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 March 5th, 2011, 09:11 AM #1 Member   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, 09:36 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Stellar Numbers It seems your numbers follow the pattern: $12T_{n-1}+1$ where $T_n=\frac{n}{2}(n+1)$ is the nth triangular number, giving: $12$$\frac{n-1}{2}\(n$$\)+1=6n(n-1)+1=6n^2-6n+1$
 March 5th, 2011, 09:45 AM #3 Member   Joined: Jun 2010 Posts: 38 Thanks: 0 Re: Stellar Numbers Can I ask why did you use the triangular numbers formula?
 March 5th, 2011, 09:51 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 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, 09:54 AM   #5
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Re: Stellar Numbers

Quote:
 Originally Posted by MarkFL 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.
Wow you are a math god

 March 5th, 2011, 10:35 AM #6 Member   Joined: Jun 2010 Posts: 38 Thanks: 0 Re: Stellar Numbers One more question is that why did you use (n-1) What was the reasoning behind it?
 March 5th, 2011, 10:46 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 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: $S_n=12T_n+1$ $S_n=12$$\frac{n}{2}(n+1)$$+1=6n(n+1)+1=6n^2+6n+1$ It just depends on how you define it. We could also define the stellar numbers recursively: $S_n=S_{n-1}+12n$ where $S_0=1$
 March 5th, 2011, 12:55 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 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) $S_{n+1}=S_n+12(n+1)$ (2) $S_n=S_{n-1}+12n$ Subtract (2) from (1) to obtain: $S_{n+1}-S_n=S_n-S_{n-1}+12(n+1)-12n$ (3) $S_{n+1}=2S_n-S_{n-1}+12$ (4) $S_n=2S_{n-1}-S_{n-2}+12$ Subtract (4) from (3) to obtain: $S_{n+1}=3S_n-3S_{n-1}+S_{n-2}$ which we may write as: (5) $S_n=3S_{n-1}-3S_{n-2}+S_{n-3}$ Now, we assume this has a solution of the form $S_n=r^n$ and substituting this into (5) we obtain: $r^n=3r^{n-1}-3r^{n-2}+r^{n-3}$ Dividing through by $r^{n-3}$ we obtain the characteristic equation: $r^3-3r^2+3r-1=$$r-1$$^3=0$ so we have the general solution: $S_n=k_1+k_2n+k_3n^2$ Now we use initial conditions to find the constants: $S_0=1=k_1$ $S_1=13=k_1+k_2+k_3$ $S_2=37=k_1+2k_2+4k_3$ From the first equation we have $k_1=1$ so the second equation becomes: $12=k_2+k_3\:\therefore\:k_2=12-k_3$ which we may substitute into the third equation: $37=1+2$$12-k_3$$+4k_3$ $12=2k_3\:\therefore\:k_3=6,k_2=6$ thus we have: $S_n=1+6n+6n^2$
 April 15th, 2011, 07:38 AM #9 Newbie   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 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 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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