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%I #15 May 08 2021 08:29:04
%S 3,5,1,9,2,6,9,17,4,1,17,3,17,17,1,33,8,2,33,33,3,24,33,22,33,33,2,33,
%T 33,22,33,65,16,4,65,1,65,65,22,52,65,22,65,12,1,65,65,11,65,52,3,40,
%U 65,1,12,65,11,65,65,11,65,65,1,129,32,8,129,2,11,39
%N a(n) is the smallest positive integer such that the binary representation of n*a(n) is a "binary square" (i.e., a term of A020330).
%C a(n) exists because if n has t bits, then (2^t+1)*n is a binary square.
%H Alois P. Heinz, <a href="/A320386/b320386.txt">Table of n, a(n) for n = 1..8191</a>
%e a(5) = 2 because 5 is not a binary square, but 5*2 = 10 is (its binary representation is 1010).
%p a:= proc(n) local k; for k while not (s-> (l->
%p l::even and s[1..l/2]=s[l/2+1..l])(length(s)))(
%p convert(convert(k*n, binary), string)) do od; k
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Oct 12 2018
%o (PARI) is(n) = my(L=#binary(n)\2); n>>L==bitand(n,2^L-1); \\ A020330
%o a(n) = my(k=1); while (!is(k*n), k++); k; \\ _Michel Marcus_, Oct 12 2018
%Y Cf. A020330.
%K nonn,base
%O 1,1
%A _Jeffrey Shallit_, Oct 12 2018
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