OFFSET
1,1
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
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]
EXAMPLE
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, ...
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 *)
PROG
(Magma) s:=func<n|Max(PrimeDivisors(n))>; [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
(PARI) is(n) = {if(!isprime(n) || n==2, return(0)); my(pm1 = n-1, f = factor(pm1)[, 1]); (pm1 /= (f[#f]*1<<valuation(pm1, 2)*3^valuation(pm1, 3)))>1} \\ David A. Corneth, Dec 13 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Dec 10 2020
EXTENSIONS
More terms from Amiram Eldar, Dec 11 2020
STATUS
approved