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April 16th, 2011, 11:01 AM  #1 
Newbie Joined: Oct 2009 Posts: 14 Thanks: 0  Number of terms in a sequence of primes
Dear all, I'm trying to figure out how many terms are in this sequence: , where is an odd prime. I can't see the answer immediately, so I tried some smaller sequences: i) For (ie) : There are terms. ii) For (ie) : There are terms. So for terms, I reasoned that there'd be terms. But the correct answer is . Where have I gotten wrong? This may seem an easy question, but it's from a more complicated question so I thought to post it here. Thank you very much1 
April 16th, 2011, 11:42 AM  #2 
Senior Member Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0  Re: Number of terms in a sequence of primes
this is actually an arithmetic progression question, p being an odd prime doesn't interfere in the answer. dividing the sequence by 2 yields , which has obviously terms, the same number of terms of the original sequence. Where did "k" come from anyways? 
April 17th, 2011, 08:31 AM  #3 
Newbie Joined: Oct 2009 Posts: 14 Thanks: 0  Re: Number of terms in a sequence of primes
Thanks for your response, proglote. So why does the solution say that there are terms? And I used to illustrate my reasoning when we had terms (so that I didn't have to write ). In any case, I asked the original question since I am trying to figure out these two steps for a question on number theory: 
April 17th, 2011, 10:41 AM  #4 
Member Joined: Jul 2010 Posts: 44 Thanks: 0  Re: Number of terms in a sequence of primes
The solution could possibly be taking into account zero as the first term. Otherwise you've got it all right I believe

April 17th, 2011, 11:17 AM  #5 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Number of terms in a sequence of primes 
April 19th, 2011, 01:00 PM  #6  
Newbie Joined: Oct 2009 Posts: 14 Thanks: 0  Re: Number of terms in a sequence of primes Quote:
@mrtamborineman10: Could you please explain what you mean by zero as the first term? We have: , where the RHS starts with . Where would the 0 appear? @Hoempa: We haven't done double factorials so that confuses me a bit. Could you explain without double factorials?  

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