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 2009-07-05, 19:56 #1 mart_r     Dec 2008 you know...around... 2×52×13 Posts By-products 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 10x10-square 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 - I don't know why, but it appears to be that prime quadruplets (p+0,2,6,8) can only be simultaneously full-period-primes in base 19 and higher To be continued. Later. Attached Thumbnails
 2009-07-06, 16:03 #2 mart_r     Dec 2008 you know...around... 2·52·13 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 12x12-square, 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 And I don't know whether it's possible to get a fully filled one of this size. 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
 2009-07-10, 09:56 #3 mart_r     Dec 2008 you know...around... 2×52×13 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 "semi-generalized fermat primes": for n odd, (n^(2^m)+1)/2 ?
 2009-07-17, 20:02 #4 mart_r     Dec 2008 you know...around... 12128 Posts Turn a date plus time into a number like this: YYYYMMDDHHMMSS then there are "super-quadruplets": four primes in an interval of seven seconds - and these are the next three: 2010-08-07 18:59:57 / 2010-08-07 18:59:59 / 2010-08-07 19:00:01 / 2010-08-07 19:00:03 2011-05-21 00:59:57 / 2011-05-21 00:59:59 / 2011-05-21 01:00:01 / 2011-05-21 01:00:03 2012-01-09 15:59:57 / 2012-01-09 15:59:59 / 2012-01-09 16:00:01 / 2012-01-09 16:00:03 (and the previous one was 2009-01-01 02:59:57 / 2009-01-01 02:59:59 / 2009-01-01 03:00:01 / 2009-01-01 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 "GFP8-Quintuplets": 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 GFP4-Octuplets, look for "10332305196" in the OEIS. Last fiddled with by mart_r on 2009-07-17 at 20:14
 2009-07-25, 12:47 #5 mart_r     Dec 2008 you know...around... 2·52·13 Posts (3^3-1)/2=13 is prime (3^13-1)/2=797161 is prime (3^797161-1)/2=143055179...566042801 has no factors < 6.2*10^12 (using only BASIC and long integers <=2^31-1) The aliquot sequence for 22179589964132019724649435809540255084187350060120875 takes 101 steps to reach 1. Can someone find a longer (odd) sequence?
2009-07-25, 17:35   #6
10metreh

Nov 2008

1001000100102 Posts

Quote:
 Originally Posted by mart_r (3^3-1)/2=13 is prime (3^13-1)/2=797161 is prime (3^797161-1)/2=143055179...566042801 has no factors < 6.2*10^12 (using only BASIC and long integers <=2^31-1) The aliquot sequence for 22179589964132019724649435809540255084187350060120875 takes 101 steps to reach 1. Can someone find a longer (odd) sequence?
Have a look in the Aliquot Sequences forum, there's a thread devoted to odd sequences.

2009-07-26, 11:06   #7
mart_r

Dec 2008
you know...around...

2·52·13 Posts

Quote:
 Originally Posted by 10metreh Have a look in the Aliquot Sequences forum, there's a thread devoted to odd sequences.
Oh, okay. Thanks.
I have to check the aliquot section more often.

2009-07-27, 16:01   #8
wblipp

"William"
May 2003
New Haven

236110 Posts

Quote:
 Originally Posted by mart_r (3^3-1)/2=13 is prime (3^13-1)/2=797161 is prime (3^797161-1)/2=143055179...566042801 has no factors < 6.2*10^12 (using only BASIC and long integers <=2^31-1)
Unfortunately, PFGW show (3^797161-1)/2 is composite (be sure to use the b flag to change the default PRP base from 3).

N-1 has lots of cyclotomic factors. If N had been PRP it would have fun to attempt an N-1 proof, although it would take substantial luck to find enough factors.

No factors through 4.5e+13

 2009-07-27, 16:16 #9 kar_bon     Mar 2006 Germany 288910 Posts here are all factors < 1e7 for ((3^797161-1)/2)-1.
2009-07-27, 18:32   #10
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by kar_bon here are all factors < 1e7 for ((3^797161-1)/2)-1.
2 * (3^797161-1)/2 - 1
= 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 2009-07-27 at 19:20

 2009-07-28, 00:33 #11 CRGreathouse     Aug 2006 32·5·7·19 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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