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March 5th, 2011, 10:11 AM  #1 
Joined: Jun 2010 Posts: 38 Thanks: 0  Stellar Numbers
The first four stages for a star with 6 vertices have 1,13,37,73 number of dots. I have found a pattern that if I subtract one from each term it's a multiple of 12. So I need to figure out an equation for it. Thanks 
March 5th, 2011, 10:36 AM  #2 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,087 Thanks: 406 Math Focus: The calculus  Re: Stellar Numbers
It seems your numbers follow the pattern: where is the nth triangular number, giving: 
March 5th, 2011, 10:45 AM  #3 
Joined: Jun 2010 Posts: 38 Thanks: 0  Re: Stellar Numbers
Can I ask why did you use the triangular numbers formula?

March 5th, 2011, 10:51 AM  #4 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,087 Thanks: 406 Math Focus: The calculus  Re: Stellar Numbers
You observed that each number was some multiple of 12 plus 1, so I used this to write the numbers as: 12(0) + 1, 12(1) + 1, 12(3) + 1, 12(6) + 1 and I recognized the sequence 0, 1, 3, 6 as being the triangular numbers. 
March 5th, 2011, 10:54 AM  #5  
Joined: Jun 2010 Posts: 38 Thanks: 0  Re: Stellar Numbers Quote:
 
March 5th, 2011, 11:35 AM  #6 
Joined: Jun 2010 Posts: 38 Thanks: 0  Re: Stellar Numbers
One more question is that why did you use (n1) What was the reasoning behind it? 
March 5th, 2011, 11:46 AM  #7 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,087 Thanks: 406 Math Focus: The calculus  Re: Stellar Numbers
Because the first stellar number was a function of the zeroth triangular number, and the second stellar number was a function of the first triangular number, etc. We could say 1 is the zeroth stellar number and 13 is the first, etc. Then we would have: It just depends on how you define it. We could also define the stellar numbers recursively: where 
March 5th, 2011, 01:55 PM  #8 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,087 Thanks: 406 Math Focus: The calculus  Re: Stellar Numbers
Armed only with the recursive formula, we could find the closed form as follows: We need to transform the relation into a linear homogeneous relation with constant coefficients. So we may write: (1) (2) Subtract (2) from (1) to obtain: (3) (4) Subtract (4) from (3) to obtain: which we may write as: (5) Now, we assume this has a solution of the form and substituting this into (5) we obtain: Dividing through by we obtain the characteristic equation: so we have the general solution: Now we use initial conditions to find the constants: From the first equation we have so the second equation becomes: which we may substitute into the third equation: thus we have: 
April 15th, 2011, 07:38 AM  #9 
Joined: Apr 2011 Posts: 1 Thanks: 0  Re: Stellar Numbers
To MarkFL: Hi, I saw the work that you did on Stellar Numbers on this forum and I had a question about them and you seem to be very knowledgeable in this subject! As part of a 'math essay', I am required to write an introductory paragraph to Stellar Numbers and I was just wondering if you happened to know what real life applications triangular/stellar numbers have? Itd be awesome if you could somehow help me with that, thanks! =) 
April 15th, 2011, 12:18 PM  #10 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,087 Thanks: 406 Math Focus: The calculus  Re: Stellar Numbers
To be completely honest, I had never heard of stellar numbers before this topic. I just happened to recognize that the stellar numbers appeared to be a function of triangular numbers, which come from the summation of consecutive natural numbers beginning with 1.


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