cp's OEIS Frontend

A308707 a(n) = gcd(n, phi(n) + sigma(n)), where phi is A000010 and sigma is A000203.

%I A308707 #51 Sep 08 2022 08:46:21
%S A308707 1,2,3,1,5,2,7,1,1,2,11,4,13,2,1,1,17,9,19,10,1,2,23,4,1,2,1,4,29,10,
%T A308707 31,1,1,2,1,1,37,2,1,2,41,6,43,4,3,2,47,4,1,1,1,2,53,6,1,8,1,2,59,4,
%U A308707 61,2,7,1,1,2,67,2,1,14,71,3,73,2,1,4,1,6,79,2,1,2,83,4,1,2,1,44,89,6
%N A308707 a(n) = gcd(n, phi(n) + sigma(n)), where phi is A000010 and sigma is A000203.
%C A308707 If 2p = phi(p) + sigma(p), where p is A000040, then:
%C A308707 (i) primes m such that a(m-1) is equal to 1: 2, 5, 17, 37, 101, 257, 401, 577, 677, 1297, 1601, 2917, 3137, 4357, 5477, 8101, ...
%C A308707 Conjecture: ALL m are primes of the form i^2 + 1 (see A002496);
%C A308707 (ii) the smallest prime k such that a(k-1) is equal to n: 2, 3, 73, 13, 1464101, 43, 197, 113, 19, 31, 156817, 397, 9096257, 71, 405001, 387, ...
%C A308707 (iii) primes r such that a(r-1) is equal to r-1: 2, 3, 313, 23761, 3343777, 12558913, 45326161, 1178491681, ...
%C A308707 From _Bernard Schott_, Jun 23 2019: (Start)
%C A308707 There are distinct families of integers that satisfy a(k) = 1:
%C A308707 (i) k = p^q with p prime and q >= 2: A001597,
%C A308707 (ii) k = p*q with p, q primes and 2 < p < q: A046388,
%C A308707 (iii) k = 2*p^2 with p prime <> 3: A079704 \ {18},
%C A308707 (iv) conjecture: k = m^2 with m >= 1: A000290 \ {0}; if m is prime, it's not a conjecture, see (i). This conjecture is stronger than the conjecture of the 1st comment. (End)
%F A308707 a(n) = gcd(n, A065387(n)). - _Michel Marcus_, Jun 19 2019
%F A308707 a(n) = n if n = 1 or n is prime: A008578.
%F A308707 a(2*p) = 2 if p prime >= 3: A100484 \ {4}. - _Bernard Schott_, Jun 26 2019
%o A308707 (Magma) [Gcd(n, EulerPhi(n)+SumOfDivisors(n)): n in [1..100]];
%o A308707 (PARI) a(n) = gcd(n, eulerphi(n) + sigma(n)); \\ _Michel Marcus_, Jun 19 2019
%Y A308707 Cf, A000010, A000040, A000203, A000290, A001597, A002496, A008578, A050873, A046388, A065387, A308470.
%K A308707 nonn
%O A308707 1,2
%A A308707 _Juri-Stepan Gerasimov_, Jun 18 2019