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A061143 a(1) = 1; a(n) = sum of terms in the continued fraction for the square of the continued fraction [a(1); a(2), a(3), a(4),..., a(n-1)]. 0

%I #6 Apr 09 2014 10:14:37

%S 1,1,4,13,82,121,218,278,498,305,358,510,666,828,939,1189,1155,1335,

%T 1870,2101,1770,16411,1380,2082,14035,2437,1734,3079,4537,2335,3923,

%U 2750,8259,3144,2986,2803,6187,2833,5008,3251,3442,4588,3767

%N a(1) = 1; a(n) = sum of terms in the continued fraction for the square of the continued fraction [a(1); a(2), a(3), a(4),..., a(n-1)].

%e (1 +(1/(1 +1/4)))^2 = (9/4)^2 = 81/25 = 3 +1/(4 +1/6). So a(4) = 3 + 4 + 6 = 13.

%K easy,nonn

%O 1,3

%A _Leroy Quet_, May 29 2001

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Last modified May 27 03:56 EDT 2024. Contains 372847 sequences. (Running on oeis4.)