PARI's commands - Page 69

science_man_88 · 2010-08-23, 13:56

I never said $k*b!^n+1$ is of form 4x+1 you stated $k*b!^n/x^2$ for x>1= integer for some x.

this says that 4x+1 can be a $k*b!^n+1$

and same for the rest so really I was just taking your last statement that the statements above and that statement were equal.

science_man_88 · 2010-08-23, 13:59

http://www.research.att.com/~njas/se...lish&go=Search

was the best i could find with all of some group in one sequence.

science_man_88 · 2010-08-23, 14:03

note also 65 = 2^6+1=2!^6+1 = 2!^3!+1

science_man_88 · 2010-08-23, 14:15

one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?

3.14159 · 2010-08-23, 14:35

Quote:

Originally Posted by science_man_88

one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?

3, 5: Diff = 2
5, 7: Diff = 2
7, 9: Diff = 2
9, 13: Diff = 2^2
13, 17: Diff = 2^2
17, 19: Diff = 2^2.
19, 25: Diff = 6 =/= 2^n.

Strong law of small numbers at work.

science_man_88 · 2010-08-23, 14:36

Quote:

Originally Posted by 3.14159

3, 5: Diff = 2
5, 7: Diff = 2
7, 9: Diff = 2
9, 13: Diff = 2^2
13, 17: Diff = 2^2
17, 19: Diff = 2^2.
19, 25: Diff = 6 =/= 2^n.

Strong law of small numbers at work.

okay sorry I forgot I took out 19 lol but the rest all have first difference of power of 2.

3.14159 · 2010-08-23, 14:37

The sieve project is coming along.. Just that we have made zero progress.

Anyone have any spare ideas on it?

My best idea is to begin with the k-range loop, then a forprime loop which will be linked to a forstep loop, which, hopefully, will eliminate all the unwanted values. Then, it will save the remaining numbers in a file, and then I make the appropriate modifications to make the range testable, and then proceed to test.

science_man_88 · 2010-08-23, 14:43

Code:

for(k=,,forprime(_=,,forstep(-=,,_,write(,))))

something on the lines of this ?

3.14159 · 2010-08-23, 14:44

Or: I trial-divide each candidate to 1048576, and test those which have no small factors.

Quote:

Originally Posted by science_man_88

something on the lines of this ?

Yes. Although I doubt it will work as I want it to.

3.14159 · 2010-08-23, 14:46

The forstep loop will have to contain the liftmod for the primes, to eliminate bad values of k (Those which are divisible by a certain prime p.)

3.14159 · 2010-08-23, 14:50

I can print the bad values that are divisible by a certain prime p, so far. But I can't get a good enough text editor to remove those specific values.

Ex: 757.

Code:

Those would be the bad k-values for k * 11!⁴ + 1, because they form candidates divisible by 757.

Proof: 15006 * 11!⁴ + 1 = 757 * 50325944549556534016411764332893

2010-08-23, 13:56	#749
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	I never said $kb!^n+1$ is of form 4x+1 you stated $kb!^n/x^2$ for x>1= integer for some x. this says that 4x+1 can be a $kb!^n+1$ and same for the rest so really I was just taking your last statement that the statements above and that statement were equal. Last fiddled with by science_man_88 on 2010-08-23 at 14:05*

2010-08-23, 14:37	#755
3.14159 May 2010 Prime hunting commission. 2⁴×3×5×7 Posts	The sieve project is coming along.. Just that we have made zero progress. Anyone have any spare ideas on it? My best idea is to begin with the k-range loop, then a forprime loop which will be linked to a forstep loop, which, hopefully, will eliminate all the unwanted values. Then, it will save the remaining numbers in a file, and then I make the appropriate modifications to make the range testable, and then proceed to test. Last fiddled with by 3.14159 on 2010-08-23 at 14:42

2010-08-23, 14:43	#756
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	Code: for(k=,,forprime(_=,,forstep(-=,,_,write(,)))) something on the lines of this ? Last fiddled with by science_man_88 on 2010-08-23 at 14:44 Reason: forgot to use write function

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2010-08-23, 13:59	#750
science_man_88 "Forget I exist" Jul 2009 Dumbassville 10000011000000₂ Posts	http://www.research.att.com/~njas/se...lish&go=Search was the best i could find with all of some group in one sequence.

2010-08-23, 14:03	#751
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	note also 65 = 2^6+1=2!^6+1 = 2!^3!+1

2010-08-23, 14:15	#752
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?

2010-08-23, 14:46	#758
3.14159 May 2010 Prime hunting commission. 2⁴×3×5×7 Posts	The forstep loop will have to contain the liftmod for the primes, to eliminate bad values of k (Those which are divisible by a certain prime p.)