Project Euler #28

https://projecteuler.net/problem=28


The series to get the result has this pattern:
1 + [3 + 5 + 7 + 9] + [13 + 17 + 21 + 25 ] + [31 + 37 + 43 + 49] + [ 56 ...
There's always a quertet of four summands.
How many you skip between the 4 summands (difference of one summand to the next):
0    2                 4                      6                      8

x :
0    1                 2                      3                      4
f(x) := x*2 :
0    2                 4                      6                      8
h(x) := (2x+1)^2 :
1    (2+1)^2=9         (4+1)^2=25            49                     81
Note that this is always the last of the quertet.
Now we take that number 4 times because each block is:
[ h(x)-3f(x) + h(x)-2f(x) + h(x)-f(x) + h(x) ] = 4h(x)-6f(x)
g(h) := g(x-1) + 4*h(x) - 6*f(x)
1   25               101                    261                    537

That doesn't mean that it has to be recursive.
With a bit of magic (also know as mathematics) you get this formula:
g(x) := 1 + 2/3 x (13 + x (15 + 8 x))