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# Hard number sequence: 4, 8, 16, 26, 41, 57, 79, 104, 138, 184, 241, ?

3 replies to this topic

### #1 gmgm

gmgm

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Posted 21 June 2014 - 02:07 AM

Hello! I have been struggling to solve this number sequence puzzle for a while now and I have been unsuccessful so far. Let's see how you do! If you find the answer I would very much like to know how you managed to solve it!

Here is the sequence:
4, 8, 16, 26, 41, 57, 79, 104, 138, 184, 241, ?

Have fun!

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### #2 m00li

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Posted 22 June 2014 - 09:36 AM

Hello! I have been struggling to solve this number sequence puzzle for a while now and I have been unsuccessful so far. Let's see how you do! If you find the answer I would very much like to know how you managed to solve it!

Here is the sequence:
4, 8, 16, 26, 41, 57, 79, 104, 138, 184, 241, ?

Have fun!

Let S1=(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

S2=(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

S3=(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

so on till

S11=(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)

where n is any natural number

then the sequence is given by

(4/3628800)S1 -(8/362880)S2 +(16/80640)S3 -(26/30240)S4 +(41/17280)S5 -(57/14400)S6 +(79/17280)S7 -(104/30240)S8 +(138/80640)S9 -(184/362880)S10 +(241/3628800)S11

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### #3 Perhaps check it again

Perhaps check it again

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Posted 22 June 2014 - 06:35 PM

gmgm,

m00li's approach is a *possible* one, but there are an infinite number of possibilities for the twelfth term.

m00li assumed a polynomial formula.

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### #4 gmgm

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Posted 25 August 2014 - 08:51 PM

I see that many people have tried to solve this unsuccessfully. I thought it would be good to post it again, this time in the original form in which it appears in the source, just in case there is some significant information there.

4, 8, 16, 26, 41,
57, 79, 104,
138, 184,
241,
?

a = 298
z = 323
A = 324
Z = 349

The only significant thing I see there is that
Spoiler for

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