Do you use any modulus equations at all? Can you give us a hint. I have pretty much given up on the problem...

What I did is a forward and backward. I recognized, that essentially what I did is a mod 64 method done a bit different, so I returned to it. I additionally use mod 17. But all this still does not help me to find an exact solution so I also have almost given up. Nevertheless I can share a few more details which show the kind of approach. For example the solution for s=220 must have at least 69 summands.

We have x^16= 1 mod 17 for all x which are not divisible by 17, a consequence of Fermat's little theorem. We also have x^16= 1 mod 64 for all x not divisible by 2.

Because 220^16= 1 mod 17 and 220^16= 0 mod 64, we must have exactly

1+i*17 bases in our sum which are are not divisible by 17 and

0+64*j bases which are not divisible by 2.

If the number of bases is less than 64 then j=0, all bases would have to be even. But then, because 220 is also even we could divide the solution by 2^16 and would get a solution with less than 64 bases for 110^16. With the same argument this can be reduced to a solution to 55^16, which does not exist.

So we have at least 64 bases. The smallest i for which this possible is i=4, so we have at least 69 bases which are not divisible by 17 in the sum which adds up to 220^16.

Of course this also allows pruning if you search for a solution for 220 beginning with large bases working down to small bases. If for example your partial sum is 23 mod 64, the part still to do must be 41 mod 64 which means, you have still have to add at least 41 uneven bases. Then you know that this is impossible if you only have for example the numbers 1..80 left.

But even with using these ideas for pruning my current program seems not good enough to find a solution for s=220.