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

### #1

Posted 21 June 2014 - 02:07 AM

### #2

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 S_{1}=(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

S_{2}=(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

S_{3}=(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11)

so on till

S_{11=}(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

_{1}-(8/362880)S

_{2}+(16/80640)S

_{3}-(26/30240)S

_{4}+(41/17280)S

_{5}-(57/14400)S

_{6}+(79/17280)S

_{7}-(104/30240)S

_{8}+(138/80640)S

_{9}-(184/362880)S

_{10}+(241/3628800)S

_{11}

### #3

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.

### #4

Posted 25 August 2014 - 08:51 PM

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

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