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The Engineer
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Number Riddles

04/11/2010 9:11 PM

Ok, you all know the drill, here's the series:

2, 4, 9, 24, 75, 258, 931,...

what comes next in the series?

I'm counting on you guys triing to unseat that Number Riddles Champion known as Mikerho.

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#1

Re: Number Riddles

04/11/2010 10:24 PM

Champion? Don't make me laugh!

I will attempt the solution, but only as a secondary interest to DVD episodes of "24"!

In all seriousness though, thank you Roger for providing "food for the mind"

(Hopefully) More to come.

Mike

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#2

Re: Number Riddles

04/11/2010 10:41 PM

Frustration comes next. milo

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#3

Re: Number Riddles

04/13/2010 12:18 AM

Next in the serie: 2956 (+/- 35) based on an educated guess with quick polynomial (6th power) estimation as performed with Excel.

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The Engineer
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#6
In reply to #3

Re: Number Riddles

04/13/2010 7:23 AM

Incorrect.

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#4

Re: Number Riddles

04/13/2010 3:19 AM

3047 ?

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#7
In reply to #4

Re: Number Riddles

04/13/2010 7:25 AM

I'm afraid not, keep at it, you'll be glad you tried.

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#5

Re: Number Riddles

04/13/2010 6:51 AM

I know I'm close, but I've become impatient. Here's what I've found so far

each number in the series can be factored. Doing so shows that the 1st number is divisible by one, the second by 2, the third by 3, the fourth by 4, ..., the 7th by 7.

The remaining factor is as follows:

1st) (1)*2

2nd) (2)*2

3rd) (3)*3

4th) (4)*6

5th) (5)*15

6th) (6)*43

7th) (7)*133

So we have series of factors remaining after the value is divided by the value of its place in the series. The new series is as follows:

2,2,3,6,15,43,133

A pattern appears to develop, but goes to junk at the 6th and 7th positions. The pattern that I found was that

2nd')1st'*1, 2=2*1

3rd')2nd'*1.5, 3=2*1.5

4th')3rd'*2, 6=3*2

5th')4th'*2.5, 15=6*2.5

6th')pattern breaks

So there was a linearly increasing factor: 1, 1.5, 2, 2.5 that related each value to the previous up until the 6th... it all goes out the window at the 6th and on to the 7th... sigh

maybe this will help someone else figure it out.

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#10
In reply to #5

Re: Number Riddles

04/13/2010 9:46 AM

Nice effort Russsss. If you get a chance, check out this link on Pascal's Triangle, which is where the Number Riddle came from.

http://ptri1.tripod.com/

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#8

Re: Number Riddles

04/13/2010 8:35 AM

Is this correct 2,4,9,24,75,258,931,3440,12879,48630…

I can't take credit for this if it's correct or incorrect. Sent it to a fellow I know who is good with numbers.

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The Engineer
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#9
In reply to #8

Re: Number Riddles

04/13/2010 9:36 AM

Yes! That is correct, well done. The series came from Pascal's Triangle, the series is the series of largest coefficients for even powered binomial expansions (or the middle value of the even rows of Pascals Triangle). Now had I just given that as the series you all could have googled it, so I added a twist. For the nth term in the series I added n to the corresponding value. So....

1+1=2, 2+2=4, 6+3=9, 20+4=24, 70+5=75, 252+6=258, 924+7=938, 3432+8=3440=

2, 4, 9, 24, 75, 258, 938, 3440,....

To steal the explanation for Pascal's Triangle from Wikipedia:

Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion

(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.

Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:

(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + ... + an−1xyn−1 + anyn,

where the coefficients ai in this expansion are precisely the numbers on row n of Pascal's triangle. In other words, This is the binomial theorem.

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Anonymous Poster (1); Bayes (4); DonC (1); Mikerho (1); Milo (1); Randall (1); Russsss (1)

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