OFFSET
0,3
COMMENTS
The corresponding numerator sequence is A222468.
a(n) = Q(n,2) with the denominator polynomials Q of A084950. All the given formulas follow from there. The limit of the continued fraction (0 + K_{k>=1} (2/k))/2 = 1/(1+2/(2+2/(3+2/(4+... is (1/2)*sqrt(2)*BesselI(1,2*sqrt(2))/BesselI(0,2*sqrt(2)) = 0.5631786198117... See A222466 for more digits.
For a combinatorial interpretation in terms of labeled Morse codes see a comment on A084950. Here each dash has label x=2, and the dots have label j if they are at position j. Labels are multiplied and all codes on [1,2,...,n] are summed.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..449
FORMULA
Recurrence: a(n) = n*a(n-1) + 2*a(n-2), a(-1) = 0, a(0 ) =1, n >= 1.
As a sum: a(n) = Sum_{m =0..floor(n/2)} a(n-m,m)*2^m, n >= 0, with a(n,m) = (n!/m!)*binomial(n,m) = |A021009(n,m)| (Laguerre).
Explicit form: a(n) = 2*(w/2)^(n+1)*(BesselI(0,w)*BesselK(n+1,w) - BesselK(0,w)*BesselI(n+1,w)*(-1)^(n+1)) with w := -2*sqrt(2).
E.g.f.: (i*Pi*sqrt(2)/sqrt(1-z))*(BesselJ(1, 2*i*sqrt(2)*sqrt(1-z))*BesselY(0, 2*i*sqrt(2)) - BesselY(1, 2*i*sqrt(2)*sqrt(1-z))*BesselJ(0,2*i*sqrt(2))) with the imaginary unit i = sqrt(-1).
Asymptotics: lim_{n->infinity} a(n)/n! = BesselI(0,2*sqrt(2)) = 4.2523508795026...
EXAMPLE
a(4) = 4*a(3) + 2*a(2) = 4*14 + 2*4 = 64.
Continued fraction convergent: 1/(1+2/(2+2/(3+2/4))) = 9/16 = 36/64 = A222468(4)/a(4).
Morse code a(4) = 64 from the sum of all 5 labeled codes on [1,2,3,4], one with no dash, three with one dash and one with two dashes: 4! + (3*4 + 1*4 + 1*2)*2 + 2^2 = 64.
MATHEMATICA
RecurrenceTable[{a[0]==a[1]==1, a[n]==n*a[n-1]+2*a[n-2]}, a, {n, 20}] (* Harvey P. Dale, Jul 06 2017 *)
PROG
(PARI) m=30; v=concat([1, 4], vector(m-2)); for(n=3, m, v[n]=n*v[n-1] + 2*v[n-2]); concat([1], v) \\ G. C. Greubel, May 16 2018
(Magma) I:=[1, 4]; [1] cat [n le 2 select I[n] else n*Self(n -1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, May 16 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gary Detlefs and Wolfdieter Lang, Mar 21 2013
STATUS
approved