|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
|
KEYWORD
|
easy,nonn,tabf
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|