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

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,...
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).

Crossrefs

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

Sequence in context: A161075 A161114 A161049 * A161074 A161113 A161048

Adjacent sequences: A138739 A138740 A138741 * A138743 A138744 A138745

Keyword

nonn,tabf

Author

Leroy Quet, Mar 27 2008

Extensions

a(19)-a(103) from Lucas A. Brown, Apr 12 2021

Status

approved