20090705, 19:56  #1 
Dec 2008
you know...around...
2×5^{2}×13 Posts 
Byproducts and other curiosities
While "Prime Curios" gets flooded with really trivial numbers, whereas my last few submissions won't appear there, I need a space for my own calculations that might be worth a mention (or discussion) but don't deserve a separate thread each.
For general entertainment:  206780313999369083332356327764879 is the biggest prime p such that round(p/q#*r#) is always prime for 2<=r<=q, with q=83.  There are 14 primes between 7,362,853,038 and 7,362,854,037, but 76 between 14,982,264,191 and 14,982,265,190  The first byte without primes is 6D 45 6D xx  see attachment (I suppose you get the intention) Open questions:  A 10x10square where each row, each column and each of the diagonals contain no number twice  96% complete is the best I could do. 100% is impossible, but what about 97% or 98%? Code:
1234567890 8796301425 2348 56917 0915478236 7569230148 412068 573 9051743682 6873912054 34 28 5761 5687124309 To be continued. Later. 
20090706, 16:03  #2 
Dec 2008
you know...around...
2·5^{2}·13 Posts 
Because it's topical: today is the 40,000th day in MS Excel!
Also, in connection to the first open question in the above post, this 12x12square, in which each row, column and diagonal no number is seen twice, is 97,2% complete: Code:
01 02 03 04 05 06 07 08 09 10 11 12 05 12 01 02 09 11 10 04 07 03 08 06 11 10 06 07 01 02 03 09 08 04 12 05 08 09 03 06 07 11 01 02 05 10 04 04 03 10 11 08 05 06 07 01 02 09 02 08 04 05 10 09 11 06 07 03 01 09 05 11 12 04 01 02 03 10 08 06 07 07 06 08 09 10 03 04 05 12 01 02 10 01 02 06 07 09 08 12 04 11 05 03 06 07 05 10 11 12 01 02 03 09 04 08 12 11 09 08 03 04 05 06 01 02 07 10 03 04 07 01 02 08 12 10 05 06 09 11 The only soluble n x n squares that I know of are the ones where n>1 is odd and not a multiple of 3. For example n=5: 12345 34512 51234 23451 45123 
20090710, 09:56  #3 
Dec 2008
you know...around...
2×5^{2}×13 Posts 
 The 40,933th interval of one million (i.e. 40,932,000,001  40,933,000,000) contains 40,933 primes!
 13543584514*p#1 is prime for p=2...40  Has anyone investigated the idea of "semigeneralized fermat primes": for n odd, (n^(2^m)+1)/2 ? 
20090717, 20:02  #4 
Dec 2008
you know...around...
1212_{8} Posts 
Turn a date plus time into a number like this: YYYYMMDDHHMMSS
then there are "superquadruplets": four primes in an interval of seven seconds  and these are the next three: 20100807 18:59:57 / 20100807 18:59:59 / 20100807 19:00:01 / 20100807 19:00:03 20110521 00:59:57 / 20110521 00:59:59 / 20110521 01:00:01 / 20110521 01:00:03 20120109 15:59:57 / 20120109 15:59:59 / 20120109 16:00:01 / 20120109 16:00:03 (and the previous one was 20090101 02:59:57 / 20090101 02:59:59 / 20090101 03:00:01 / 20090101 03:00:03) The next one in European(?) date format would be 08.02.2010 14:59:57 / 08.02.2010 14:59:59 / 08.02.2010 15:00:01 / 08.02.2010 15:00:03 And here are two "GFP8Quintuplets": five consecutive even numbers n where n^8+1 is prime (with six consecutive even numbers, at least one would be divisible by 17): 2866672^8+1 27932594^8+1 2866674^8+1 27932596^8+1 2866676^8+1 27932598^8+1 2866678^8+1 27932600^8+1 2866680^8+1 27932602^8+1 For GFP4Octuplets, look for "10332305196" in the OEIS. Last fiddled with by mart_r on 20090717 at 20:14 
20090725, 12:47  #5 
Dec 2008
you know...around...
2·5^{2}·13 Posts 
(3^31)/2=13 is prime
(3^131)/2=797161 is prime (3^7971611)/2=143055179...566042801 has no factors < 6.2*10^12 (using only BASIC and long integers <=2^311) The aliquot sequence for 22179589964132019724649435809540255084187350060120875 takes 101 steps to reach 1. Can someone find a longer (odd) sequence? 
20090725, 17:35  #6  
Nov 2008
100100010010_{2} Posts 
Quote:


20090727, 16:01  #8  
"William"
May 2003
New Haven
2361_{10} Posts 
Quote:
N1 has lots of cyclotomic factors. If N had been PRP it would have fun to attempt an N1 proof, although it would take substantial luck to find enough factors. No factors through 4.5e+13 

20090727, 18:32  #10  
Aug 2006
3^{2}·5·7·19 Posts 
Quote:
= 3^797161  3 = 3(3^797160  1) = 3(3^199290  1)(3^199290 + 1)(3^398580 + 1) = 3(3^99645  1)(3^99645 + 1)(3^199290 + 1)(3^398580 + 1) . . . I made a reasonable amount of progress on the 3^99645  1 factor with cyclotomics. Last fiddled with by CRGreathouse on 20090727 at 19:20 

20090728, 00:33  #11 
Aug 2006
3^{2}·5·7·19 Posts 
I've been working on this a bit more on the factor database. I've split it as
(3^797161  1)/2  1 = 3(3^99645  1)(3^99645 + 1)(3^199290 + 1)(3^398580 + 1)/2 since the db seems not to like numbers with hundreds of factors. 
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