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Number of distinct exponents when n is factorized as a product of primes.
146

%I #43 Sep 01 2022 14:28:37

%S 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,

%T 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,

%U 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

%N Number of distinct exponents when n is factorized as a product of primes.

%C First term greater than 2 is a(360) = 3.

%C From _Michel Marcus_, Apr 24 2016: (Start)

%C A006939(n) gives the least m such that a(m) = n.

%C A062770 is the sequence of integers m such that a(m) = 1. (End)

%C 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

%C 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

%H Giovanni Resta, <a href="/A071625/b071625.txt">Table of n, a(n) for n = 1..10000</a>

%H E. T. Bell, <a href="http://dx.doi.org/10.1090/S0002-9904-1937-06651-1">Functions of coprime divisors of integers</a>, Bull. Amer. Math. Soc. 43 (1937), 818-822.

%H Carlo Sanna, <a href="https://doi.org/10.1007/s12044-020-0556-y">On the number of distinct exponents in the prime factorization of an integer</a>, Proceedings - Mathematical Sciences, Indian Academy of Sciences, Vol. 130, No. 1 (2020), Article 27, <a href="https://www.ias.ac.in/describe/article/pmsc/130/0027">alternative link</a>, <a href="https://arxiv.org/abs/1902.09224">arXiv preprint</a>, arXiv:1902.09224 [math.NT], 2019.

%e n = 5040 = 2^4*(3*5)^2*7, three different exponents arise:4,2 and 1; so a(5040)=3.

%t ffi[x_] := Flatten[FactorInteger[x]];

%t lf[x_] := Length[FactorInteger[x]];

%t ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}];

%t Table[Length[Union[ep[w]]], {w, 1, 256}]

%t (* Second program: *)

%t {0}~Join~Array[Length@ Union@ FactorInteger[#][[All, -1]] &, 104, 2] (* _Michael De Vlieger_, Apr 10 2019 *)

%o (PARI) a(n) = #Set(factor(n)[,2]); \\ _Michel Marcus_, Mar 12 2015

%o (Python)

%o from sympy import factorint

%o def a(n): return len(set(factorint(n).values()))

%o print([a(n) for n in range(1, 106)]) # _Michael S. Branicky_, Sep 01 2022

%Y Cf. A051903, A051904, A006939.

%Y Cf. A001221, A001222, A001615, A059404, A062770, A118914, A181819, A323014, A323022, A323023, A323024, A323025.

%K nonn

%O 1,12

%A _Labos Elemer_, May 29 2002