Re: numbpart(n, {a = k})

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Fri, Apr 17, 2026 at 08:55:23AM +0200, Ruud H.G. van Tol wrote:
> 
> For https://oeis.org/A046682, I created this code:
> 
> first(nn) = {
>   my( v = Vec( 1 / prod( k=1, nn, 1+(-x)^k, 1+x*O(x^nn) ) ) );
>   vector( nn, i, ( numbpart(i-1) + v[i] ) / 2 );
> }
> This takes about a minute to return 10k values.
> 
> Is there a faster way to initialize v?

Yes. There is a formula which is missing from the OEIS, you should probably add it:
The generating series is

Phi(-x)/Phi(x^2) + 1/Phi(x))/2

where Phi is the Euler function mentionned in https://en.wikipedia.org/wiki/Dedekind_eta_function
which admits a lacunary series and can be computed in GP using eta().

So the formula in GP is simply:

first2(nn) = Vec(eta(-x+O(x^nn))/eta(x^2+O(x^nn)) + 1/eta(x+O(x^(nn))))/2

? first2(10000);
  ***   last result computed in 75 ms.
? first2(100000);
  ***   last result computed in 3,004 ms.

Similarly if you want all numbpart up to n, you should just compute
1/eta(x+O(x^(n+1)))

> Would a variant numbpart(n, {a = k}) be interesting, like Maple has?
> With the optional a-parameter like with partitions(k, {a = k}, {n = k}).

numbpart use Rademacher formula, which is much faster than #partitions(n) but is
only valid for partitions(n,,).
I do not know fast formula for the other cases.

For example pari can compute numbpart(1000000) in 5 ms

How far Maple can go?

Cheers,
Bill.