A339464 a(n) = (prime(n)-1) / gpf(prime(n)-1) where gpf(m) is the greatest prime factor of m, A006530.

1, 2, 2, 2, 4, 8, 6, 2, 4, 6, 12, 8, 6, 2, 4, 2, 12, 6, 10, 24, 6, 2, 8, 32, 20, 6, 2, 36, 16, 18, 10, 8, 6, 4, 30, 12, 54, 2, 4, 2, 36, 10, 64, 28, 18, 30, 6, 2, 12, 8, 14, 48, 50, 128, 2, 4, 54, 12, 40, 6, 4, 18, 10, 24, 4, 30, 48, 2, 12, 32, 2, 6, 12, 54, 2

Offset: 2

Author

Bernard Schott, Dec 06 2020

Keywords

• nonn

Comments

Paul Erdős asked if there are infinitely many primes p such that (p-1)/A006530(p-1) = 2^k or = 2^q*3^r (see Richard K. Guy reference).

A074781 is the sequence of primes p such that (p-1)/A006530(p-1) = 2^k.

A339465 is the sequence of primes p such that (p-1)/A006530(p-1) = 2^q*3^r with q, r >=1.

It is not known if these two sequences are infinite.

Examples

			Prime(6) = 13 and a(6) = 12/3 = 4 = 2^2.
Prime(11) = 31 and a(11) = 30/5 = 6 = 2*3.
Prime(20) = 71 and a(20) = 70/7 =10 = 2*5.
Prime(36) = 151 and a(36) = 150/5 = 30 = 2*3*5.
		

References

• Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B46, p. 154.

Links

• MacTutor History of Mathematics, Paul Erdős.

Crossrefs

Cf. A006093 (prime(n)-1), A006530, A052126, A074781 (ratio = 2^k), A339465 (ratio = 2^q*3^r), A339466 (ratio <> 2^k and <> 2^q*3^r).

Programs

Mathematica

f[n_] := n/FactorInteger[n][[-1, 1]]; f /@ (Select[Range[3, 400], PrimeQ] - 1) (* Amiram Eldar, Dec 07 2020 *)

PARI

gpf(n) = vecmax(factor(n)[, 1]); \\ A006530
a(n) = my(x=prime(n)-1); x/gpf(x); \\ Michel Marcus, Dec 07 2020

Formula

a(n) = A006093(n)/A006530(A006093(n)).

a(n) = A052126(A006093(n)). - Michel Marcus, Dec 07 2020