This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A063896 #59 Dec 29 2022 11:29:45
%S A063896 0,1,1,3,7,31,255,8191,2097151,17179869183,36028797018963967,
%T A063896 618970019642690137449562111,
%U A063896 22300745198530623141535718272648361505980415
%N A063896 a(n) = 2^Fibonacci(n) - 1.
%C A063896 The recurrence can also be written a(n)+1 = (a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1) = log_p(a(n-1)+1) + log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fibonacci(n). Hence any other sequence p^Fibonacci(n)-1 could also serve as a valid solution to that recurrence, depending only on the value of the term a(1). - _Hieronymus Fischer_, Jun 27 2007
%C A063896 Written in binary, a(n) contains Fibonacci(n) 1's. Thus the sequence converted to base-2 is A007088(a(n)) = 0, 1, 1, 11, 111, 11111, 11111111, ... . - _Hieronymus Fischer_, Jun 27 2007
%C A063896 In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1)*b(n-2) + b(n-1) + b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - _Rahul Goswami_, Apr 15 2020
%C A063896 a(n) is also the numerator of the continued fraction [2^F(0), 2^F(1), 2^F(2), 2^F(3), ..., 2^F(n-2)] for n>0. For the denominator, see A005203. - _Chinmay Dandekar_ and _Greg Dresden_, Sep 19 2020
%H A063896 Alois P. Heinz, <a href="/A063896/b063896.txt">Table of n, a(n) for n = 0..18</a>
%H A063896 M. Tamba and Y. S. Valaulikar, <a href="http://pubs.sciepub.com/tjant/4/4/4">A nonlinear extension of Fibonacci sequence</a>, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 4, 2016.
%F A063896 The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).
%F A063896 a(n) = A000301(n) - 1. - _R. J. Mathar_, Apr 26 2007
%F A063896 a(n) = a(n-2)*2^ceiling(log_2(a(n-1))) + a(n-1) for n>1. - _Hieronymus Fischer_, Jun 27 2007
%F A063896 a(n) = A000225(A000045(n)). - _Alois P. Heinz_, Mar 19 2020
%p A063896 a:= n-> 2^(<<0|1>, <1|1>>^n)[1,2]-1:
%p A063896 seq(a(n), n=0..15); # _Alois P. Heinz_, Aug 12 2017
%t A063896 2^Fibonacci[Range[0,15]]-1 (* _Harvey P. Dale_, May 20 2014 *)
%t A063896 RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == (a[n - 1] + 1)*(a[n - 2] + 1) - 1}, a[n], {n, 0, 12}] (* _Ray Chandler_, Jul 30 2015 *)
%o A063896 (PARI) a(n) = 2^fibonacci(n) - 1 \\ _Charles R Greathouse IV_, Oct 03 2016
%Y A063896 Cf. A000045 (Fibonacci), A000225, A000301, A005203, A061107.
%Y A063896 See A131293 for a base-10 analog with Fib(n) 1's.
%K A063896 nonn,easy
%O A063896 0,4
%A A063896 _Robert G. Wilson v_, Aug 29 2001