OFFSET
1,2
COMMENTS
For any j>=0, 5*A003598(j) is a term of the sequence. - Benoit Cloitre, Mar 08 2002
From Robert Israel, Jun 29 2017: (Start)
This is a semigroup: if m and n are in the sequence, then so is m*n.
If n is in the sequence and is divisible by prime p, then so is p*n.
The only prime powers in the sequence are the powers of 5.
Conjecture: Every member of the sequence except 1 is of the form p*m where p is prime and m is in the sequence. (End)
There are infinitely many primes p that divide some term in the sequence. Proof: Define the set A as all primes p such that a k where p divides 2^(5^k) + 3^(5^k) exists is finite. Since 2^(5^(k+1)) + 3^(5^(k+1)) is 2^(5^k) + 3^(5^k) multiplied by some positive integer a. It can be verified that gcd(a, 2^(5^k) + 3^(5^k)) is 5, so 2^(5^(k+1)) + 3^(5^(k+1)) has a larger number of different prime factors than 2^(5^k) + 3^(5^k). Therefore, A is infinite. For each prime q in A, suppose that q divides 2^(5^x) + 3^(5^x) for some x, then 5^x also divides it, so 5^x*q divides it as well, hence 5^x*q is a term of the sequence. The original theorem is proved. - Yifan Xie, Nov 14 2024
LINKS
Robert Israel and Giovanni Resta, Table of n, a(n) for n = 1..1000 (first 205 terms from Robert Israel)
MAPLE
select(t -> 3 &^ t + 2 &^ t mod t = 0, [seq(i, i=1..10^6, 2)]); # Robert Israel, Jun 29 2017
PROG
(PARI) isok(n) = ((3^n+2^n) % n) == 0; \\ Michel Marcus, Jun 29 2017
(PARI) isok(n)=(Mod(2, n)^n+Mod(3, n)^n)==0; \\ significantly more efficient
for(n=1, 10^6, if(isok(n), print1(n, ", "))); \\ Joerg Arndt, Aug 13 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved