A367266 Numbers k such that there exists i >= 1 such that k divides 3^3^i + 1.

1, 2, 4, 7, 14, 19, 28, 37, 38, 74, 76, 133, 148, 163, 259, 266, 326, 487, 518, 532, 652, 703, 974, 1036, 1141, 1297, 1406, 1459, 1948, 2282, 2594, 2812, 2917, 2918, 3097, 3409, 4564, 4921, 5188, 5834, 5836, 6031, 6194, 6818, 9079, 9253, 9842, 10213, 11668, 12062, 12388, 13636, 18019, 18158, 18506, 19441, 19684, 19927

Offset: 1

Author

Jianing Song, Nov 11 2023

Keywords

• nonn

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.

Examples

			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.
		

Crossrefs

Cf. A094358 (squarefree divisors of 2^2^i - 1), A357265 (divisors of 3^3^i - 1), A014127.

The subsequence of primes is given by A367649.

Programs

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)