My Math Forum Integer fun...

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 May 4th, 2012, 05:40 PM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Integer fun... p = product of 2 consecutive integers n-1 and n. s = sum of m consecutive integers, the first being n+1. s = p Example ( n = 12, m = 8 ): p = 11 * 12 = 132 s = 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 132 If m = 4,541,160 then what's n ?
 May 4th, 2012, 07:42 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Integer fun... With a bit of algebra, we find: $n(m)=\frac{(m+1)+\sqrt{(m+1)(3m+1)}}{2}$ and so: $n(4541160)=6203341$
 May 4th, 2012, 08:16 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Integer fun... Yepper!!
 May 4th, 2012, 10:23 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Integer fun... I was interested in which values of m lead to an integral value of n, so I wrote a short program on my calculator, and found the sequence: $0,\,8,\,120,\,1680,\,23408,\,326040,\,4541160,\,.. .$ With some good old trial-and-error, I found this sequence can be generated with the recursion: $M_{n+3}=15M_{n+2}-15M_{n+1}+M_{n}$ where: $M_0=0,\,M_1=8,M_2=120$ and hence, the closed form is: $M_n=\frac{1}{6}$$\(2+\sqrt{3}$$$$7-4\sqrt{3}$$^n+$$2-\sqrt{3}$$$$7+4\sqrt{3}$$^n-4\)$
 May 5th, 2012, 05:36 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Integer fun... Nice! The negative solution also valid; right? If m = 4,541,160 then what's n ? n = 6,203,341 or -1,662,180 Using my m = 8 example: n = 12 or n = -3 p = -4 * -3 = 12 s = -2 + -1 + 0 + 1 + 2 + 3 + 5 + 5 = 12
 May 5th, 2012, 07:42 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Integer fun... I considered the negative solution, but decided to discard it, but I see no problem with it.
 May 5th, 2012, 08:48 AM #7 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Integer fun... Ah, this is my kind of math! Thanks for a nice thread!
May 5th, 2012, 10:34 AM   #8
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Re: Integer fun...

Quote:
 Originally Posted by MarkFL I was interested in which values of m lead to an integral value of n, so I wrote a short program on my calculator, and found the sequence: $0,\,8,\,120,\,1680,\,23408,\,326040,\,4541160,\,.. .$ With some good old trial-and-error, I found this sequence can be generated with the recursion: $M_{n+3}=15M_{n+2}-15M_{n+1}+M_{n}$
Interesting ... $M_n+1$ and $3M_n+1$ are both perfect squares.

 May 5th, 2012, 11:04 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Integer fun... 97512 [97513 + 97514 +...+ 235415] 235416 97512 * 235416 = 97513 + 97514 +...+ 235415 = 22,955,884,992 Did that in my head watching the New York : Washington hockey game
May 6th, 2012, 12:08 AM   #10
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Re: Integer fun...

Hello MarkFL:

I think it should be:

[attachment=0:19nksj9y]n(m).jpg[/attachment:19nksj9y]

How do you think ?
Attached Images
 n(m).jpg (6.1 KB, 166 views)

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