A138742 - OEIS

%I #18 Oct 10 2022 07:55:44

%S 1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,2,4,23,1,1,2,2,9,1,90,1,14,5,2,1,1,2,2,

%T 7,1,2,4,5,1,2,4,1,8,32,2,1,8,3,1,2,1,8,5,2,3,1,1,2,2,8,11,4,3,3,2,3,

%U 4,3,8,1,6,22,4,2,1,1,1,1,1,5,1,1,2,2,1,11,1,4,3,3,97,3,1,1,4,1,1,3,87,5,2,7,3

%N Let r_1 = 1. Let r_{m+1} = r_1 + 1/(r_2 + 1/(r_3 +...(r_{m-1} + 1/r_m)...)), a continued fraction of rational terms. Then row n of this irregular array contains the simple continued fraction terms of r_n.

%C The number of terms in the continued fraction of r_n is A138743(n).

%C The continued fraction of the limit, as n->infinity, of r_n is sequence A053978.

%C r_n = A064845(n)/A064846(n).

%H Lucas A. Brown, <a href="https://github.com/lucasaugustus/oeis/blob/main/A138742%2B3%2B4.py">A138742+3+4.py</a>

%e {r_n}: 1, 1, 2, 5/3, 31/18, 1231/720,...

%e r_5 = 31/18, for instance, equals the simple continued fraction 1+ 1/(1 + 1/(2 + 1/(1 + 1/(1 +1/2)))). So row 5 is (1,1,2,1,1,2).

%Y Cf. A053978, A064845, A064846, A138743, A138744.

%K nonn,tabf

%O 1,3

%A _Leroy Quet_, Mar 27 2008

%E a(19)-a(103) from _Lucas A. Brown_, Apr 12 2021