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A308707
a(n) = gcd(n, phi(n) + sigma(n)), where phi is A000010 and sigma is A000203.
0
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, 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, 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
OFFSET
1,2
COMMENTS
If 2p = phi(p) + sigma(p), where p is A000040, then:
(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, ...
Conjecture: ALL m are primes of the form i^2 + 1 (see A002496);
(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, ...
(iii) primes r such that a(r-1) is equal to r-1: 2, 3, 313, 23761, 3343777, 12558913, 45326161, 1178491681, ...
From Bernard Schott, Jun 23 2019: (Start)
There are distinct families of integers that satisfy a(k) = 1:
(i) k = p^q with p prime and q >= 2: A001597,
(ii) k = p*q with p, q primes and 2 < p < q: A046388,
(iii) k = 2*p^2 with p prime <> 3: A079704 \ {18},
(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)
FORMULA
a(n) = gcd(n, A065387(n)). - Michel Marcus, Jun 19 2019
a(n) = n if n = 1 or n is prime: A008578.
a(2*p) = 2 if p prime >= 3: A100484 \ {4}. - Bernard Schott, Jun 26 2019
PROG
(Magma) [Gcd(n, EulerPhi(n)+SumOfDivisors(n)): n in [1..100]];
(PARI) a(n) = gcd(n, eulerphi(n) + sigma(n)); \\ Michel Marcus, Jun 19 2019
KEYWORD
nonn
AUTHOR
STATUS
approved