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

1,2

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.

Links

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

Thomas Bloom, Erdős Problem #913.

P. Erdős, Miscellaneous problems in number theory, Congr. Numer. (1982), 25-45. See p. 28.

Erdős problems database contributors, Erdős problems database, maintained by Thomas Bloom and Terence Tao.

Example

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

Mathematica

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

Prog

(PARI) is(n) = { my(e = factor(n*(n+1))[, 2]); #Set(e) == #e; }
(Python) from sympy import factorint
def is_A359747(n): return n>0 and len(e:=factorint(n*(n+1)).values()) == len(set(e)) # David Radcliffe, Sep 10 2025

Crossrefs

Subsequence of A130091 and A342028.

A359748 is a subsequence.

A245639 is a subsequence.

Cf. A001694, A002378, A060355.

Sequence in context: A219019 A306612 A117587 * A244930 A130420 A214096

Adjacent sequences: A359744 A359745 A359746 * A359748 A359749 A359750

Keyword

nonn

Author

Amiram Eldar, Jan 13 2023

Status

approved