A045576 Numbers k that divide 3^k + 2^k.

1, 5, 25, 55, 125, 275, 605, 625, 1375, 3025, 3125, 6655, 6875, 15125, 15625, 30025, 31375, 33275, 34375, 73205, 75625, 78125, 150125, 156875, 166375, 171875, 330275, 345125, 366025, 378125, 390625, 439835, 750625, 784375, 805255, 831875

Offset: 1

Author

David W. Wilson

Keywords

• nonn

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)

Crossrefs

Cf. A003598, A007689.

Programs

Maple

select(t -> 3 &^ t + 2 &^ t mod t = 0, [seq(i,i=1..10^6,2)]); # Robert Israel, Jun 29 2017

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