Can someone help me understand this number sequence?

Question

In the link below there is a number sequence and I do not know how it is put together. It is also arranged in an irregular triangle on the page, I tried looking up how the "Abramowitz and Stegun order" is arranged but did not come up with clear results.

https://oeis.org/A036043

Answer 1

Looking at Abramowitz and Stegun page 831, the order seems to be shorter partitions first, with smaller numbers before larger numbers.

So for example, one of the rows is the length of partitions of $5$ with the values being $1, 2, 2, 3, 3, 4, 5$. These correspond to something like

• $5$ of length $1$
• $1+4$ of length $2$
• $2+3$ of length $2$
• $1+1+3$ of length $3$
• $1+2+2$ of length $3$
• $1+1+1+2$ of length $4$
• $1+1+1+1+1$ of length $5$

Answer 2

Ok. First of all, lets rewrite the sequence:

   n\k   1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16   17    18   19   20    21   22 ...  


   1     1 

   2     1  2   

   3     1  2  3         

   4     1  2  2  3  3  4       

   5     1  2  2  3  3  4  5    

   6     1  2  2  2  3  3  3  4  4  5  6    

   7     1  2  2  2  3  3  3  3  4  4  4  5  5  6  7   

   8     1  2  2  2  2  3  3  3  3  3  4  4  4  4  4  5  5  5  6  6  7  8    

   9     1  2  2  2  2  3  3  3  3  3  3  3  4  4  4  4  4  4  5  5  5  5  5  6  6  6  7  7  8  9

  10     1  2  2  2  2  2  3  3  3  3  3  3  3  3  4  4  4  4  4  4  4  4  4  5  5  5  5  5  5  5  6  6  6  6  6   7   7   7   8   8   9  10

As you can see, each number in sequence is function of $n$ and $k$: $f(n,k)$.

Take number $n$ and find all possible sums of natural numbers that are equal no $n$. For example $5=5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1$. Sort these sums, so that the one that has less summands would be always earlier, and among sums with same number of summands one that has bigger summand in it would be earlier(I've already done it). Lets call for each such sum partition. $f(n,k)$ is number of summands in $k^{th}$ partition $n$