login
a(n) = log_2(A368473(n)).
1

%I #7 May 04 2024 05:37:15

%S 0,0,0,1,0,0,0,1,0,0,1,0,0,0,2,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,2,0,0,

%T 0,0,0,0,1,1,0,0,2,1,1,0,1,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,0,0,0,1,1,0,

%U 0,0,2,2,0,0,1,0,0,0,0,1,0,1,0,0,0,0,1

%N a(n) = log_2(A368473(n)).

%C The first position of k, for k = 0, 1, ..., is 1, 4, 15, 126, 1134, ..., which is the position of A085629(2^k) in A138302.

%H Amiram Eldar, <a href="/A372505/b372505.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = log_2(A005361(A138302(n))).

%t f[n_] := Module[{p = Times @@ FactorInteger[n][[;; , 2]], e}, e = IntegerExponent[p, 2]; If[p == 2^e, e, Nothing]]; Array[f, 150]

%o (PARI) lista(kmax) = {my(p, e); for(k = 1, kmax, p = vecprod(factor(k)[, 2]); e = valuation(p, 2); if(p >> e == 1, print1(e, ", ")));}

%Y Cf. A005361, A085629, A138302, A368473.

%Y Cf. A369934, A370078, A372467, A372469.

%K nonn,easy

%O 1,15

%A _Amiram Eldar_, May 04 2024