cp's OEIS Frontend

A114552 Odd-indexed bisection of A114551, which is the continued fraction expansion of Sum_{n>=0} 1/A112373(n); also, a(n) = A112373(n+1)/A112373(n) for n>=0.

%I A114552 #16 Aug 26 2022 07:23:39
%S A114552 1,2,6,78,73086,4999703411742,1710009514450915230711940280907486,
%T A114552 1000118217480414644596377710149364954622541926875237885435586055466747491994623223953758
%N A114552 Odd-indexed bisection of A114551, which is the continued fraction expansion of Sum_{n>=0} 1/A112373(n); also, a(n) = A112373(n+1)/A112373(n) for n>=0.
%C A114552 A112373 is defined by the recurrence: let b(n) = A112373(n), then
%C A114552 b(n) =(b(n-1)^3 + b(n-1)^2)/b(n-2) for n>=2 with b(0)=b(1)=1.
%C A114552 The next term 58499...45086 has 228 digits and is too large to include.
%H A114552 Andrew N. W. Hone, <a href="http://arxiv.org/abs/1507.00063">Curious continued fractions, nonlinear recurrences and transcendental numbers</a>, arXiv:1507.00063 [math.NT], 2015  and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Hone/hone3.html">J. Int. Seq. 18 (2015) # 15.8.4</a>.
%F A114552 a(n) = a(n-1)*(1 + a(0)*a(1)*a(2)*...*a(n-1)) for n>=1, with a(0)=1.
%F A114552 a(n) = a(n-1)*(1 - a(n-1) + a(n-1)^2/a(n-2)) for n>=2, with a(0)=1, a(1)=2.
%e A114552 a(5) = 4999703411742 = 73086*(1 + 1*2*6*78*73086).
%e A114552 a(5) = a(4)*(1 - a(4) + a(4)^2/a(3)) = 73086*(1 - 73086 + 73086^2/78).
%o A114552 (PARI) a(n)=if(n==0,1,if(n==1,2,a(n-1)*(1-a(n-1)+a(n-1)^2/a(n-2))))
%Y A114552 Cf. A112373, A114550 (constant), A114551 (continued fraction).
%K A114552 nonn
%O A114552 0,2
%A A114552 _Paul D. Hanna_, Dec 08 2005
%E A114552 a(7) corrected by _Georg Fischer_, Aug 26 2022