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

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Profile JeppeSNProject donor
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Message 132682 - Posted: 8 Sep 2019 | 8:47:22 UTC
Last modified: 8 Sep 2019 | 8:54:33 UTC

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

Profile JeppeSNProject donor
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Joined: 5 Apr 14
Posts: 973
ID: 306875
Credit: 11,517,616
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321 LLR Silver: Earned 100,000 credits (360,928)Cullen LLR Bronze: Earned 10,000 credits (98,851)ESP LLR Silver: Earned 100,000 credits (139,922)Generalized Cullen/Woodall LLR Bronze: Earned 10,000 credits (35,236)PPS LLR Ruby: Earned 2,000,000 credits (2,486,479)PSP LLR Silver: Earned 100,000 credits (212,242)SoB LLR Silver: Earned 100,000 credits (237,390)SR5 LLR Bronze: Earned 10,000 credits (16,010)SGS LLR Bronze: Earned 10,000 credits (32,929)TRP LLR Bronze: Earned 10,000 credits (71,060)Woodall LLR Silver: Earned 100,000 credits (109,455)321 Sieve Silver: Earned 100,000 credits (101,851)PSA Turquoise: Earned 5,000,000 credits (7,614,290)
Message 132684 - Posted: 8 Sep 2019 | 8:51:41 UTC
Last modified: 8 Sep 2019 | 8:56:29 UTC

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

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

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