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A374327
The maximal exponent in the prime factorization of the numbers whose maximal exponent in their prime factorization is a power of 2.
5
1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2
OFFSET
1,3
COMMENTS
First differs from {A369933(n+1), n>=1} at n = 378.
The first occurrence of 2^k, for k = 0, 1, ..., is at 1, 3, 14, 224, 57307, ..., which is the position of 2^(2^k) at A369938.
LINKS
FORMULA
a(n) = 2^A374328(n).
a(n) = A051903(A369938(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=0} 2^k * d(k) / Sum_{k>=0} d(k) = 1.41151462942556759486..., where d(k) = 1/zeta(2^k+1) - 1/zeta(2^k) for k>=1, and d(0) = 1/zeta(2).
MATHEMATICA
f[n_] := Module[{e = If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]}, If[e == 2^IntegerExponent[e, 2], e, Nothing]]; Array[f, 150]
PROG
(PARI) lista(kmax) = {my(e); for(k = 2, kmax, e = vecmax(factor(k)[, 2]); if(e >> valuation(e, 2) == 1, print1(e, ", "))); }
CROSSREFS
Similar sequences: A374324, A374325, A374326, A374328.
Sequence in context: A305393 A259154 A369933 * A368473 A106035 A293811
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Jul 04 2024
STATUS
approved