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The maximal exponent in the prime factorization of the largest divisor of n whose number of divisors is a power of 2.
4

%I #9 May 08 2024 08:52:54

%S 0,1,1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,3,1,1,1,1,3,1,1,

%T 1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,1,1,1,3,1,3,1,1,1,1,1,1,1,3,1,1,1,1,

%U 1,1,1,3,1,1,1,1,1,1,1,3,3,1,1,1,1,1,1

%N The maximal exponent in the prime factorization of the largest divisor of n whose number of divisors is a power of 2.

%C First differs from A331273 at n = 32.

%C Differs from A368247 at n = 1, 128, 216, 256, 384, 432, 512, ... .

%C All the terms are of the form 2^k-1 (A000225).

%H Amiram Eldar, <a href="/A372604/b372604.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A051903(A372379(n)).

%F a(n) = A092323(A051903(n)+1).

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{i>=1} 2^i * (1 - 1/zeta(2^(i+1)-1)) = 1.36955053734097783559... .

%e 4 has 3 divisors, 1, 2 and 4. The number of divisors of 4 is 3, which is not a power of 2. The number of divisors of 2 is 2, which is a power of 2. Therefore, A372379(4) = 2 and a(4) = A051903(2) = 1.

%t f[n_] := 2^Floor[Log2[n + 1]] - 1; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

%o (PARI) s(n) = 2^exponent(n+1) - 1;

%o a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

%Y Cf. A000225, A051903, A092323, A372379, A372466.

%Y Cf. A331273, A368247.

%Y Similar sequences: A007424, A368781, A372601, A372602, A372603.

%K nonn,easy

%O 1,8

%A _Amiram Eldar_, May 07 2024