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

%I A339464 #34 Dec 10 2020 04:41:51
%S A339464 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 A339464 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 A339464 54,12,40,6,4,18,10,24,4,30,48,2,12,32,2,6,12,54,2
%N A339464 a(n) = (prime(n)-1) / gpf(prime(n)-1) where gpf(m) is the greatest prime factor of m, A006530.
%C A339464 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 A339464 A074781 is the sequence of primes p such that (p-1)/A006530(p-1) = 2^k.
%C A339464 A339465 is the sequence of primes p such that (p-1)/A006530(p-1) = 2^q*3^r with q, r >=1.
%C A339464 It is not known if these two sequences are infinite.
%D A339464 Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B46, p. 154.
%H A339464 MacTutor History of Mathematics, <a href="https://mathshistory.st-andrews.ac.uk/Biographies/Erdos/">Paul Erdős</a>.
%F A339464 a(n) = A006093(n)/A006530(A006093(n)).
%F A339464 a(n) = A052126(A006093(n)). - _Michel Marcus_, Dec 07 2020
%e A339464 Prime(6) = 13 and a(6) = 12/3 = 4 = 2^2.
%e A339464 Prime(11) = 31 and a(11) = 30/5 = 6 = 2*3.
%e A339464 Prime(20) = 71 and a(20) = 70/7 =10 = 2*5.
%e A339464 Prime(36) = 151 and a(36) = 150/5 = 30 = 2*3*5.
%t A339464 f[n_] := n/FactorInteger[n][[-1, 1]]; f /@ (Select[Range[3, 400], PrimeQ] - 1) (* _Amiram Eldar_, Dec 07 2020 *)
%o A339464 (PARI) gpf(n) = vecmax(factor(n)[, 1]); \\ A006530
%o A339464 a(n) = my(x=prime(n)-1); x/gpf(x); \\ _Michel Marcus_, Dec 07 2020
%Y A339464 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 A339464 nonn
%O A339464 2,2
%A A339464 _Bernard Schott_, Dec 06 2020