OFFSET
1,1
COMMENTS
The sum of the series 6/a(1)+1/a(2)+1/a(3)+... is a transcendental number, and has a continued fraction expansion whose coefficients are given explicitly in terms of the sequence a(n) and the ratios a(n+1)/a(n).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..8
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
Andrew N. W. Hone, Continued fractions for some transcendental numbers, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.
FORMULA
The sequence is generated by taking a(n+1)=b(n-1)*a(n)*(1+n*a(n)), b(n)=a(n+1)/a(n) for n>=1 with initial values a(1)=7,b(0)=2. Alternatively, if a(1)=7,a(2)=112 are given then a(n+1)*a(n-1)=a(n)^2*(1+n*a(n)) for n>=2.
Sum_{n>=1} 1/a(n) = -5/7 + A270137. - Amiram Eldar, Nov 20 2020
MATHEMATICA
a[1] = 7; a[2] = 112;
a[n_] := a[n] = (a[n-1]^2 (1+(n-1)a[n-1]))/a[n-2];
Array[a, 5] (* Jean-François Alcover, Dec 16 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Andrew Hone, Mar 11 2016
STATUS
approved