Finding a monotonically increasing sequence of integers I'm trying to find a monotonically increasing sequence of integers, the sequence should be increasing as slow as possible and it has couple of additional requirements:


*

*$n_{i+1} \gt n_i$

*$n_{i} + n_{j+1} \gt n_{i+1} + n_{j}$ where $i+1 < j$


As an example we may consider a sequence of positive integers.
$$
\left\{
n
\right\}^{4}_{n=1}
$$
If $i=1$ and $j=3$, $1+4 = 2+3$ which violates 2nd requirement.
The best what I was able to find so far is Fibonacci sequence which seems to work for both requirements if $i \ge 2$.
 A: If I interpret the "smallest memory" as a sequence that as many terms can be stored for a given memory size as possible. In other words, the sequence should be increasing as slow as possible while meeting the requirements. The OP is further 'implying' that (yeah, this question is really not clear at all) a step size smaller than 1 may take more memory, which means that he is thinking to store data in integer type, not in floating point.
As the sequence should meet $$a_n > a_{n-1}+a_{n-2}-a_{n-3},$$ the slowest sequence will be $$a_n = a_{n-1}+a_{n-2}-a_{n-3}+1$$
So $a_0,a_1,a_2\dots$ will be:
$$0,1,2,4,6,9,12,16,20,25,30,36,42\cdots$$
And it is observed that the steps are:
$$1,1,2,2,3,3,4,4,5,5,6,6\cdots$$
Therefore,
\begin{align}
&a_{n}-a_{n-1}=\lfloor \frac {n+1}2\rfloor, \quad \text{or}\\
&a_{2n}-a_{2n-1}=n\\
&a_{2n+1}-a_{2n}=n+1\\
&a_{2n}-a_{2n-2}=2n\\
&a_{2n}=\sum_{k=1}^n2k=n(n+1)\\
&a_{2n+1}=n(n+1)+n+1=(n+1)^2\\
\end{align}
Or, simply
$$a_n=\lfloor\frac {n+1}2\rfloor\lfloor\frac {n+2}2\rfloor$$
To check if this sequence meets your 2nd requirements (obviously it is increasing so it meets the 1st requirement), if
$$j>i+1 \rightarrow j\ge i+2$$
then
$$a_{j+1}-a_j=\lfloor \frac {j+2}2\rfloor>a_{i+1}-a_i=\lfloor \frac {i+2}2\rfloor$$
