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Your mission, if you choose to accept it, is three parts: A) Figure out the pattern; B) Give an eleven-number sequence beginning with 2 that follows the same pattern; and C) Give an eleven-letter sequence beginning with α (alpha).

A)  1, 3, 34, 5, 56, 10, 160, 4, 42, 8, 118
B)  2, ?,  ?, ?,  ?,  ?,   ?, ?,  ?, ?,   ?
C)  α, ?,  ?, ?,  ?,  ?,   ?, ?,  ?, ?,   ?

Hint:

For part C, you may need multiple letters in a given spot in the sequence.

Hint 2:

Each number in the sequence is not necessarily directly related to the previous number.

Hint 3:

Think about aspects of each number individually as opposed to aspects of how they fit into the sequence as a whole.

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  • $\begingroup$ Would you like to tell us whether these are length-11 sequences or length-11 initial segments of longer (perhaps infinite) sequences? That is, would it have made sense to ask us to continue the first sequence? $\endgroup$ – Gareth McCaughan Jun 17 '16 at 12:07
  • $\begingroup$ Yes. It is absolutely possible to continue these sequences ad infinitum. Depending on the starting number, it's theoretically possible to continue the sequence backwards, as well, though with 1 and 2 it is impossible (hint hint). $\endgroup$ – user24580 Jun 17 '16 at 17:50
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The sequence A is obtained in the following way:

1. Start with a number (1 in the example). The sequence is now 1.
2. Add to the sequence the number of letters of the number (1=one, so add 3). The sequence is now 1,3.
3. Add the sum of the numerical values of the letters of the same number (o=15, n=14, e=5, so add 34). The sequence is now 1, 3, 34.
4. Apply 2. and 3. again to the first number in the sequence to which they have not yet been applied. Repeat.

Thus the sequence B is

2, 3, 58, 5, 56, 10, 115, 4, 42, 8, 118.

As for C,

I can think of several ways to do it, but I'll try this one (keeping in mind that the Greek alphabet has 24 letters): in step 2., use the number of letters in the English way to write the Greek letter ("alpha" has 5 letters), and convert again to Greek letters (if α is the digit 0, then ζ is 5). Of course, we count in base 24 here!
In step 3., same thing, again in base 24 converted to Greek letters.

This gives

α, ζ, βν, ε, γε, η, γπ, θ, δτ, ν, ζη.

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  • $\begingroup$ Absolutely correct! Well done. :) $\endgroup$ – user24580 Jun 17 '16 at 18:39
1
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(Community evidence locker— feel free to add or correct)

Too good-looking to resist sharing, whether pertinent or not, the factors of sequence A are quite reminiscent of a binomial distribution, with a lot of balanced 7-liness.

                               2
                               4
                               5
                               8
                      2       10       2
                      4       16       3
                      7       20       6
                      8       32       7
               2     14   2   40      14   2    2
              17     28   5   80   2  21   4   59

    A)  1, 3, 34, 5, 56, 10, 160,  4, 42,  8, 118


    i   1  2   3  4   5   6    7   8   9  10   11
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  • $\begingroup$ ... I have absolutely no idea what I'm looking at here. :/ $\endgroup$ – user24580 Jun 14 '16 at 3:36
  • $\begingroup$ The columns show the factors of each element in sequence A. I'll try to make that clearer. $\endgroup$ – humn Jun 14 '16 at 3:38
  • $\begingroup$ Got it. Yeah, that has absolutely nothing to do with anything. But +1 for an interesting lead. $\endgroup$ – user24580 Jun 14 '16 at 3:42
  • $\begingroup$ Oky doke, I'll leave it alone for now and let these comments serve as a guardrail for solvers. Very intriguing puzzle. $\endgroup$ – humn Jun 14 '16 at 3:48

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