cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A367648 Primes p such that the multiplicative order of 3 modulo p is a power of 3.

Original entry on oeis.org

2, 13, 109, 433, 757, 3889, 8209, 17497, 52489, 58321, 70957, 1190701, 1705861, 2598157, 6627097, 13463173, 57395629, 23245229341, 79320757897, 1069604540569, 1631815099669, 5774114968057, 8635817966221, 23765922477217, 43781455818469, 307283335691329
Offset: 1

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Author

Jianing Song, Nov 25 2023

Keywords

Comments

Prime factors of numbers of the form 3^3^i - 1: p divides 3^3^i - 1 if and only if the multiplicative order of 3 modulo p is a power of 3 not exceeding 3^i.

Examples

			13 is a term since the multiplicative order of 3 modulo 13 is 3 = 3^1, which means that 13 is a factor of 3^3^1 - 1.
109 is a term since the multiplicative order of 3 modulo 109 is 27 = 3^3, which means that 109 is a factor of 3^3^3 - 1.
		

Crossrefs

Subsequence of A367265.
Cf. A023394 (ord(2,p) being a power of 2, prime factors of numbers of the form 2^2^i - 1 (or of the form 2^2^i + 1)), A367649 (ord(3,p) being 2 times a power of 3, prime factors of numbers of the form 3^3^i + 1).

Programs

  • PARI
    isA367648(n) = isprime(n) && (n!=3) && isprimepower(3*znorder(Mod(3,n)))

Extensions

a(18)-a(19) from Michel Marcus, Nov 27 2023
a(20)-a(25) from Max Alekseyev, Jul 22 2024
a(26) from Jinyuan Wang, Jan 29 2025

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

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 11 2023

Keywords

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)
Showing 1-2 of 2 results.