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A064171 a(1) = 1; a(n+1) = product of numerator and denominator of the continued fraction, [a(1); a(2), a(3),..., a(n)]. 0

%I #9 Jun 17 2023 22:54:04

%S 1,1,2,15,3542,44438728965,87757630046458789424678387797052,

%T 675856755205061509378984088665012353791755857887475951486745195889516813384613294753875414807401

%N a(1) = 1; a(n+1) = product of numerator and denominator of the continued fraction, [a(1); a(2), a(3),..., a(n)].

%C Numerator and denominator in definition have no common divisors >1.

%e [a(1); a(2), a(3)] = 1 + 1/(1 + 1/2) = 5/3. So a(4) = 5 * 3 = 15.

%K nonn

%O 1,3

%A _Leroy Quet_, Sep 19 2001

%E a(8) from _Sean A. Irvine_, Jun 17 2023

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Last modified April 24 07:52 EDT 2024. Contains 371922 sequences. (Running on oeis4.)