If on odd factor F divides N, then 10^(N/F) + 1 is a divisor of 10^N + 1. 3 divides 6, so 10^2+1 = 101 divides 10^6+1 = 1000001. Corollary: If 10^N+1 is prime, N has no odd factors >1. Corollary: If 10^N+1 is prime, N is a power of 2. Heuristic: The chance that N is prime is about 1/logN. Caveat: This needs to be adjusted for N being non-random (e.g. odd). Consequence: The expected number of primes in any sequence that grows much faster than exponentially is finite. Empirically, no formula for N has been found that raises the probability much above a constant K times 1/logN. So there might be an infinity of Mersenne primes 2^N-1, but we probably know all the Fermat primes 2^2^N+1. There may be an infinity of repunit primes (111...111), but we probably know all the 10^N+1 primes: 2, 11, and 101. Rich ----- Quoting Bill Gosper <billgosper@gmail.com>:
On 2017-09-25 15:36, Dan Asimov wrote:
Decided to list these just for the heck of it. When's the next prime after 101? Mathematica says its ? 10^2^20+1. --rwg
Some amusing almost-patterns for 10^0 + 1 through 10^19 + 1:
1 = 1
11 = 11
101 = 101
1001 = 7 * 11 * 13
10001 = 73 * 137
100001 = 11 * 9091
1000001 = 101 * 9901
10000001 = 11 * 909091
100000001 = 17 * 5882353
1000000001 = 7 * 11 * 13 * 19 * 52579
10000000001 = 101 * 3541 * 27961
100000000001 = 11 * 11 * 23 * 4093 * 8779
1000000000001 = 73 * 137 * 99990001
10000000000001 = 11 * 859 * 1058313049
100000000000001 = 29 * 101 * 281 * 121499449
1000000000000001 = 7 * 11 * 13 * 211 * 241 * 2161 * 9091
10000000000000001 = 353 * 449 * 641 * 1409 * 69857
100000000000000001 = 11 * 103 * 4013 * 21993833369
1000000000000000001 = 101*9901*999999000001
10000000000000000001 = 11*909090909090909091
10^20 + 1 is too large for my software.
?Dan
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