%I #16 Oct 17 2019 15:46:50
%S 0,1,2,3,4,5,6,7,8,9,10,143,12,143,14,15,16,17,18,475,20,21,286,667,
%T 24,475,286,27,28,667,30,31,32,33,34,245,36,1517,950,741,40,1517,42,
%U 2279,572,45,1334,2867,48,245,950,51,572,2279,54,3245,56,741,1334,3245
%N a(n) = lcm(n, r(n)) where r(n) = A030101(n) corresponds to the binary reversal of n.
%H Rémy Sigrist, <a href="/A326302/b326302.txt">Table of n, a(n) for n = 0..8192</a>
%F a(n) >= n with equality iff n belongs to A057890.
%F a(n) < A062383(n)*n. - _Robert Israel_, Oct 17 2019
%e For n = 35:
%e - the binary reversal of 35 ("100011" in binary) is 49 ("110001" in binary),
%e - hence a(35) = lcm(35, 49) = lcm(5*7, 7^2) = 5*7^2 = 245.
%p f:= proc(n) local L,j;
%p L:= convert(n,base,2);
%p ilcm(n, add(2^(j-1)*L[-j],j=1..nops(L)))
%p end proc:
%p map(f, [$0..100]);
%o (PARI) a(n, base=2) = lcm(n, fromdigits(Vecrev(digits(n, base)), base))
%Y Cf. A030101, A057890 (fixed points), A161825 (GCD variant), A068634 (decimal variant).
%Y Cf. A062383.
%K nonn,base,look
%O 0,3
%A _Rémy Sigrist_, Oct 17 2019