A343431 Part of n composed of prime factors of the form 6k-1.

1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 11, 1, 1, 1, 5, 1, 17, 1, 1, 5, 1, 11, 23, 1, 25, 1, 1, 1, 29, 5, 1, 1, 11, 17, 5, 1, 1, 1, 1, 5, 41, 1, 1, 11, 5, 23, 47, 1, 1, 25, 17, 1, 53, 1, 55, 1, 1, 29, 59, 5, 1, 1, 1, 1, 5, 11, 1, 17, 23, 5, 71, 1, 1, 1, 25, 1, 11, 1, 1, 5, 1, 41, 83, 1, 85, 1, 29, 11, 89, 5

Offset: 1

Author

Peter Munn, Apr 15 2021

Keywords

• nonn
• easy
• mult

Comments

Completely multiplicative with a(p) = p if p is of the form 6k-1 and a(p) = 1 otherwise.

Largest term of A259548 that divides n.

Crossrefs

Equivalent sequence for distinct prime factors: A170825.

Equivalent sequences for prime factors of other forms: A000265 (2k+1), A343430 (3k-1), A170818 (4k+1), A097706 (4k-1), A248909 (6k+1), A065330 (6k+/-1), A065331 (<= 3), A355582 (<= 5).

Range of terms: A259548.

Programs

Mathematica

f[p_, e_] := If[Mod[p, 6] == 5, p^e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* after Amiram Eldar at A248909 *)

PARI

a(n) = {my(f = factor(n)); for (i=1, #f~, if ((f[i, 1] + 1) % 6, f[i, 1] = 1); ); factorback(f); } \\ after Michel Marcus at A248909

Python

from math import prod
from sympy import factorint
def A343431(n): return prod(p**e for p, e in factorint(n).items() if not (p+1)%6) # Chai Wah Wu, Dec 26 2022

Formula

a(n) = n / A065331(n) / A248909(n) = A065330(n) / A248909(n).