A071625 Number of distinct exponents when n is factorized as a product of primes.
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
Keywords
Examples
n = 5040 = 2^4*(3*5)^2*7, three different exponents arise:4,2 and 1; so a(5040)=3.
Links
- Giovanni Resta, Table of n, a(n) for n = 1..10000
- 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.
Crossrefs
Programs
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Maple
# Using function 'PrimeSignature' from A124010. a := n -> nops(convert(PrimeSignature(n), set)): seq(a(n), n = 1..105); # Peter Luschny, Jun 15 2025
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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 *)
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PARI
a(n) = #Set(factor(n)[,2]); \\ Michel Marcus, Mar 12 2015
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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
Comments