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%I #62 Sep 08 2022 08:46:24
%S 1,2,2,5,8,7,16,3,20,6,9,16,13,30,22,53,76,65,142,104,19,124,36,161,
%T 198,20,219,80,15,32,3,12,4,17,22,20,43,64,27,92,30,41,72,19,92,28,
%U 121,150,136,287,424,89,514,302,817,1120,969,110,108,73,182,128,311
%N a(n) = numerator((a(n-1) + a(n-2) + 1)/a(n-1)), with a(1)=1, a(2)=2.
%C As n goes to infinity, 32 < (a(n+1) + a(n))/(a(n+1) - a(n)) < 32.35 (conjectured).
%C This sequence is the continued fraction expansion of ~ 1.4072426692639398147... (calculated to 19 digits).
%C The sum of the reciprocals of this sequence is ~ 4.97804721273463... (calculated to 14 digits).
%C The average reduction 0.621 < a(n)/(a(n-1)+a(n-2)+1) < 0.622 (conjectured); the average reduction is the average of the individual reductions (The change from the numerator to the numerator in the simplest form):
%C a(3)/(a(2)+a(1)+1) -> 2/(2+1+1) -> 0.5
%C a(4)/(a(3)+a(2)+1) -> 5/(2+2+1) -> 1
%C a(5)/(a(4)+a(3)+1) -> 8/(5+2+1) -> 1
%C a(6)/(a(5)+a(4)+1) -> 7/(8+5+1) -> 0.5
%C a(7)/(a(6)+a(5)+1) -> 16/(7+8+1) -> 1
%C a(8)/(a(7)+a(6)+1) -> 3/(16+7+1) -> 0.125
%C a(9)/(a(8)+a(7)+1) -> 20/(3+16+1) -> 1
%C a(10)/(a(9)+a(8)+1) -> 6/(20+3+1) -> 0.25
%C That number is the average of these to ~ 100000 terms (There was some fluctuation to take into account).
%H Vimal Vinod, <a href="/A327597/b327597.txt">Table of n, a(n) for n = 1..10000</a>
%F It appears that this sequence's growth can be approximated by a(n) ~ (1 + 1/c)^n where 17.8 < c < 18.5.
%e a(1) = 1.
%e a(2) = 2.
%e a(3) = numerator((1 + 2 + 1)/2) -> numerator(2/1) = 2.
%e a(4) = numerator((2 + 2 + 1)/2) -> numerator(5/2) = 5.
%e a(5) = numerator((2 + 5 + 1)/5) -> numerator(8/5) = 8.
%e a(6) = numerator((5 + 8 + 1)/8) -> numerator(7/4) = 7.
%t Nest[Append[#, Numerator[(#2 + #1 + 1)/#2] & @@ #[[-2 ;; -1]]] &, {1, 2}, 61] (* _Michael De Vlieger_, Sep 30 2019 *)
%o (Python)
%o from fractions import Fraction
%o num_terms = 100
%o S = [1, 2]
%o for n in range(num_terms-2):
%o s = Fraction((S[n]+S[n+1]+1),S[n+1]).numerator
%o S.append(s)
%o print(S) # Should print the sequence to the length specified.
%o (PARI) lista(nn) = {my(xa = 1, ya = 2, za); print1(xa, ", ", ya, ", "); for (n=3, nn, za = numerator((ya + xa + 1)/ya); print1(za, ", "); xa = ya; ya = za;);} \\ _Michel Marcus_, Sep 24 2019
%o (Magma) a:=[1,2]; [n le 2 select a[n] else Numerator((Self(n-1) + Self(n-2) + 1)/Self(n-1)):n in [1..64]]; // _Marius A. Burtea_, Sep 27 2019
%K nonn
%O 1,2
%A _Vimal Vinod_, Sep 18 2019