
%C Cf. A006040 and A228229.
%H Seiichi Manyama, <a href="/A228230/b228230.txt">Table of n, a(n) for n = 0..269</a>
%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html">Modified Bessel Function of the First Kind</a>
%F a(n) = (1/2^n)*n!*(n + 1)!*Sum_{k = 0..n} 2^k/(k!*(k + 1)!).
%F a(n) = n!*(n+1)!*(the coefficient of x^n*y^(n+1) in the expansion of exp(x + y)/(1 - x*y/2).
%F Generating function: (1/(1 - x/2))*(1/sqrt(x))*BesselI(1, 2*sqrt(x)) = Sum_{n >= 0} a(n)*x^n/(n!*(n + 1)!).
%F Defining recurrence equation: a(n) = (1/2)*n*(n + 1)*a(n-1) + 1 with a(0) = 1.
%F Alternative recurrence equation: a(0) = 1, a(1) = 2, and for n >= 2, a(n) = ((1/2)*n*(n+1) + 1)*a(n-1) - (1/2)*n*(n - 1)*a(n-2).
%F The sequence b(n) := (1/2^n)*n!*(n + 1)! satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. It follows that we have the finite continued fraction expansion a(n) = (1/2^n)* n!*(n + 1)!*(1 + 1/(1 - 1/(4 - 3/(7 - ... - 1/2*n*(n - 1)/(1/2*n*(n + 1) + 1))))). Taking the limit yields the continued fraction expansion (1/sqrt(2))*BesselI(1,2*sqrt(2)) = Sum_{k >= 0} 2^k/(k!*(k + 1)!) = 1 + 1/(1 - 1/(4 - 3/(7 - 6/(11 - ... - (1/2)*n*(n - 1)/((1/2)*n*(n + 1) + 1 - ...))))) = 2.394833097....
%p #A228230
%p a:=proc(n) option remember
%p if n = 0 then 1 else 1/2n(n+1)thisproc(n-1) + 1
%p fi
%p end:
%p seq(a(n), n = 0..20);
%Y Cf. A006040, A228229.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Aug 19 2013
%E Typo in the first formula corrected by _Vaclav Kotesovec_, Jul 02 2015
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