A375670 The maximum exponent in the prime factorization of the largest 5-rough divisor of n.

0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 2, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 2, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1

Offset: 1

Author

Amiram Eldar, Aug 23 2024

Keywords

• nonn
• easy

Comments

First differs from A106799 at n = 35.

The largest exponent among the exponents of the primes that are larger than 3 in the prime factorization of n.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A003586, A007310 (5-rough numbers), A051903, A065330, A106799, A375669.

Programs

Mathematica

a[n_] := Module[{m = n / Times@@({2,3}^IntegerExponent[n,{2,3}])}, If[m == 1, 0, Max[FactorInteger[m][[;; , 2]]]]]; Array[a, 100]

PARI

a(n) = {my(m = n >> valuation(n, 2)/3^valuation(n, 3)); if(m == 1, 0,vecmax(factor(m)[,2]));}

Formula

a(n) = A051903(A065330(n)).

a(n) = 0 if and only if n is a 3-smooth number (A003586).

a(n) = 1 if and only if n is a product of a squarefree 5-rough number larger than 1 and a 3-smooth number.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} k * d(k) = 1.1034178389191320571029... , where d(k) is the asymptotic density of the occurrences of k in this sequence: d(1) = 3/(2*zeta(2)), and d(k) = (1/zeta(k+1)) / ((1-1/2^(k+1))*(1-1/3^(k+1))) - (1/zeta(k)) / ((1-1/2^k)*(1-1/3^k)) for k >= 2.

In general, the asymptotic mean of the maximum exponent in the prime factorization of the largest p-rough divisor of n is Sum_{k>=1} k * d(k), where d(1) = 1/(zeta(2) * f(p, 2)), d(k) = 1/(zeta(k+1) * f(p, k+1)) - 1/(zeta(k) * f(p, k)) for k >= 2, and f(p, m) = Product_{q prime < p} (1-1/q^m).