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Find closed form of a sequence $2,5,11,23,\dots$

How to get generating function for this sequence (closed form)?

Explicit form is $f(x)=2+5x+11x^2+23x^3+\cdots$

Is it possible to get to geometric series representation?

I tried to derive the series multiple times, but that doesn't help.

Could someone give a hint?

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    $\begingroup$ How does the series continue? Is it f(n) = 2f(n-1) + 1? $\endgroup$
    – arbitrary username
    Nov 24, 2015 at 16:18

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Hint Adding $1$ to each term of the sequence gives $$3, 6, 12, 24, \ldots,$$ and dividing each term of this new series by $3$ gives $$1, 2, 4, 8, \ldots .$$

On the other hand, $1 + x + x^2 + x^3 + \cdots$ is the series for the function $x \mapsto \frac{1}{1 - x}$.

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    $\begingroup$ The n-th term of one sequence with the same first four terms is $(n^3-3n^2+8n-2)/2$. $\endgroup$
    – jbuddenh
    Nov 24, 2015 at 16:23
  • $\begingroup$ Yes, of course, there are infinitely many sequences with these particular four terms, and OEIS alone finds 57 of them: oeis.org/… . One is the sequence for which $a_1 := 2$ and $a_{k + 1}$ is the smallest prime $> 2 a_k$. But given the apparent context of the problem, one probably ought to be able to compute the generating function by hand, and so the sequence I hint at seems to be the one intended by the source of the problem. $\endgroup$
    – Travis Willse
    Nov 24, 2015 at 16:31
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The consecutive differences are $3,6,12,\cdots$

So, $T_r-T{(r-1)}=3\cdot2^{r-1}, r\ge2$

Let $T_n=a2^{n+1}+F_n$

$3\cdot2^{r-1}=a2^{r+1}+F_r-\{a2^r+F_{r-1}\}=a2^r+F_r-F_{r-1}$

Set $3\cdot2^{r-1}=a2^r\iff a=\dfrac32$

Can you take it from here?

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The $n$-th term is $$f(n)=2^n+2^{n-1}-1=3\times 2^{n-1}-1$$

It satisfies the relation $f(n+1)=3\times 2^n-1=2(3\times 2^{n-1}-1)+1=2f(n)+1$ and $f(1)=2$

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  • $\begingroup$ That's what WA gives... $\endgroup$
    – lhf
    Nov 24, 2015 at 17:18
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$2, 5, 11, 23, 47, \cdots$ Well, I think $a_n = 3 \cdot 2^{n - 1} - 1$ here.

OK, let me elaborate a bit. Observe that $d_n := a_{n + 1} - a_{n} = 3 \cdot 2^{n - 1}$. Since $$a_n - a_1 = \sum_{k = 1}^{n - 1} d_{k} = 3 \sum_{k = 1}^{n - 1} 2^{k - 1} = 3 \cdot (2^{n - 1} - 1) = 3 \cdot 2^{n - 1} - 3.$$ Since $a_1 = 2$, we have $a_n = 3 \cdot 2^{n - 1} - 1$.

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