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

2,2

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.

References

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

Links

Table of n, a(n) for n=2..76.

MacTutor History of Mathematics, Paul Erdős.

Formula

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

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

Example

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.

Mathematica

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

Prog

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

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).

Sequence in context: A244458 A286613 A229061 * A278224 A161421 A306337

Adjacent sequences: A339461 A339462 A339463 * A339465 A339466 A339467

Keyword

nonn

Author

Bernard Schott, Dec 06 2020

Status

approved