Due to my lack of mathematical training, I'm having a hard time phrasing this question, so please bear with me.
Let '$a_{(1,j)}$' define the following series: $a_{(1,1)}=1$, $a_{(1,n)}=a_{(1,(n-1))}+2$. Let $a_{(i,j)}$ define the following arithmetic progression for $i>1$: $a_{(i,1)}=a_{((i-1),1)}+1+i$, $a_{(i,n)}=a_{(i,(n-1))}+2\cdot i$. Constructing a triangle whose diagonals consist of elements of arithmetic progressions defined along this line, we get (assuming I got everything right):
1 4 3 8 8 5 13 14 12 7 19 21 20 16 9 26 29 29 26 20 11 34 38 39 37 32 24 13 43 48 50 49 45 38 28 15 . . .
Where the elements of the ith diagonal (starting from the top) are the elements of $a_{(i,j)}$.
My questions about this peculiar creature are as follows:
- Can we find an explicit formula for the row sums of this triangle, and prove that it is valid? An emphasis on the 'proof' part because using semi-intuitive techniques from high school (that's as far as I go) I've found the following formula, which seems to work: $\frac{1}{2}(n^3+2n^2-n)$. But of course I have no proof that this formula is inductively correct.
- Assuming the formula I found earlier is correct, would the fact that it is a polynomial of 3rd degree hold for other 'triangles of arithmetic progressions'? (This interests me because of the seeming connection between the number '3' and our 'triangle' of sorts.)
Thank you for your help.