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So basically the title is the question. Is there a sequence that summed up minus the last entry is equal to 2 times last entry. After a long time spend trying I don't even know if this is possible. For example we have the following sequence:

$\text{1 2 3 6 12 24 48 ... }$

Which is a sequence that summed up minus the last entry is equal to 1 times the last entry.

$\text{(1 + 2 + 3 + 6 + 12 + 24 + 48) - 48 = 48}$

PS: First post, so if I did anything wrong please tell me.

(Edit: I am looking for an infinite sequence)

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    What about the sequences $2,1$ or $1,1,1$? – 2017-02-18
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    It seems heavily implied he is looking for an infinite sequence, not a sequence of finite length... – 2017-02-18
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    Yeah sorry, I am looking for an infinite sequence – 2017-02-18
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    @Fawad my interpretation is that he is looking for an infinite sequence $(a)_n$ satisfying the property that $\sum\limits_{k=1}^{n-1} a_k =2a_n$ for all values of $n$, which of course could be rearranged to give a recurrence relation definition for the sequence $a_n:= \frac{1}{2}\sum\limits_{k=1}^{n-1} a_k$. He just gave a specific value of $n$ as an example but the sequence continues past that. The trivial sequence $0,0,0,\dots$ works. Otherwise, you'll have to make an exception for the first entry in the sequence... – 2017-02-18
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    So, assuming we get to ignore the condition for $n=1$, take a seed value... say $a_1=1$, then that implies $a_2=\frac{1}{2}a_1=\frac{1}{2}$, $a_3=\frac{1}{2}(a_1+a_2)=\frac{3}{4}$, $a_4=\frac{1}{2}(a_1+a_2+a_3)=\frac{9}{8}$ etc... you can continue building the sequence recursively for as long as you like. – 2017-02-18
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    @JMoravitz it's better if you would answer it :-) – 2017-02-18
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    @JMoravitz Thanks, that's the awnser! – 2017-02-18

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To go a little farther than the comments, your condition is $$2a_n = \sum_{k=0}^{n-1} a_k$$ (unlike the comments, I am calling the first element of the sequence $a_0$). Note that $$\sum_{k=0}^{n-1} a_k = a_{n-1} + \sum_{k=0}^{n-2} a_k$$ and $$2a_{n-1} = \sum_{k=0}^{n-2} a_k$$ So $$2a_n = a_{n-1} + 2a_{n-1} = 3a_{n-1}\\a_n =\left(3\over2\right)a_{n-1}$$

Hence your sequence is $$a_n = \left(3\over 2\right)^na_0$$

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    @WiebeKuper In particular, there is no (non-zero) infinite sequence of *integers* with this property, since eventually $a_0$ will not be divisible by $2^n$ for some $n$. You didn't state that you are only looking for integers in your question, but that is one possible interpretation that would explain why you couldn't find any examples. – 2017-02-18
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    @ErickWong Yeah probably – 2017-02-18