A359747 Numbers k such that k*(k+1) has in its canonical prime factorization mutually distinct exponents.

1, 3, 4, 7, 8, 16, 24, 27, 31, 48, 63, 71, 72, 107, 108, 124, 127, 199, 242, 243, 256, 400, 431, 432, 499, 512, 576, 647, 783, 863, 967, 971, 1024, 1151, 1152, 1372, 1567, 1600, 1999, 2187, 2311, 2400, 2591, 2592, 2887, 2916, 3087, 3136, 3456, 3887, 3888, 3968, 4000

Offset: 1

Author

Amiram Eldar, Jan 13 2023

Keywords

• nonn

Comments

Equivalently, numbers k such that A002378(k) = k*(k+1) is a term of A130091.

Equivalently, numbers k such that the multisets of exponents in the prime factorizations of k and k+1 are disjoint and each have distinct elements.

Either k or k+1 is a powerful number (A001694). Except for k=8, are there terms k such that both k and k+1 are powerful (i.e., terms that are also in A060355)? None of the terms A060355(n) for n = 2..39 is in this sequence.

Examples

			3 is a term since 3*4 = 12 = 2^2 * 3^1 has 2 distinct exponents in its prime factorization: 1 and 3.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..1000

Crossrefs

Subsequence of A130091 and A342028.

A359748 is a subsequence.

Cf. A001694, A002378, A060355.

Programs

Mathematica

q[n_] := UnsameQ @@ (FactorInteger[n*(n+1)][[;; , 2]]); Select[Range[4000], q]

PARI

is(n) = { my(e = factor(n*(n+1))[, 2]); #Set(e) == #e; }