A328016 Numbers k such that k, k+1, ... k+6 are all cubefree (A004709).

1, 9, 17, 33, 41, 57, 65, 73, 89, 97, 113, 137, 145, 153, 169, 177, 193, 201, 209, 217, 225, 233, 257, 273, 281, 289, 305, 313, 329, 353, 361, 385, 393, 409, 417, 425, 433, 441, 449, 465, 473, 489, 505, 521, 529, 545, 553, 569, 577, 585, 601, 609, 633, 641, 649, 657

Offset: 1

Author

Amiram Eldar, Oct 01 2019

Keywords

• nonn
• easy

Comments

There cannot be 8 consecutive cubefree numbers since one of them must be divisible by 8 = 2^3.

All the terms are congruent to 1 mod 8.

The asymptotic density of this sequence is A328017.

Examples

			9 is in the sequence since the numbers 9, 10, ... 15 are all cubefree.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Leon Mirsky, Arithmetical pattern problems relating to divisibility by rth powers, Proceedings of the London Mathematical Society, Vol. s2-50, No. 1 (1949), pp. 497-508.

Crossrefs

Cf. A004709, A007675, A194002, A325058, A328017.

Programs

Mathematica

cubeFreeQ[n_] := FreeQ[FactorInteger[n], {, k /; k > 2}]; aQ[n_] := AllTrue[n + Range[0, 6], cubeFreeQ]; Select[Range[650], aQ]