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A051903
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Maximum exponent in the prime factorization of n.
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266
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0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2, 1, 1, 1, 3, 1
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OFFSET
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1,4
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COMMENTS
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Maximum number of invariant factors among abelian groups of order n. - Álvar Ibeas, Nov 01 2014
a(n) is the highest of the frequencies of the parts of the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product(p_j-th prime, j=1..r) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24) = 3; indeed, the partition having Heinz number 24 = 2*2*2*3 is [1,1,1,2], where the distinct parts 1 and 2 have frequencies 3 and 1, respectively. - Emeric Deutsch, Jun 04 2015
a(n) is the smallest k such that b^(phi(n)+k) == b^k (mod n) for all b.
The Euler phi function can be replaced by the Carmichael lambda function.
Problems:
(*) Are there composite numbers n > 4 such that n == a(n) (mod phi(n))? By Lehmer's totient conjecture, there are no such squarefree numbers.
(**) Are there odd numbers n such that a(n) > 1 and n == a(n) (mod lambda(n))? These are odd numbers n such that a(n) > 1 and b^n == b^a(n) (mod n) for all b.
(***) Are there odd numbers n such that a(n) > 1 and n == a(n) (mod ord_{n}(2))? These are odd numbers n such that a(n) > 1 and 2^n == 2^a(n) (mod n).
Note: if (***) do not exist, then (**) do not exist. (End)
Niven (1969) proved that the asymptotic mean of this sequence is 1 + Sum_{j>=2} 1 - (1/zeta(j)) (A033150). - Amiram Eldar, Jul 10 2020
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LINKS
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FORMULA
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Conjecture: a(n) = a(A003557(n)) + 1. This relation together with a(1) = 0 defines the sequence. - Velin Yanev, Sep 02 2017
This conjecture seems very easily provable to me: if the factorization of n is p1^k1 * p2^k2 * ... * pm^km, then the factorization of the largest squarefree divisor of n is p1 * p2 * ... * pm. So the factorization of A003557(n) is p1^(k1-1) * p2^(k2-1) * ... * pm^(km-1) if exponents of zero are allowed, or with the product terms that have an exponent of zero removed if they're not (if that results in an empty product, consider it to be 1 as usual).
The formula then follows from the fact that provided all ki >= 1, Max(k1, k2, ..., km) = Max(k1-1, k2-1, ..., km-1) + 1, and Max(k1-1, k2-1, ..., km-1) is not altered by removing the ki-1 values that are 0, provided we treat the empty Max() as being 0. That proves the formula and the provisos about empty products and Max() correspond to a(1) = 0.
Also, for any n, applying the formula Max(k1, k2, ..., km) times to n = p1^k1 * p2^k2 * ... * pm^km reduces all the exponents to zero, i.e., to the case a(1) = 0, so that case and the formula generate the sequence. (End)
Sum_{k=1..n} (-1)^k * a(k) ~ c * n, where c = Sum_{k>=2} 1/((2^k-1)*zeta(k)) = 0.44541445377638761933... . - Amiram Eldar, Jul 28 2024
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EXAMPLE
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For n = 72 = 2^3*3^2, a(72) = max(exponents) = max(3,2) = 3.
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MAPLE
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a := 0 ;
for f in ifactors(n)[2] do
a := max(a, op(2, f)) ;
end do:
a ;
# second Maple program:
a:= n-> max(0, seq(i[2], i=ifactors(n)[2])):
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MATHEMATICA
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Table[If[n == 1, 0, Max @@ Last /@ FactorInteger[n]], {n, 100}] (* Ray Chandler, Jan 24 2006 *)
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PROG
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(Haskell)
a051903 1 = 0
(Python)
from sympy import factorint
return max(factorint(n).values()) if n > 1 else 0
(Scheme, with memoization-macro definec)
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CROSSREFS
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Average value is A033150 = 1.7052....
Cf. A002322, A005361, A008479, A028234, A051904, A052409, A067029, A091050, A129132, A327295, A328310, A329885.
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KEYWORD
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nonn,easy,changed
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AUTHOR
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STATUS
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approved
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