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%I #16 Oct 11 2017 12:36:17
%S 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,
%T 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,
%U 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
%N 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.
%C The continued fractions, for rows 3 and up, each have a final term >= 2.
%C The number of terms in the n-th row is A126337(n).
%C The sum of the reciprocals of the terms in rows 1 through n is A126338(n)/A126339(n).
%H Michael De Vlieger, <a href="/A126336/b126336.txt">Table of n, a(n) for n = 1..11269</a> (rows 1 <= n <= 80).
%e 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).
%t f[l_List] := Append[l, ContinuedFraction[Plus @@ (1/# &) /@ Flatten[l]]]; Flatten@ Nest[f, {{1}}, 15] (* _Ray Chandler_, Dec 26 2006 *)
%Y Cf. A126337, A126338, A126339.
%K easy,nonn,tabf
%O 1,3
%A _Leroy Quet_, Dec 25 2006
%E Extended by _Ray Chandler_, Dec 26 2006
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