What is the next no. , cant solve this one

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- Mar 13th 2016, 08:24 AM #1

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- Mar 13th 2016, 09:32 AM #2

- Mar 13th 2016, 05:09 PM #3

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## Re: 1,3,7,25,103,?

321 is the next number

Here's what I did.

Plotted the points (1, 1), (2, 3), (3, 7), (4, 25) and (5, 103).

Turns out that a quartic function fits this data perfectly (R^2=1)

Then two options from there.

Method 1: Look at a table of differences so that the 4th row is constant (34) and complete the table backwards to get 321

Method 2: Sub x=6 into the quartic function found (by regression on a GC) and get 321.

The quartic equation is (17/12)* x^4 - (73/6)* x^3 + (463/12)* x^2 - (299/6) * x + 23

- Mar 13th 2016, 06:22 PM #4
## Re: 1,3,7,25,103,?

The problem with this is that we can use any polynomial fit larger than degree 4 and still get an answer...321 is the only possible fit for a quartic, but you can do a quintic, etc. fit as well. Unless there is some kind of pattern that can be spotted the problem is basically impossible. Plato gave a link that has a huge list of known series but it obviously doesn't cover all of them.

-Dan

Addendum: As another example I can easily come up with a next number: We have 1, 3, 7, 25, 103. 103 is the first term, followed by a 0, followed by the second term. So the next term could easily be the second term followed by a 0 followed by third term: 307.

- Mar 13th 2016, 06:45 PM #5

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- Mar 13th 2016, 08:00 PM #6

- Mar 13th 2016, 09:37 PM #7

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- Mar 13th 2016, 10:32 PM #8

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- Mar 13th 2016, 11:43 PM #9

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- Mar 14th 2016, 04:30 AM #10
## Re: 1,3,7,25,103,?

- Mar 14th 2016, 05:26 AM #11

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- Aug 28th 2016, 12:11 AM #12

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- Aug 28th 2016, 05:52 AM #13

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- Aug 29th 2016, 05:23 AM #14

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- Aug 31st 2016, 10:48 AM #15

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