A114551 Continued fraction expansion of the constant (A114550) equal to Sum_{n>=0} 1/A112373(n) such that the partial quotients satisfy a(2n) = A112373(n) for n > 0 and a(2n+1) = A112373(n+1)/A112373(n) for n >= 0.

2, 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742, 342022190843338960032, 1710009514450915230711940280907486, 584861200495456320274313200204390612579749188443599552

Offset: 0

Author

Paul D. Hanna, Dec 08 2005

Keywords

• cofr
• nonn

Comments

A112373 is defined by the recurrence: let b(n) = A112373(n), then

b(n) = (b(n-1)^3 + b(n-1)^2)/b(n-2) for n >= 2 with b(0)=b(1)=1.

Examples

			2.584401724019776724812076147153331342112382090467969...
= Sum_{n>=0} 1/A112373(n) = 1/1 + 1/1 + 1/2 + 1/12 + 1/936 + 1/68408496 + ...
= [2;1,1,2,2,6,12,78,936,73086,68408496,...] (continued fraction).
The recurrence of partial quotients is demonstrated by:
(odd-index) a(7) = 78 = a(6)*a(5) + a(5) = 12*6 + 6;
(even-index) a(8) = 936 = a(7)*a(6) = 78*12.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..20
• Taras Goy and Mark Shattuck, Determinant Formulas of Some Hessenberg Matrices with Jacobsthal Entries Jacobsthal Entries, Applications and Appl. Math. (2021) Vol. 16, Issue 1, Art. 10.
• A. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015 and J. Int. Seq. 18 (2015) # 15.8.4.

Crossrefs

Cf. A112373, A114550 (constant), A114552 (bisection).

Programs

Mathematica

a[0] = 2; a[1] = a[2] = 1;
a[n_] := a[n] = a[n-1] a[n-2] + Mod[n, 2] a[n-2];
a /@ Range[0, 14] (* Jean-François Alcover, Oct 01 2019 *)

PARI

a(n)=if(n<0,0,if(n<3,[2,1,1][n+1],a(n-1)*a(n-2)+(n%2)*a(n-2)))

Formula

a(2n) = a(2n-1)*a(2n-2) for n>=2, a(2n+1) = a(2n)*a(2n-1) + a(2n-1) for n>=1, with a(0)=2, a(1)=a(2)=1. - Jeffrey Shallit