A228230 Recurrence a(n) = (1/2)*n*(n+1)*a(n-1) + 1 with a(0) = 1.

1, 2, 7, 43, 431, 6466, 135787, 3802037, 136873333, 6159299986, 338761499231, 22358258949247, 1743944198041267, 158698922021755298, 16663386812284306291, 1999606417474116754921, 271946472776479878669257, 41607810334801421436396322, 7114935567251043065623771063

Offset: 0

Author

Peter Bala, Aug 19 2013

Keywords

• nonn
• easy

Comments

Cf. A006040 and A228229.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..269
• E. W. Weisstein, Modified Bessel Function of the First Kind

Crossrefs

Cf. A006040, A228229.

Programs

Maple

#A228230
a:=proc(n) option remember
if n = 0 then 1 else 1/2n(n+1)thisproc(n-1) + 1
fi
end:
seq(a(n), n = 0..20);

Formula

a(n) = (1/2^n)*n!*(n + 1)!*Sum_{k = 0..n} 2^k/(k!*(k + 1)!).

a(n) = n!*(n+1)!*(the coefficient of x^n*y^(n+1) in the expansion of exp(x + y)/(1 - x*y/2)).

G.f.: (1/(1 - x/2))*(1/sqrt(x))*BesselI(1, 2*sqrt(x)) = Sum_{n >= 0} a(n)*x^n/(n!*(n + 1)!).

Defining recurrence equation: a(n) = (1/2)*n*(n + 1)*a(n-1) + 1 with a(0) = 1.

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).

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....

Extensions

Typo in the first formula corrected by Vaclav Kotesovec, Jul 02 2015