login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A071625
Number of distinct exponents when n is factorized as a product of primes.
146
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1
OFFSET
1,12
COMMENTS
First term greater than 2 is a(360) = 3.
From Michel Marcus, Apr 24 2016: (Start)
A006939(n) gives the least m such that a(m) = n.
A062770 is the sequence of integers m such that a(m) = 1. (End)
We define the k-th omega of n to be Omega(red^{k-1}(n)) where Omega = A001222 and red^{k} is the k-th functional iteration of A181819. The first two omegas are A001222 and A001221, while this sequence is the third, and A323022 is the fourth. The zeroth omega is not uniquely determined from prime signature, but one possible choice is A056239 (sum of prime indices). - Gus Wiseman, Jan 02 2019
Sanna (2020) proved that for each k>=1, the sequence of numbers n with A071625(n) = k has an asymptotic density A_k = (6/Pi^2) * Sum_{n>=1, n squarefree} rho_k(n)/psi(n), where psi is the Dedekind psi function (A001615), and rho_k(n) is defined by rho_1(n) = 1 if n = 1 and 0 otherwise, rho_{k+1}(n) = 0 if n = 1 and (1/(n-1)) * Sum_{d|n, d<n} rho_k(d) otherwise. - Amiram Eldar, Oct 18 2020
LINKS
E. T. Bell, Functions of coprime divisors of integers, Bull. Amer. Math. Soc. 43 (1937), 818-822.
Carlo Sanna, On the number of distinct exponents in the prime factorization of an integer, Proceedings - Mathematical Sciences, Indian Academy of Sciences, Vol. 130, No. 1 (2020), Article 27, alternative link, arXiv preprint, arXiv:1902.09224 [math.NT], 2019.
EXAMPLE
n = 5040 = 2^4*(3*5)^2*7, three different exponents arise:4,2 and 1; so a(5040)=3.
MATHEMATICA
ffi[x_] := Flatten[FactorInteger[x]];
lf[x_] := Length[FactorInteger[x]];
ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}];
Table[Length[Union[ep[w]]], {w, 1, 256}]
(* Second program: *)
{0}~Join~Array[Length@ Union@ FactorInteger[#][[All, -1]] &, 104, 2] (* Michael De Vlieger, Apr 10 2019 *)
PROG
(PARI) a(n) = #Set(factor(n)[, 2]); \\ Michel Marcus, Mar 12 2015
(Python)
from sympy import factorint
def a(n): return len(set(factorint(n).values()))
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Sep 01 2022
KEYWORD
nonn
AUTHOR
Labos Elemer, May 29 2002
STATUS
approved