I have programs below for the cases n=3 and n=4. If n is increasing by 1, then there is one loop more. As you can see, that additional loop has always the structure:
x[i+1] from ceil(((n-2)*24-d[i])/(n-i)) to x[i]
whereas d[i] is the sum of the values of x before.
Now I want to write "nequaln":=proc(k), which goes through all the values for n from 3 to k and produces a list [a_3,a_4,a_5,...,a_n].
I guess that I need to program a kind of dynamical loop, but I failed completely. May someone help me? That would be very kind.
for u from ceil((3-2)*24/(3-0)) to (3-2)*24-3+1 do
for v from ceil(((3-2)*24-u)/(3-1)) to u do
if (3-2)*24-u-v>=1 then
for u from ceil((4-2)*24/(4-0)) to (4-2)*24-4+1 do
for v from ceil(((4-2)*24-u)/(4-1)) to u do
for w from ceil(((4-2)*24-u-v)/(4-2)) to v do
if (4-2)*24-u-v-w>=1 then