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

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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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3 answers to this question

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Posted · Report post

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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Posted · Report post

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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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

it is suggested that the next number is greater than or equal to 298, which confirms the superlinear trend which is apparent from the other 11 members anyway.

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