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%I #29 Jul 16 2021 05:22:44
%S 0,1,3,4,7,8,9,13,14,15,16,22,23,24,25,32,33,34,35,36,44,47,48,49,58,
%T 59,60,62,63,64,74,75,78,79,80,81,95,96,98,99,100,114,119,120,121,135,
%U 136,138,140,141,142,143,144,160,162,164,167,168,169,185,187,189,192
%N Squares and numbers k such that the continued fraction expansion of sqrt(k) is multiplicative.
%C If we consider each square k as having a continued fraction expansion c of all zeros after c(0) = sqrt(k)-1, then the continued fraction expansion of sqrt(k) for each square is trivially multiplicative.
%C For nonsquares, c(1) must be 1 and so k must satisfy m + 1/2 < sqrt(k) <= m+1, for some integer m.
%e The continued fraction of sqrt(22) is c = (4; 1, 2, 4, 2, 1, 8, ...) = A010126, which is multiplicative with c(2^e) = 2, c(3^e) = 4, c(p^e) = 1 otherwise.
%Y Union of A000290 and A108575.
%Y Continued fraction expansions: A040001, A010121, A040005, etc.
%K nonn,easy
%O 1,3
%A _Mitch Harris_, Jun 18 2005
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