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

0,1

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.

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.

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

Example

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.

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 *)

Prog

(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)))

Crossrefs

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

Sequence in context: A359652 A179974 A246402 * A191620 A214751 A306512

Adjacent sequences: A114548 A114549 A114550 * A114552 A114553 A114554

Keyword

cofr,nonn

Author

Paul D. Hanna, Dec 08 2005

Status

approved