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).


2 Answers 2


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.

  • $\begingroup$ I generated all the sequences for small $n$'s and found a bit different numbers: 3, 9, 35, 201, 1827, 27337, 692003, 30251721... And this is A125792. $\endgroup$ Nov 3, 2012 at 2:27
  • $\begingroup$ Seems to be a bit of disagreement at $n=4$. Perhaps you could both check your calculations. $\endgroup$ Nov 3, 2012 at 4:55
  • $\begingroup$ @GerryMyerson: Found the error and fixed $\endgroup$ Nov 3, 2012 at 6:09

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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .