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

On 2026-04-17 15:14, Ruud H.G. van Tol wrote:
> On 2026-04-17 13:29, Bill Allombert wrote:
>> On Fri, Apr 17, 2026 at 12:03:17PM +0200, Bill Allombert wrote:

>> [...]
>>
>> first3(nn) =
>> {
>>    my(E = eta(x+O(x^(nn))), Ei=1/E);
>>    Vec(subst(E,x,-x)*subst(Ei+O(x^(nn\2)),x,x^2) + Ei)/2;
>> }
>
> Test:
> ? #first3(10001)
> # 10000
>
> first4(nn) = {
>    my(E = eta(x+O(x^(nn))), Ei = 1/E);
>    Vec( subst(E, x, -x) * subst( Ei + O(x^( nn\2 + 1 )), x, x^2) + Ei)/2;
> }
>
> Test:
> ? first2(10001) == first4(10001)
> # 1
>
> That is just great, thanks!

I now made it:

first5(nn)= {
    my
    ( x= 'x
    , E= eta(x+O(x^(nn)))
    , Ei= 1/E
    , ps= (subst(E, x, -x) * subst(Ei + O(x^(nn\2 + 1)), x, x^2) + Ei)/2
    );
    Vec(ps);
}

to hide any x in the environment, as gp-coders can mess of that up.

See https://oeis.org/history/view?seq=A046682&v=9999
for the latest state.

- - - - - - -

Then also mini-fied that to

first6(nn)= {
   my
   ( x= 'x
   , E= eta(x + O(x^nn))
   , Ei= 1/E
   , ps= (subst(E, x, -x) * subst(Ei, x, x^2) + Ei) / 2
   );
   Vec(ps);
}

Test:
? my(nn=9999); first5(nn) == first6(nn)
% 1

but first6 is less fast than first5.

-- Ruud