OFFSET
1,1
COMMENTS
The sequence is inspired by problem 3, Balkan Mathematical Olympiad 27 April - 2 May 2024, Varna, Bulgaria, (see link).
The sequence is infinite because numbers of the form m = 3^(4*k + 1) + 2 are divisible by 5 = 3^1 + 2.
Numbers 3^k + 2 such that for some m < k, k == m (mod A298827(m)). - Robert Israel, Sep 15 2025
LINKS
Robert Israel, Table of n, a(n) for n = 1..933
Balkan Mathematical Olympiad, 2024, Problems
FORMULA
a(n) = 3^A388145(n) + 2. - Robert Israel, Sep 15 2025
EXAMPLE
245 = 3^5 + 2 and 245 = 49*5 = 49 * (3^1 + 2), so 245 is a term.
2189 = 3^7 + 2 and 2189 = 11*199 = 199 * (3^2 + 2), so 2189 is a term.
129140165 = 3^17 + 2 and 129140165 = 5*25828033 = (3^1 + 2)*25828033 or 129140165 = 11*11740015 = (3^2 + 2)*11740015, so 129140165 is a term.
MAPLE
map(t -> 3^t+2, select(k -> ormap(j -> 3 &^ k + 2 mod (3^j+2) = 0, [$1 .. k-1]), [$1..50])); # Robert Israel, Sep 15 2025
PROG
(Magma) f:=func<n|PrimeDivisors(n) eq [3]>; [n:n in [3^a+2:a in [1..50]]|exists{d: d in Divisors(n)|d ne n and f(d-2) }];
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius A. Burtea, Sep 15 2024
STATUS
approved