Hi,

I improved the code to be able to display the grid but I c= annot display the coefficients.
Thank you=

Fabrice

\starttext

\startluacode
=C2=A0 =C2=A0 =C2=A0function Bino= m(n,k)
=C2=A0 =C2=A0 =C2=A0 =C2=A0if k > n then
=C2=A0 =C2=A0 =C2= =A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0return 0
=C2=A0 =C2=A0 =C2=A0 =C2=A0elsei= f (n =3D=3D 0 or k =3D=3D 0) then
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0 =C2=A0return 1
=C2=A0 =C2=A0 =C2=A0 =C2=A0else
=C2=A0 =C2=A0 =C2= =A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0return (n*Binom(n-1,k-1))/k
=C2=A0 =C2=A0= =C2=A0 =C2=A0end
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0
=C2=A0 =C2=A0 en= d

=C2=A0 =C2=A0 context.startxtable({"align=3D{middle,lohi}, wi= dth=3D1.2cm,offset=3D0.8ex,bodyfont=3D9pt"})
=C2=A0 =C2=A0 for i = =3D 1, 10 do
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 context.startxrow({"= ;background=3Dcolor,backgroundcolor=3Dcyan"})
=C2=A0 =C2=A0 =C2=A0 = =C2=A0 =C2=A0 =C2=A0 for j =3D 1, i do
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 context.startxcell()
=C2=A0 =C2=A0 =C2= =A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0\ctxlua{context(= Binom(i,j))}
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0 context.stopxcell()
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0= end
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 context.stopxrow()
=C2=A0 =C2= =A0 end
=C2=A0 =C2=A0context.stopxtable()
=C2=A0 =C2=A0 =C2=A0 =C2=A0= =C2=A0 =C2=A0
\stopluacode

\stoptext

Le=C2=A0sam. 18 juil. 2020 =C3=A0=C2=A020:32, Wolfgang Schuster <wolfgang.s= chuster.lists@gmail.com> a =C3=A9crit=C2=A0:

Fabrice Couvreur schrieb am 18.07.2020 um 20:05:

=20
Hello,
I want to display the values of Pascal's triangle in a table.
Unfortunately my knowledge of Lua is not sufficient to run this code.
Thank you.
Fabrice
<= div class=3D"gmail_default" style=3D"font-family:verdana,sans-serif">
\starttext

\startluacode
=C2=A0 =C2=A0 =C2=A0function Bin= om(n,k)
=C2=A0 =C2=A0 =C2=A0 =C2=A0if k > n then
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0= =C2=A0 =C2=A0return 0
=C2=A0 =C2=A0 =C2=A0 =C2=A0elseif (n =3D=3D 0 or k =3D=3D 0) then
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0ret= urn 1
=C2=A0 =C2=A0 =C2=A0 =C2=A0else
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2= =A0 =C2=A0=20 =C2=A0return (n*Binom(n-1,k-1))/k
=C2=A0 =C2=A0 =C2=A0end
=C2=A0 =C2= =A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0=20 =C2=A0context(Binom(n,k))

Are you sure you want to use context(...) here?

<= span lang=3D"en">=C2=A0 =C2=A0 end

=C2=A0 =C2=A0 co= ntext.startxtable({"align=3D{middle,lohi},=20 width=3D1.2cm,offset=3D0.8ex,bodyfont=3D9pt"})
=C2=A0 =C2=A0=20 context.startxrow({"background=3Dcolor,backgroundcolor=3Dgreen"})=
=C2=A0 =C2=A0 for i =3D 0, 10 do
=C2=A0 =C2=A0 =C2=A0 for j =3D 0, i do
=C2=A0 =C2=A0 = =C2=A0 =C2=A0 =C2=A0 context.startxcell()
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 context(Binom(i,j))
=C2=A0 =C2=A0 = =C2=A0 =C2=A0 =C2=A0 context.stopxcell()
=C2=A0 =C2=A0 =C2=A0 end
=C2=A0 =C2=A0 end
=C2=A0 =C2=A0 context.stopxrow()=

context.stopxtable()

Wolfgang

___________________________________________________________________________= ________
If your question is of interest to others as well, please add an entry to t= he Wiki!

maillist : ntg-cont= ext@ntg.nl / http://www.ntg.nl/mailman/listinfo/nt= g-context
webpage=C2=A0 : http://www.pragma-ade.nl / http://context.aanhet.net=
archive=C2=A0 : https://bitbucket.org/phg/context-m= irror/commits/
wiki=C2=A0 =C2=A0 =C2=A0: http://contextgarden.net
___________________________________________________________________________= ________