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%I #12 Oct 22 2023 17:01:15
%S 1,1,1,1,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,
%T 1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,1,2,1,4,
%U 1,2,1,8,1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,8,1,2,1,4
%N Highest power of 2 dividing n which is < sqrt(n), for n >= 2; a(1) = 1.
%t Join[{1}, Table[Last[Select[Divisors[n], # < Sqrt[n] && IntegerQ[Log[2, #]] &]], {n, 2, 100}]]
%t a[n_] := 2^Min[IntegerExponent[n, 2], Ceiling[Log2[n]/2] - 1]; a[1] = 1; Array[a, 100] (* _Amiram Eldar_, Oct 19 2023 *)
%o (PARI) a(n) = if (n==1, 1, vecmax(select(x->((x^2 < n) && (2^logint(x,2)==x)), divisors(n)))); \\ _Michel Marcus_, Oct 19 2023
%Y Cf. A006519, A033676, A135517, A363827.
%K nonn
%O 1,6
%A _Ilya Gutkovskiy_, Oct 19 2023