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 A114551 Continued fraction expansion of the constant (A114550) equal to the sum 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. 5
 2, 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742, 342022190843338960032, 1710009514450915230711940280907486, 584861200495456320274313200204390612579749188443599552 (list; graph; refs; listen; history; text; internal format)
 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 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: A046772 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

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Last modified February 23 15:49 EST 2020. Contains 332168 sequences. (Running on oeis4.)