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I am trying to figure out several sequences about simple numbers.

Given simple sequences following x^2 T times.

2^2(6 times) => 002 004 008 016 032 064

3^2(5 times) => 003 009 027 081 243

Then, I have to transform it into a sequence of sums of the previous numbers.

2 4 8 16 32 64 become 2 6 14 30 62 126

My question is: Given a sum of a sequence and the T(times), how to I figure out(GENERIC WAY) the x from x^2?

Example: Given 126 as the sum of the sequence, knowing it was interacted 6 times, what is the x^2 6 times in sequence, and then summed all the number, gives as result 126.

Answer: 2.


1 2 3 4 5 6 7

Sequence 2^2 7 times:

2 4 8 16 32 64


2 + 4 + 8 + 16 + 32 + 64 => 126

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Nice question! Do you know anything about "geometric series" (geometric sequences, geometric progressions)? If not, look them up, and see if you can work out how they relate to your question. – Billy Dec 23 '12 at 5:16
"Sequence 2^2 6 times" doesn't make sense. You mean either "Sequence of 6 iterates of the function $x\mapsto2x$" or "First 6 terms of the sequence 2^n". – Greg Martin Dec 23 '12 at 6:58
up vote 2 down vote accepted

Your question is, given $n$ and also $$k=a+a^2+...+a^n,$$ find the value of the initial term $a$. The sum on the right is $$k=\frac{a^{n+1}-a}{a-1},$$ so that you seek the solution(s) to $$a^{n+1}-(k+1)a+k=0.$$ If the given $n$ is 2 or more this is at least a cubic equation, and somewhat difficult to solve directly.

Provided the sequences are integers, you could use that $a$ is a factor of $k$, and so search for the solution $a$ among the factors of $k$.

EDIT: Actually 1 is a solution of the equation, and you could rule that out easily since if $a=1$ then $n=k$. On factoring out $x-a$ one just arrives back at the original equation $$k=a+a^2+...+a^n,$$ which is of degree $n$ and also difficult to solve directly.

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