A138742 - OEIS

A138742

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.

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, 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, 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

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OFFSET

1,3

COMMENTS

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

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

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

LINKS

Table of n, a(n) for n=1..103.

Lucas A. Brown, A138742+3+4.py

EXAMPLE

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