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#1
03-08-2010, 11:30 PM
 s3b4k Junior Member Join Date: Feb 2010 Posts: 38
discrete math- sequence question

I need to come up with a rule or a formula for the following:

1,2,2,2,3,3,3,3,3,5,5,5,5,5,5,5.....
#2
03-09-2010, 12:10 AM
 someonesdad Senior Member Join Date: Jul 2009 Location: Northwest USA Posts: 1,584

Hint: 2*n + 1 and primes
#3
03-09-2010, 12:22 AM
 jpanhalt E-book Developer Join Date: Jan 2008 Location: Ohio, USA(GMT-5) Posts: 3,669

There may be more than one pattern that gives that initial sequence. One solution is: the digits 1,2,3,and 5 are primes (positive). The repeat frequencies seem to follow just the odd numbers. The next three repeat frequencies would be 9, 11 and 13. That is, the next two sequences would be nine 7's, then eleven 11's.

John
 The Following User Says Thank You to jpanhalt For This Useful Post: s3b4k (03-11-2010)
#4
03-09-2010, 12:45 AM
 s3b4k Junior Member Join Date: Feb 2010 Posts: 38

ok thanks a lot for the help, but is there an exact formula i can use, or is it only solvable by explaining it?
#5
03-11-2010, 12:34 PM
 Markd77 Senior Member Join Date: Sep 2009 Location: Birmingham Posts: 2,784 Blog Entries: 1

1 isn't normally considered a prime if that makes any difference.
#6
03-11-2010, 12:48 PM
 jpanhalt E-book Developer Join Date: Jan 2008 Location: Ohio, USA(GMT-5) Posts: 3,669

Sorry. Problem easily solved in the description.

John
#7
03-11-2010, 03:09 PM
 Markd77 Senior Member Join Date: Sep 2009 Location: Birmingham Posts: 2,784 Blog Entries: 1

I searched for the sequence on google and found this. Still no idea what it is.
Quote:
 For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. (a) 3, 6, 11, 18, 27, 38, 51, 66, 83, (b) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8, 8, 8, 8, 8,
#8
03-11-2010, 03:59 PM
 jpanhalt E-book Developer Join Date: Jan 2008 Location: Ohio, USA(GMT-5) Posts: 3,669

For (b) the sequence starts with 1, and if there are at least 2 previous term, then the subsequent term is the sum of the previous two terms, otherwise it is 2 (or alternatively, the previous term is doubled). The replicates of each term is as in the first problem.

I am too tired to look at (a) now. Perhaps this question should have its own thread.

John

Edit: For (a) here's a reference: http://mathworld.wolfram.com/Near-SquarePrime.html

Last edited by jpanhalt; 03-11-2010 at 05:24 PM. Reason: added reference
#9
03-11-2010, 11:19 PM
 s3b4k Junior Member Join Date: Feb 2010 Posts: 38

this question is driving me nuts, are the next three terms going to be 8,8,8 or 7,7,7. because it good either be prime numbers or just the sum of the previous numbers, but if its the sum of the previous numbers how do you get the two there
#10
03-11-2010, 11:32 PM
 jpanhalt E-book Developer Join Date: Jan 2008 Location: Ohio, USA(GMT-5) Posts: 3,669

Quote:
 Originally Posted by s3b4k this question is driving me nuts, are the next three terms going to be 8,8,8 or 7,7,7. because it good either be prime numbers or just the sum of the previous numbers, but if its the sum of the previous numbers how do you get the two there
There is not enough information to know for sure. The sum of the previous numbers (Fibonacci, 8) may be what your teacher wants. In either case, you have to make an explanation for the first number (or absence thereof).

John

PS, when I went to school 1 was still prime. But then, it was a public school.
 The Following User Says Thank You to jpanhalt For This Useful Post: s3b4k (03-11-2010)

 Tags discrete, math, question, sequence

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