| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
In Crux Mathematicorum 9(1983)45, problem 813, Charles W. Trigg notes that the following numbers are all prime: 31 331 3331 33331 333331 3333331 and asks how long this pattern continues. He also wants to know the first prime number of this form that occurs after a composite number. I was able to come up with the following results: Let 3(k)1 denote the number consisting of k 3's followed by a 1. Then 3(7)1 is prime. 3(8)1 = 17 X 19607843 3(9)1 = 673 X 4952947 3(10)1 = 307 X 108577633 3(11)1 = 19 X 83 X 211371803 3(12)1 = 523 X 3049 X 2090353 3(13)1 = 607 X 1511 X 1997 X 18199 3(14)1 = 181 X (1841620626151) 3(15)1 = 199 X (16750418760469) 3(16)1 = 31 X (1075268817204301) All the factors shown are prime except for the ones in parentheses which are not known (by me) to be prime. I was unable to determine if 3(17)1 is prime or composite. Anyone care to search any further?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 21.1 | AURORA::HALLYB | Wed Feb 01 1984 00:42 | 10 | ||
It isn't too hard to determine that p = 3(17)1 is a "probable prime" by checking out A^(p-1) (mod p) for various bases A. I've done this for some fun A (=47, =74353, =132049) and found the result = 1 in all cases. So we go on the assumption that 3(17)1 is probably a prime. Stan, I assume you got this far rather easily. Does this mean our factoring capabilities so limited that we can't check out an 18-digit number? Or did you just try a simple direct search? John | |||||
| 21.2 | HARE::STAN | Thu May 31 1984 23:36 | 6 | ||
Hallyburton's factoring program (see note 73) was unable to factor 3(17)1 either. It came up with more factors to 3(16)1 than I had. 3(16)1=31 * 1499 * (717324094199). It also found 3(18)1 = 1009 * 1303427 * (2534550017). | |||||
| 21.3 | AURORA::HALLYB | Sun Jun 03 1984 22:39 | 21 | ||
N = 3(17)1 is prime. Someday I'll automate this; here's how it goes: n0: n1: N-1 = 3(17)0 = 2 * 3 * 5 * 2071723 * 5363222357 n0 - 1 = 2 * 3 * 17 * 19 * 1069 all factors prime n1 - 1 = 2 * 2 * 17 * 389 * 202753 all factors prime [1] For every d | (n0-1), d > 1, it turns out that 2^((n0-1)/d) > 1 (mod n0) and 2^(n0-1) = 1 (mod n0). Hence n0 is prime. [2] For every d | (n1-1), d > 1, it turns out that 2^((n1-1)/d) > 1 (mod n1) and 2^(n1-1) = 1 (mod n1). Hence n1 is prime. [1] and [2] above allow us to proceed and say: For every d | (N-1), d > 1, it turns out that 2^((N-1)/d) > 1 (mod N) and 2^(N-1) = 1 (mod N). Hence N is prime. Reference: Knuth Vol II, pp. 347-350. | |||||
| 21.4 | More 3's | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Fri Nov 15 1991 15:21 | 13 |
>It also found 3(18)1 = 1009 * 1303427 * (2534550017). 3(19)1 = 29 * 29 * 1039 * 3389 * 11256299321 3(20)1 = 23 * 164844923 * 87917500639 3(21)1 = 177943 * 18732590398798117 3(22)1 = 61 * 179 * 241049 * 12664572810301 3(23)1 = 312929 * 2228959 * 477893202221 3(24)1 = 17*821593951*238656128290493 | |||||