A222472 Numerator sequence of the n-th convergent of the continued fraction 1/(1+3/(2+3/(3+3/(4+...

1, 2, 9, 42, 237, 1548, 11547, 97020, 907821, 9369270, 105785433, 1297533006, 17185285377, 244486594296, 3718854770571, 60235136112024, 1035153878216121, 18813475216226250, 360561490742947113, 7267670240507621010

Offset: 1

Author

Gary Detlefs and Wolfdieter Lang, Mar 09 2013

Keywords

• nonn
• easy

Comments

The corresponding denominator sequence is A213190.

a(n) = Phat(n,3) with the numerator polynomials Phat of A221913. All the given formulas follow from there and from the comments given under A084950. The limit of the continued fraction (0 + K_{k=1..oo} (3/k))/3 = 1/(1+3/(2+3/(3+3/(4+... is (1/3)*sqrt(3)*BesselI(1,2*sqrt(3))/BesselI(0,2*sqrt(3)) = 0.484516174987404...

For a combinatorial interpretation in terms of labeled Morse codes see a comment on A221913. Here each dash has label x=3, and the dots have label j if they are at position j. Labels are multiplied and all codes on positions [2,...,n+1] are summed.

Examples

			a(4) = 4*a(3) + 2*a(2) = 4*9 + 3*2 = 42.
Continued fraction convergent: 1/(1+3/(2+3/(3+3/4))) = 14/29 = 42/87 = a(4)/A213190(4).
Morse code: a(5) = 237 from the sum of all 5 labeled codes on [2,3,4,5], one with no dash, three with one dash and one with two dashes: 5!/1 + (4*5 + 2*5 + 2*3)*(3) +3^2 = 237.
		

Crossrefs

Cf. A084950, A221913, A222467, A001040(n+1) (x=1), A058798 (x=-1).

Programs

PARI

a=vector(50); a[1]=1;a[2]=2; for(n=3, #a, a[n]=n*a[n-1]+3*a[n-2]); a \\ Altug Alkan, Apr 20 2018

Formula

Recurrence: a(n) = n*a(n-1) + 3*a(n-2), with a(-1) = 1/3, a(0) = 0, n >= 1.

As a sum: a(n) = Sum_{m=0..floor(n/2)} b(n-m,m)*3^m, n >= 1, with b(n,m) = binomial(n,m)*(n+1)!/(m+1)! = |A066667(n,m)| (Laguerre coefficients, parameter alpha =1).

Explicit form: a(n) = -2*(sqrt(3))^n*(BesselK(1, -2*sqrt(3))*BesselI(n+1, -2*sqrt(3)) + (-1)^(n+1)*BesselI(1,-2*sqrt(3))*BesselK(n+1,-2*sqrt(3))).

E.g.f.: Pi*(BesselJ(1, 2*I*sqrt(3)*sqrt(1-z))*BesselY(1, 2*I*sqrt(3)) - BesselY(1, (2*I)*sqrt(3)*sqrt(1-z))*BesselJ(1, 2*I*sqrt(3)))/sqrt(1-z). Here Phat(0,x) = 0.

Asymptotics: lim_{n->oo} a(n)/n! = BesselI(1,2*sqrt(3))/(sqrt(3)) = 3.468649618760...