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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: Exciting Problem''
by lynx on 2008-07-09 11:55:01 |
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| I computed by hand (and a calculator) in the following way:
Take the number of positive integers of the form 3k+2 less or equal than X=6008819575 which is (X-1)/3 (i.e. [(X-2)/3]+1, where [y] is the integer part of y),
subtract by the number of positive integers of the form p(3k+e_p) (\leq X) with p be prime factor of X (i.e. p=5,11,17,23,29,41 or 47) and e_p=1 (if p=2mod3) or e_p=2 (if p=1mod3) (this means that p(3k+e_p)=2mod3). Fortunally e_p=1 for all p, thus the number of pos. int. of the form p(3k+1) less equal than X is
\sum_{p=5,11,17,...,47}((X/p)+1)/3
(note that ((X/p)+1)/3=[((X/p)-1)/3]+1)
then we sum the number of positive integers of the form pq(3k+2) (\leq X) with p and q be two distinct prime factors of X which is equal to
\sum_{pq=5*11,5*17,...,41*47}((X/pq)-1)/3
Then we subtract by
\sum_{pqr=5*11*17,...,29*41*47}((X/pqr)+1)/3
and go on, at the end we subtract by
((X/5*11*...*47)+1)/3
so the final result is
(1/3)*(X(1-1/5)(1-1/11)(1-1/23)...(1-1/47) -2^7)=
(1/3)*(5*4*10*22*...*46 - 128)=
1209002624
I made the mistake of considering [(X-2)/3]+1 equal to (X+2)/3 instead (X-1)/3.
I recognize this explanation is a mess but I hope you are able to understand it after think on it. |
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