

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.


4



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
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OFFSET

1,3


COMMENTS

The continued fractions, for rows 3 and up, each have a final term >= 2.
The number of terms in the nth 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 * A352631 A134446 A125749
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



