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%I #22 Jul 18 2021 11:34:03
%S 3,7,8,13,14,15,22,23,24,32,33,34,35,44,47,48,58,59,60,62,63,74,75,78,
%T 79,80,95,96,98,99,114,119,120,135,136,138,140,141,142,143,160,162,
%U 164,167,168,185,187,189,192,194,195,215,219,220,222,223,224,248,252,254
%N Numbers k such that the continued fraction for sqrt(k) is multiplicative.
%C m^2 - m < a(n) < m^2 for some m.
%e The continued fraction for sqrt(7) is c = (2, 1, 1, 1, 4, 1, 1, 1, 4, ...) = A010121. If we index starting at 0, so that c(0) = 2, c is multiplicative (the value at c(0) is immaterial). Hence 7 is in the sequence.
%Y Cf. A010121, A109054 (with squares included).
%K nonn
%O 1,1
%A _David W. Wilson_, Jun 10 2005