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While this is current, any other mathematicians here want
to comment on whether this sequence is provably infinite? Perhaps by a density heuristic argument? |
[QUOTE=davar55;241537]While this is current, any other mathematicians here want
to comment on whether this sequence is provably infinite? Perhaps by a density heuristic argument?[/QUOTE] It's heuristically infinite with the n-th term about 2.3n * 10^n. I doubt it can be proven infinite with present technology, but a conditional proof on the k-tuple conjecture seems plausible (though it would give bounds wildly out of proportion with the true size of the terms). |
[QUOTE=davar55;241534]Well, so far the sequence of sum(p^p)mod(10^n)
(I think it needs a better name) is: 11 751 1129 361649 361649 12462809 12462809 1273183931 1273183931 Any bets that the tenth and eleventh terms are equal, and whether the twelfth breaks THAT pattern?[/QUOTE] I don't know what you're doing obviously because all I came up with that your description could be is: [CODE]a=0;for(n=1,100,a=a+prime(n)^prime(n);print(a%(10^n))) [/CODE] |
The next step would be to filter out to only values = 0.
But I think you need a double loop. |
[QUOTE=davar55;241584]The next step would be to filter out to only values = 0.
But I think you need a double loop.[/QUOTE] [CODE]a=0;for(m=1,2,for(n=1,1000,a=a+prime(n)^prime(n);if(a%(10^m)==0,print(n":"m)));a=0) [/CODE] like this ? |
[QUOTE=CRGreathouse;241528]Five, actually -- it will be fourth and fifth on your list:
11, 751, 1129, 361649, 361649, 12462809, 12462809, 1273183931, 1273183931. Would someone check me on these? The repeating entries are freaking me out.[/QUOTE] Verified the first 7 terms: 11, 751, 1129, 361649, 361649, 12462809, 12462809. |
with an outer loop I got 11 661 for the start.
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[QUOTE=science_man_88;241605][CODE]a=0;for(m=1,2,for(n=1,1000,a=a+prime(n)^prime(n);if(a%(10^m)==0,print(n":"m)));a=0)
[/CODE] like this ?[/QUOTE] As a rule of thumb, any time you write for(n=1, _, ... prime(n) ...) you should be writing forprime(p=2, prime(_), ... p ...) instead. |
[QUOTE=CRGreathouse;241693]As a rule of thumb, any time you write
for(n=1, _, ... prime(n) ...) you should be writing forprime(p=2, prime(_), ... p ...) instead.[/QUOTE] flaws: 1) isn't always faster 2) results of our codes don't line up so why improve performance of something that gives wrong supposedly inaccurate results 3) I'm already in a bad mood as my sister just called to ask for any mail for here then told me to goto the end of our driveway to check any newer mail and it's almost 9 at night then got mad when i said do I have to, because they technically shouldn't be sending anything here for her or the thing they have planned for my mom.so I'm kinda not in the caring mood. |
[QUOTE=science_man_88;241706]3) I'm already in a bad mood as my sister just called to ask for any mail for here then told me to goto the end of our driveway to check any newer mail and it's almost 9 at night then got mad when i said do I have to, because they technically shouldn't be sending anything here for her or the thing they have planned for my mom.so I'm kinda not in the caring mood.[/QUOTE]
Well, that's what I get for trying to help improve your programming skills. :smile: [QUOTE=science_man_88;241706]1) isn't always faster[/QUOTE] It's pretty much faster in all cases. prime() is very inefficient, it typically just goes through the primes until it hits the right one. for(n=1,10,prime(n)) goes through numbers in this order: 2, 2, 3, 2, 3, 5, 2, 3, 5, 7, 2, 3, 5, 7, 11, 2, 3, 5, 7, 11, 13, 2, 3, 5, 7, 11, 13, 17, 2, 3, 5, 7, 11, 13, 17, 19, 2, 3, 5, 7, 11, 13, 17, 19, 23, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 for a total of 10(10+1)/2 = 55 numbers. forprime(p=2,29,p) goes straight through: 2, 3, 5, 7, 11, 13, 17, 19, 23. forprime(p=2, prime(10), p) goes through the primes once to discover the tenth prime, then again for the actual loop. [QUOTE=science_man_88;241706]2) results of our codes don't line up so why improve performance of something that gives wrong supposedly inaccurate results[/QUOTE] Post an example. The only times that [code]forprime(p=2,prime(floor(N)), f(p))[/code] should differ from [code]for(n=1,N, f(prime(n)))[/code] are[list][*] When n or p are used in f (in which case you might just need to change variables!)[*] When N is not a t_INT, t_REAL, or t_FRAC, in which case both give errors, but the errors are different[*] When you run out of primes (in which case forprime tells you upfront and for() only tells you when you run out)[*] When you look at the time taken (essentially, forprime is always faster)[/list] |
p^2 or p^p?
[QUOTE=davar55;241534]Well, so far the sequence of sum(p^p)mod(10^n)
(I think it needs a better name) is: 11 751 1129 361649 361649 12462809 12462809 1273183931 1273183931 Any bets that the tenth and eleventh terms are equal, and whether the twelfth breaks THAT pattern?[/QUOTE] Isn't this the sum of P^2 for p prime, NOT p^p? This thread has me confused... |
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