A365871 Starts of runs of 3 consecutive integers whose exponent of least prime factor in their prime factorization is even.

475, 1519, 2223, 2275, 3283, 4475, 4923, 4975, 5823, 6723, 6811, 7299, 7675, 8107, 8379, 8523, 8955, 9475, 10323, 10467, 11275, 12427, 12463, 12591, 13075, 13867, 13923, 14355, 15631, 15723, 16675, 18027, 18275, 18475, 18767, 19323, 19375, 19647, 22075, 22831

Offset: 1

Author

Amiram Eldar, Sep 21 2023

Keywords

• nonn
• easy

Comments

Numbers k such that k, k+1 and k+2 are all terms of A365869.

Numbers of the form 4*k+2 are not terms of A365869. Therefore, there are no runs of 4 or more consecutive integers, and all the terms of this sequence are of the form 4*k+3.

The numbers of terms not exceeding 10^k, for k = 3, 4, ..., are 1, 18, 195, 1952, 19542, 195514, 1955859, 19560453, 195611458, ... . Apparently, the asymptotic density of this sequence exists and equals 0.001956... .

Examples

			475 is a term since the exponent of the prime factor 5 in the factorization 475 = 5^2 * 19 is 2, which is even, the exponent of the prime factor 2 in the factorization 476 = 2^2 * 7 * 17 is 2, which is even, and the exponent of the prime factor 3 in the factorization 477 = 3^2 * 53 is also 2, which is even.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A067029.

Subsequence of A004767, A365869 and A365870.

Programs

Mathematica

Select[4 * Range[6000] + 3, AllTrue[# + {0, 1, 2}, EvenQ[FactorInteger[#1][[1, -1]]] &] &]

PARI

is(n) = !(factor(n)[1,2]%2);
lista(kmax) = forstep(k = 3, kmax, 4, if(is(k) && is(k+1) && is(k+2), print1(k, ", ")));