Where a[t] = bn

d[t - 1] - d = b(n-1)

>>6885

In (e, 1) where a[t] = bn,

d[t - 1] - d = b(n-1).

>>6561

Correction:

a[p-1-t]

>>6969

Correcting my correction

My x to t, t to x calculation was wrong, and because of that a[p - t - 1] worked fine for me, not a[p + 1 - t].

So to add more usefulness and to keep things in one place:

Odd e: x = 2t - 1

Even e: x = 2(t-1)

Since we're dealing with series here, I'm adding a few things regarding them.

To sum all the a's in a column, row 1:

odd e: n(e/2 + 1) + 2((n-1)n(n+1)/3)

even e: ne/2 + 2(n(n+1)(2n+1)/6)

The sum of d's in a column, row 1:

even e: n(e/2 + 1) + 2(n(n+1)(2n+1)/6)

odd e: ne/2 + 2((n-1)n(n+1)/3)

To get a partial sum, that is a sum between nth and nth + k (for some nth and k)

s1 = sum(e, k)

s2 = sum(e, nth - 1)

return s1 - s2