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Fermat Divisor Search :
Currently known Fermat divisors, sorted by k
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In Fermat Divisor Search we focus on special k that are either small (generally gives high chance of dividing a Fermat number) or has special properties that makes in attractive (see Ravi Fernando's post).
Here is a table giving, for each of the k we consider, the n values that lead to a Fermat divisor. It is mostly just copy/paste from a post I did in the main "Fermat Divisor Search" thread.
Each row is of the form:
k: (all known n that make k*2^n+1 a Fermat divisor)

3: 41, 209, 157169, 213321, 303093, 382449, 2145353, 2478785
5: 7, 25, 39, 75, 127, 1947, 3313, 23473, 125413
7: 14, 120, 290, 320, 95330, 2167800
9: 67, 9431, 461081, 2543551
11: 18759, 960901
13: 20, 114296
15: 229
17: 147, 747, 6539
19: 6838, 9450, 23290
21: 41, 276, 94801
23:
25: 2141884
27: 455, 672007
29: 57, 231, 2027, 4727
31:
33: 18766
35:
37: 16
39: 13, 113549
41:
43:
45:
47:
49:

1323:
2187:
3125: 149
3267:
3375:
19683:
/JeppeSN  


If you take the first n value of each k row, and skip no odd k, you get A215540 (41, 7, 14, 67, 18759, 20, …). If we could find a Fermat divisor for k=23, we would close a "hole" in that sequence and be able to extend it.
The first odd k that does not have three terms in its row yet, is k=11. To submit a row to OEIS, you need at least three terms, generally.
/JeppeSN  


Updated with Scott Brown's find (underlined below):
k: (all known n that make k*2^n+1 a Fermat divisor)

3: 41, 209, 157169, 213321, 303093, 382449, 2145353, 2478785
5: 7, 25, 39, 75, 127, 1947, 3313, 23473, 125413
7: 14, 120, 290, 320, 95330, 2167800
9: 67, 9431, 461081, 2543551
11: 18759, 960901
13: 20, 114296, 5523860
15: 229
17: 147, 747, 6539
19: 6838, 9450, 23290
21: 41, 276, 94801
23:
25: 2141884
27: 455, 672007
29: 57, 231, 2027, 4727
31:
33: 18766
35:
37: 16
39: 13, 113549
41:
43:
45:
47:
49:
/JeppeSN
 


Update: Ryan Propper found a huge prime (almost 5.5 million digits), and Serge Batalov found that it divides a Fermat; 7*2^18233956 + 1 divides F(18233954). I have no information about whether there is a "hole" in the k=7 sequence. The new term is underlined below:
k: (all known n that make k*2^n+1 a Fermat divisor)

3: 41, 209, 157169, 213321, 303093, 382449, 2145353, 2478785
5: 7, 25, 39, 75, 127, 1947, 3313, 23473, 125413
7: 14, 120, 290, 320, 95330, 2167800, [...?], 18233956
9: 67, 9431, 461081, 2543551
11: 18759, 960901
13: 20, 114296, 5523860
15: 229
17: 147, 747, 6539
19: 6838, 9450, 23290
21: 41, 276, 94801
23:
25: 2141884
27: 455, 672007
29: 57, 231, 2027, 4727
31:
33: 18766
35:
37: 16
39: 13, 113549
41:
43:
45:
47:
49:
/JeppeSN
 


Update: tng found one and got an F badge: 27*2^7963247 + 1 divides F(7963245). The new term is also underlined below.
Congratulations, tng! To appear on https://www.primegrid.com/primes/primes.php?project=ALL&factors=F&only=ONLY
k: (all known n that make k*2^n+1 a Fermat divisor)

3: 41, 209, 157169, 213321, 303093, 382449, 2145353, 2478785
5: 7, 25, 39, 75, 127, 1947, 3313, 23473, 125413
7: 14, 120, 290, 320, 95330, 2167800, [...?], 18233956
9: 67, 9431, 461081, 2543551
11: 18759, 960901
13: 20, 114296, 5523860
15: 229
17: 147, 747, 6539
19: 6838, 9450, 23290
21: 41, 276, 94801
23:
25: 2141884
27: 455, 672007, 7963247
29: 57, 231, 2027, 4727
31:
33: 18766
35:
37: 16
39: 13, 113549
41:
43:
45:
47:
49:
/JeppeSN
 

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Message boards :
Fermat Divisor Search :
Currently known Fermat divisors, sorted by k 