Is there a sequence that summed up minus the last entry is equal to 2 times the last entry 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)

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 $$a_n = \left(3\over 2\right)^na_0$$
