OFFSET
1,2
COMMENTS
Note that 3^3^i + 1 divides 3^3^(i+1) + 1, so this sequence is also numbers k such that k divides 3^3^i + 1 for all sufficiently large i.
Also numbers k such that there exists i >= 1 such that k divides 3^^i + 1, where 3^^i = 3^3^...^3 (i times) = A014220(i-1).
Write k = 2^{e_0} * Product_{j=1..r} (p_j)^(e_j), then k is a term if and only if e_0 <= 2, and ord(3,(p_j)^(e_j)) is 2 times a power of 3 for every 1 <= j <= r, where ord(a,k) is the multiplicative order of a modulo k: 3^3^i == -1 (mod k) if and only if 3^3^i == -1 (mod 2^{e_0}), and 3^3^i == -1 (mod (p_j)^(e_j)) for every 1 <= j <= r. This is in turn equivalent to e_0 <= 2, and ord(3,(p_j)^(e_j)) being even, and 3^i == ord(3,(p_j)^(e_j))/2 (mod ord(3,(p_j)^(e_j))) for every 1 <= j <= r. As a result, such i exists if and only if e_0 <= 2, and ord(3,(p_j)^(e_j)) is 2 times a power of 3 for every 1 <= j <= r. In other words, each term is a product of a number in {1,2,4} and odd prime powers q such that ord(3,q) is 2 times a power of 3.
If an term k is not squarefree, then it is divisible by p^2, where p is a Wieferich prime to base 3 (A014127) such that ord(3,p) is 2 times a power of 3. No such p is known.
EXAMPLE
Suppose that q is an odd prime power such that ord(3,q) = 2*3^e. e = 1 gives q = 7; e = 2 gives q = 19; e = 3 gives q = 19441 and q = 19927; e = 4 gives q = 163, 1297, 208657, 224209, 5879415781.
PROG
(PARI) isA357266_primepower(q) = if(q%2==0, q%8!=0, my(d); (q%3!=0) && ((d=znorder(Mod(3, q)))%2==0) && isprimepower(3*d/2))
isA357266(n) = my(f=factor(n)); for(i=1, #f~, if(!isA357266_primepower(f[i, 1]^f[i, 2]), return(0))); return(1)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Nov 11 2023
STATUS
approved