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


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#1 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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