A365868 Starts of runs of 3 consecutive integers that are divisible by the cube of their least prime factor.

1375, 16119, 25623, 28375, 52623, 55375, 57967, 79623, 82375, 90207, 94471, 106623, 109375, 129623, 133623, 134431, 136375, 160623, 163375, 164295, 187623, 190375, 206143, 214623, 217375, 241623, 244375, 268623, 271375, 280231, 295623, 298375, 312471, 322623, 325375

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 A365866.

Numbers of the form 4*k+2 are not terms of A365866. Therefore, there are no runs of 4 or more consecutive integers. Since the middle integer in each triple is divisible by 8, all the terms of this sequence are of the form 8*k+7.

The numbers of terms not exceeding 10^k, for k = 4, 5, ..., are , 1, 11, 109, 1092, 10899, 109125, 1091335, 10912897, ... . Apparently, the asymptotic density of this sequence exists and equals 0.0001091... .

Examples

			1375 is a term since 5 is the least prime factor of 1375 and 1375 is divisible by 5^3 = 125, 2 is the least prime factor of 1376 and 1376 is divisible by 2^3 = 8, and 3 is the least prime factor of 1377 and 1377 is divisible by 3^3 = 27.
		

Links

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

Crossrefs

Cf. A067029.

Subsequence of A004771, A068140, A365866 and A365867.

Programs

Mathematica

Select[8 * Range[41000] + 7, AllTrue[# + {0, 1, 2}, FactorInteger[#1][[1, -1]] >= 3 &] &]

PARI

is(n) = factor(n)[1,2] >= 3;
lista(kmax) = forstep(k = 7, kmax, 8, if(is(k) && is(k+1) && is(k+2), print1(k, ", ")));