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A368781
The maximal exponent in the unique factorization of n in terms of distinct "Fermi-Dirac primes".
8
0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1
OFFSET
1,4
COMMENTS
First differs from A335428 at n = 36. Differs from A050377, A344417 and A347437 at n = 1 and then at n = 36.
In the unique factorization of n in terms of distinct "Fermi-Dirac primes", n is represented as a product of prime powers (A246655) whose exponents are powers of 2 (A000079). a(n) is the maximal exponent of these prime powers (and not the maximal exponent of the exponents that are powers of 2). Thus, a(n) is a power of 2 for n >= 2.
LINKS
FORMULA
a(n) = A053644(A051903(n)).
a(n) = 2^(A299090(n)-1) for n >= 2.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=1} (2^(k-1) * (1 - 1/zeta(2^k))) = 1.56056154773294953123... .
a(n) = A051903(A353897(n)). - Amiram Eldar, May 07 2024
EXAMPLE
For n = 972 = 2^2 * 3^5, the unique factorization of 972 in terms of distinct "Fermi-Dirac primes" is 2^(2^1) * 3^(2^0) * 3^(2^2). Therefore, a(972) = 2^2 = 4.
MATHEMATICA
a[n_] := 2^Floor[Log2[Max[FactorInteger[n][[;; , 2]]]]]; a[1] = 0; Array[a, 100]
PROG
(PARI) a(n) = if(n > 1, 2^exponent(vecmax(factor(n)[, 2])), 0);
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Jan 05 2024
STATUS
approved