Number of superincreasing sequences of natural numbers Let us define a superincreasing sequence of natural numbers $b_1, b_2, \ldots,b_n$:
$$
b_{i+1} > \sum_{j=1}^i b_j
$$
or, as we talk about natural numbers:
$$
b_{i+1} \geq 1+ \sum_{j=1}^i b_j.
$$
What is the number of such sequences with the following property:
$$
\sum_{j=1}^n b_j \leq 2^{n+1}-1.
$$
A simple lower-bound estimate is $n+1$ - this is the number of superincreasing sequences which consist of powers of 2, i.e. $n$-combinations of set $\{1,2,4,\ldots,2^n\}$.
In case this is complicated to find simple exact formula, estimates will be interesting too.
UPD 1. Program generation showed the following results for the first values of $n$: 3, 9, 35, 201, 1827, 27337, 692003, 30251721... This is OEIS A125792.
UPD 2. It seems the number I was looking for is discussed in the paper V. Bakoev, 
Algorithmic approach to counting of certain types $m$-ary partitions (also available here).
 A: I find $\begin {array} {c c} n& \text{sequences}\\1&3\\2&9\\3&35\\4&201 \end {array}$
which is OEIS A125972.  The counting is suggestive.  For $n=3$ there are $9$ beginning $1,2$; $7$ beginning $1,3$; down to $1$ beginning wit $1,6$ for a total of $25$ beginning with $1$.  Similarly there are $9$ beginning with $2$ and $1$ beginning with $3$.  For $n=4$ there are $9^2$ beginning $1,2$; $7^2$ beginning $1,3$; down to $1^2$ beginning $1,6$ for $165$ beginning with $1$, and again $95^2$ beginning $2,3$ and so on.  It clearly needs some induction.
A: In fact, the sequence is described in details of A125792. The number I was looking for is the same as the number of partitions of $2^n$ into powers of $2$, excluding the trivial partition $2^n=2^n$ (note there by Valentin Bakoev).
Perhaps the easiest practical recurrent formula is the following:
$$
T(n,k) = T(n,k-1) + T(n-1,2k)
$$
$$
T(n,0) = T(0,k) = 1
$$
Then the answer to the original problem is $T(n,2)$.
One can find even more discussion and references in the description of A002577 which is $T(n,2) + 1$ (in our notation).
