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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.

Times:

1 2 3 4 5 6 7

Sequence 2^2 7 times:

2 4 8 16 32 64

Sum:

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

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    "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".2012-12-23

1 Answers 1

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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.