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Recurrence a(n) = n*(n + 1)*a(n-1) + 1 with a(0) = 1.
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%I #18 Jan 08 2024 04:36:46

%S 1,3,19,229,4581,137431,5772103,323237769,23273119369,2094580743211,

%T 230403881753211,30413312391423853,4744476733062121069,

%U 863494765417306034559,181333900737634267257391,43520136177032224141773841,11837477040152764966562484753

%N Recurrence a(n) = n*(n + 1)*a(n-1) + 1 with a(0) = 1.

%C Main subdiagonal (and main superdiagonal) of A099597. Cf. A006040 and A228230.

%H Seiichi Manyama, <a href="/A228229/b228229.txt">Table of n, a(n) for n = 0..252</a>

%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html">Modified Bessel Function of the First Kind</a>

%F a(n) = n!*(n + 1)!*sum {k = 0..n} 1/(k!*(k + 1)!).

%F Generating function: 1/(1 - x)*1/sqrt(x)*BesselI(1, 2*sqrt(x)) = sum {n >= 0} a(n)*x^n/(n!*(n + 1)!).

%F Defining recurrence equation: a(n) = n*(n + 1)*a(n-1) + 1 with a(0) = 1.

%F Alternative recurrence equation: a(0) = 1, a(1) = 3, and for n >= 2, a(n) = (n*(n + 1) + 1)*a(n-1) - n*(n - 1)*a(n-2).

%F The sequence b(n) := n!*(n + 1)! satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 2. It follows that we have the finite continued fraction expansion a(n) = n!*(n + 1)!*(1/(1 - 1/(3 - 2/(7 - 6/(13 - … - n*(n - 1)/(n^2 + n + 1)))))). Taking the limit yields the continued fraction expansion for the modified Bessel function value BesselI(1,2) = sum {k = 0..inf} 1/(k!*(k + 1)!) = 1/(1 - 1/(3 - 2/(7 - 6/(13 - ...- n*(n - 1)/(n^2 + n + 1 - ...))))) = 1.590636... (see A096789).

%F a(n) ~ BesselI(1,2) * n!*(n+1)!. - _Vaclav Kotesovec_, May 06 2015

%p A228229 :=proc(n) option remember

%p if n = 0 then 1

%p else n*(n+1)*procname(n-1) + 1

%p end if:

%p end proc:

%p seq(A228229(n), n = 0..20);

%t RecurrenceTable[{a[n] == n*(n + 1)*a[n-1] + 1, a[0] == 1},a,{n,0,20}] (* _Vaclav Kotesovec_, May 06 2015 *)

%Y Cf. A006040, A096789, A099597, A228230, A368837, A368838.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Aug 19 2013