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#155 | |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
Quote:
that: Code:
for(y=1,#mersenne,if(isprime(sum(X=1,#mersenne-(#mersenne-y),sumdigits(mersenne[X]^2))),print1(y","))) Code:
for(y=1,#mersenne,if(isprime(sum(X=1,#mersenne-(#mersenne-y),sumdigits(mersenne[X]))),print1(y","))) |
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#156 |
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Aug 2006
10111010110112 Posts |
How often do you think this would happen by chance?
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#157 |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
Something tells me you know? I could give a guess with 1 or 2 double checks I know I can calculate it approximately in the odd placed Mersenne numbers which encompass most of the Mersenne primes. 6*(A002450)+1 = odd Mersenne numbers last I checked. using this fact and :
Code:
(09:04)>for(x=1,120,print1(sumdigits(6*x+1)",")) 7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1, Code:
0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, if we plot the sumdigits based on modulo 3 we get: Code:
(09:19)>c=vector(100);c[1]==0;for(i=2,#c,c[i]=c[i-1]*4+1) (09:21)>c %447 = [0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, (09:21)>c=c%3 %448 = [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, Last fiddled with by science_man_88 on 2011-01-03 at 13:25 |
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#158 | |
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Aug 2006
3·1,993 Posts |
I'm sure I could come up with something. But really, until you have an idea of how often something would happen by chance you shouldn't say things like:
Quote:
It looks like you're examining the sum of the Mersenne exponents mod 9 (using 9 rather than 0 when 9 | n). Other than the 2nd member, these are {1, 2, 4, 5, 7, 8} and heuristically these would each occur with equal probability, for an average value of 4.5. The expected number of primes is then something like [code]> sum(n=1,40,1/log(4.5*n)) %1 = 10.03235117441056250421202288[/url] which is a bit higher (!) than observed but well within expected variance. So I would conclude that the number of primes you see is not unusual. |
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#159 |
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"Forget I exist"
Jul 2009
Dumbassville
26×131 Posts |
I've tried something weird instead of looking for a Vn(P,Q) to fit the exponents of mersenne primes I went another way and said lets replace the 0 and 1 of Un(P,Q) with 2 and 3 and continue the best I've done so far is:
Code:
(10:43)>v=vector(30);v[1]=2;v[2]=3;for(i=3,#v,v[i]=(-1)*v[i-1]-(-4)*v[i-2]);v; (10:43)>v %841 = [2, 3, 5, 7, 13, 15, 37, 23, 125, -33, 533, -665, 2797, -5457, 16645, - |
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#160 |
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"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts |
I think I have an idea on how to extend it the problem is to figure out what it's helpfulness could be. Likely 0.
Last fiddled with by science_man_88 on 2011-01-08 at 18:48 |
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#161 |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
there's a method involved here anyone want to guess what it is ? No it's not just completely unrelated random guesses.
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#162 |
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"Forget I exist"
Jul 2009
Dumbassville
203008 Posts |
obviously sick of guessing:
this sequence I'm trying for is the Mersenne exponents in Lucas sequences as defined at http://en.wikipedia.org/wiki/Lucas_sequence to get it exact we have to stray from the definition. But in my guessing and thoughts on it I came across the fact that, if you have say 2 and 3 as the start numbers(I did) once you find one way to get a next value you like there are ways to keep it while changing the next value after it. for example to get the [2,3,5] I originally found P=3 Q=2 and starting with [2,3] you can get 5 on the end with those values, to keep it all you have to do it change P by 2 in this case and move Q by 3 you get P=1 Q=-1 if you move down then you get P=-1 Q=-4, which is the one I have now it still keeps the 5 then you alter it's path to change it to what you like. the hard part is I haven't found a way to stay in integer P and Q to get the Mersenne exponents. However if we break away that part and create a new set like Lucas sequences we can likely get it with decimal P and Q. all we should have to do then is find a way to predict a range of P and Q values on a line that I've found to go with my values, and we can then calculate limits to what the next in the sequence must be, if in fact it is a sequence. |
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#163 |
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May 2004
New York City
102138 Posts |
There's a thread (over two years old) in Math or Puzzles here in which I
presented a conjecture of mine relating Fib and Lucas sequences. I'll look for it later if you can't find it. It needed a big numeric test that no one pursued here. You might be interested. |
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#164 | |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
Quote:
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#165 |
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May 2004
New York City
5×7×112 Posts |
Yes indeed you found it !
The relationship to primes might be mentioned in an earlier thread in which I presented this conjecture out of context. |
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