A339466 Primes p such that gpf((p - 1)/gpf(p - 1)) > 3, where gpf(m) is the greatest prime factor of m, A006530.

71, 101, 131, 151, 191, 197, 211, 239, 251, 281, 311, 331, 401, 419, 421, 431, 443, 461, 463, 491, 521, 547, 571, 599, 601, 617, 631, 647, 659, 661, 677, 683, 691, 701, 727, 743, 751, 761, 821, 827, 859, 881, 883, 911, 941, 947, 953, 967, 971, 991, 1013, 1021

Offset: 1

Author

Bernard Schott, Dec 10 2020

Keywords

• nonn

Comments

Paul Erdős asked if there are infinitely many primes p such that (p-1)/gpf(p-1) = 2^k or = 2^q * 3^r (see Richard K. Guy reference). This sequence lists the primes p that do not satisfy these two previous relations.

Replacing in the definition gpf by lpf (A020639) leads to A122259. In fact this sequence is a subsequence of A122259. - Peter Luschny, Dec 13 2020

Examples

			71 is prime, 70/7 = 10 = 2*5 hence 71 is a term.
101 is prime, 100/5 = 20 = 2^2*5 hence 101 is a term.
151 is prime, 150/5 = 30 = 2*3*5 hence 151 is a term.
The first few quotients obtained are: 10, 20, 10, 30, 10, 28, 30, 14, 50, 40, ...
		

References

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

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• P. Erdős and C. Pomerance, On the largest prime factors of n and n+1, Aequationes Math. 17 (1978), p. 311-321. [alternate link]

Crossrefs

Cf. A006093, A006530, A052126, A339464.

Cf. A074781 (ratio=2^k), A339465 (ratio=2^q*3^r), A339463 (ratio=2^q*5^r).

Cf. A122259.

Programs

Magma

s:=func; [p:p in PrimesInInterval(3,1100)|( not 3 in PrimeDivisors(a) and #PrimeDivisors(a) ge 2) or ( 3 in PrimeDivisors(a) and #PrimeDivisors(a) ge 3) where a is (p-1) div s(p-1)]; // Marius A. Burtea, Dec 10 2020

Maple

alias(pf = NumberTheory:-PrimeFactors): gpf := n -> max(pf(n)):
is_a := n -> isprime(n) and gpf((n-1)/gpf(n-1)) > 3:
select(is_a, [$5..1021]); # Peter Luschny, Dec 13 2020

Mathematica

q[n_] := Module[{f = FactorInteger[n - 1]}, (Length[f] == 1 && f[[1, 1]] == 2) || (Length[f] == 2 && f[[1, 1]] == 2 && f[[2, 2]] == 1) || (Length[f] == 2 && f[[2, 1]] == 3 && f[[2, 2]] > 1) || (Length[f] == 3 && f[[2, 1]] == 3 && f[[3, 2]] == 1)]; Select[Range[3, 1000], PrimeQ[#] && ! q[#] &] (* Amiram Eldar, Dec 10 2020 *)

PARI

is(n) = {if(!isprime(n) || n==2, return(0)); my(pm1 = n-1, f = factor(pm1)[,1]); (pm1 /= (f[#f]*1<1} \\ David A. Corneth, Dec 13 2020

Extensions

More terms from Amiram Eldar, Dec 11 2020