%I #34 Dec 10 2020 04:41:51

%S 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,

%T 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,

%U 54,12,40,6,4,18,10,24,4,30,48,2,12,32,2,6,12,54,2

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

%C 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).

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

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

%C It is not known if these two sequences are infinite.

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

%H MacTutor History of Mathematics, <a href="https://mathshistory.st-andrews.ac.uk/Biographies/Erdos/">Paul Erdős</a>.

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

%F a(n) = A052126(A006093(n)). - _Michel Marcus_, Dec 07 2020

%e Prime(6) = 13 and a(6) = 12/3 = 4 = 2^2.

%e Prime(11) = 31 and a(11) = 30/5 = 6 = 2*3.

%e Prime(20) = 71 and a(20) = 70/7 =10 = 2*5.

%e Prime(36) = 151 and a(36) = 150/5 = 30 = 2*3*5.

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

%o (PARI) gpf(n) = vecmax(factor(n)[, 1]); \\ A006530

%o a(n) = my(x=prime(n)-1); x/gpf(x); \\ _Michel Marcus_, Dec 07 2020

%Y 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).

%K nonn

%O 2,2

%A _Bernard Schott_, Dec 06 2020