A051903 Maximum exponent in the prime factorization of n.

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

Offset: 1

Author

Labos Elemer, Dec 16 1999

Keywords

• nonn
• easy

Comments

Smallest number of factors of all factorizations of n into squarefree numbers, see also A128651, A001055. - Reinhard Zumkeller, Mar 30 2007

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

From Thomas Ordowski, Dec 02 2019: (Start)

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

Examples

			For n = 72 = 2^3*3^2, a(72) = max(exponents) = max(3,2) = 3.
		

Links

• Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from T. D. Noe)
• Benjamin Merlin Bumpus and Zoltan A. Kocsis, Spined categories: generalizing tree-width beyond graphs, arXiv:2104.01841 [math.CO], 2021.
• Cao Hui-Zhong, The Asymptotic Formulas Related to Exponents in Factoring Integers, Math. Balkanica, Vol. 5 (1991), Fasc. 2.
• Ivan Niven, Averages of Exponents in Factoring Integers, Proc. Amer. Math. Soc., Vol. 22, No. 2 (1969), pp. 356-360.
• Eric Weisstein's World of Mathematics, Niven's Constant.
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Average value is A033150 = 1.7052....

Cf. A002322, A005361, A008479, A028234, A051904, A052409, A067029, A091050, A129132, A327295, A328310, A329885.

Programs

Haskell

a051903 1 = 0
a051903 n = maximum $ a124010_row n -- Reinhard Zumkeller, May 27 2012

Maple

A051903 := proc(n)
        a := 0 ;
        for f in ifactors(n)[2] do
                a := max(a,op(2,f)) ;
        end do:
        a ;
end proc: # R. J. Mathar, Apr 03 2012
# second Maple program:
a:= n-> max(0, seq(i[2], i=ifactors(n)[2])):
seq(a(n), n=1..120);  # Alois P. Heinz, May 09 2020

Mathematica

Table[If[n == 1, 0, Max @@ Last /@ FactorInteger[n]], {n, 100}] (* Ray Chandler, Jan 24 2006 *)

PARI

a(n)=if(n>1,vecmax(factor(n)[,2]),0) \\ Charles R Greathouse IV, Oct 30 2012

Python

from sympy import factorint
def A051903(n):
    return max(factorint(n).values()) if n > 1 else 0
# Chai Wah Wu, Jan 03 2015

Scheme

;; With memoization-macro definec.
(definec (A051903 n) (if (= 1 n) 0 (max (A067029 n) (A051903 (A028234 n))))) ;; Antti Karttunen, Aug 08 2016

Formula

a(n) = max_{k=1..A001221(n)} A124010(n,k). - Reinhard Zumkeller, Aug 27 2011

a(1) = 0; for n > 1, a(n) = max(A067029(n), a(A028234(n))). - Antti Karttunen, Aug 08 2016

Conjecture: a(n) = a(A003557(n)) + 1. This relation together with a(1) = 0 defines the sequence. - Velin Yanev, Sep 02 2017

Comment from David J. Seal, Sep 18 2017: (Start)

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

a(n) <= log(n)/log(2). - Hal M. Switkay, Jul 03 2025