cp's OEIS Frontend

A289259 Numbers k such that k^2 divides 2^k + 3^k.

%I A289259 #52 Jun 02 2024 05:28:10
%S A289259 1,5,55,1971145,3061355,109715901845,340799222665
%N A289259 Numbers k such that k^2 divides 2^k + 3^k.
%C A289259 If k is in the sequence and p is a prime factor, coprime to k, of 2^k + 3^k, then k*p is in the sequence.
%C A289259 55 = 5 * 11
%C A289259 1971145 = 5 * 11 * 35839
%C A289259 3061355 = 5 * 11 * 55661
%C A289259 109715901845 = 5 * 11 * 35839 * 55661
%C A289259 340799222665 = 5 * 11 * 55661 * 111323
%C A289259 See Known Terms link for additional terms.
%C A289259 From _Felix Fröhlich_, Jun 29 2017: (Start)
%C A289259 For k in the sequence, A220235(k) = 0.
%C A289259 Subsequence of A045576. (End)
%H A289259 Robert Israel and Ray Chandler, <a href="/A289259/a289259_4.txt">Known Terms</a>
%H A289259 A. Velampalli et al., Mathematics StackExchange, <a href="https://math.stackexchange.com/questions/2340794/">Can you prove or disprove that there exist infinitely many integers n such that n^2 divides 2^n+3^n?</a>
%e A289259 2^5 + 3^5 = 275 is divisible by 5^2, so 5 is in the sequence.
%p A289259 select(t -> 2&^t + 3&^t mod t^2 = 0, [$1..10^6]);
%o A289259 (PARI) is(n) = Mod(2, n^2)^n==-3^n \\ _Felix Fröhlich_, Jun 29 2017
%o A289259 (PARI) is(n) = Mod(2,n^2)^n+Mod(3,n^2)^n==0 \\ _Charles R Greathouse IV_, Jun 29 2017
%Y A289259 Cf. A007689, A045576, A220235.
%K A289259 nonn,more
%O A289259 1,2
%A A289259 _Robert Israel_, Jun 29 2017
%E A289259 a(6)-a(7) confirmed as next terms by _Ray Chandler_, Jul 02 2017
%E A289259 Known terms updated and moved to a-file by _Ray Chandler_, Jul 03 2017