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Old 2009-07-05, 19:56   #1
mart_r
 
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Default 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.
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Old 2009-07-06, 16:03   #2
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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
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Old 2009-07-10, 09:56   #3
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- 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 ?
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Old 2009-07-17, 20:02   #4
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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
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Old 2009-07-25, 12:47   #5
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(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?
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Old 2009-07-25, 17:35   #6
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Quote:
Originally Posted by mart_r View Post
(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.
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Old 2009-07-26, 11:06   #7
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Quote:
Originally Posted by 10metreh View Post
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.
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Old 2009-07-27, 16:01   #8
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Quote:
Originally Posted by mart_r View Post
(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
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Old 2009-07-27, 16:16   #9
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here are all factors < 1e7 for ((3^797161-1)/2)-1.
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Old 2009-07-27, 18:32   #10
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Quote:
Originally Posted by kar_bon View Post
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
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Old 2009-07-28, 00:33   #11
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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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