A126336 Irregular table where the first row is (1). Row n is the continued fraction terms of the rational equal to the sum of the reciprocals of all the terms in the previous rows.

1, 1, 2, 2, 2, 3, 2, 4, 3, 4, 1, 11, 6, 3, 1, 7, 2, 8, 2, 2, 92, 9, 1, 1, 6, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 22, 3, 5, 2, 3, 1, 1, 15, 3, 4, 2, 26, 1, 7, 12, 1, 7, 2, 1, 26, 1, 1, 4, 33, 13, 1, 5, 1, 13, 1, 8, 1, 13, 1, 18, 2, 39, 4, 1, 2, 10, 6, 1, 4, 1, 20, 43, 1, 1, 3, 1, 21, 1, 1, 2, 2, 49, 1

Offset

1,3

Comments

The continued fractions, for rows 3 and up, each have a final term >= 2.

The number of terms in the n-th row is A126337(n).

The sum of the reciprocals of the terms in rows 1 through n is A126338(n)/A126339(n).

Links

Michael De Vlieger, Table of n, a(n) for n = 1..11269 (rows 1 <= n <= 80).

Example

The sum of the reciprocals of the terms of the first 6 rows is 1 + 1 + 1/2 + 1/2 + 1/2 + 1/3 + 1/2 + 1/4 + 1/3 = 59/12. 59/12 equals the continued fraction 4 + 1/(1 + 1/11). So row 7 is (4,1,11).

Mathematica

f[l_List] := Append[l, ContinuedFraction[Plus @@ (1/# &) /@ Flatten[l]]]; Flatten@ Nest[f, {{1}}, 15] (* Ray Chandler, Dec 26 2006 *)

Crossrefs

Cf. A126337, A126338, A126339.

Sequence in context: A321576 A304101 A278636 * A364421 A352631 A134446

Adjacent sequences: A126333 A126334 A126335 * A126337 A126338 A126339

Keyword

easy,nonn,tabf

Author

Leroy Quet, Dec 25 2006

Extensions

Extended by Ray Chandler, Dec 26 2006

Status

approved