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Ratios of successive terms of A270121.
2

%I #11 Dec 16 2018 03:26:45

%S 2,16,3600,4354563600,30582275103386435842563600,

%T 8210597206238423089030527211173884373017314096293353169731842563600

%N Ratios of successive terms of A270121.

%C The ratios A270121(n+1)/A270121(n) generate the sequence for n>=1.

%H A. N. W. Hone, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Hone/hone3.html">Curious continued fractions, nonlinear recurrences and transcendental numbers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.

%H A. N. W. Hone, <a href="http://arxiv.org/abs/1509.05019">Continued fractions for some transcendental numbers</a>, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.

%F A270121 is generated by a recurrence of second order: if A270121(n)=x(n) then x(n+1)*x(n-1)=x(n)^2*(1+n*x(n)) for n>=1 with x(1)=7,x(2)=112.

%t (* b = A270121 *)

%t b[1] = 7; b[2] = 112;

%t b[n_] := b[n] = (b[n - 1]^2 (1 + (n - 1) b[n - 1]))/b[n - 2];

%t a[0] = 2; a[n_] := b[n + 1]/b[n];

%t Table[a[n], {n, 0, 5}] (* _Jean-François Alcover_, Dec 16 2018 *)

%K nonn

%O 0,1

%A _Andrew Hone_, Mar 11 2016