Pascal's triangle using q-basic

You could just use the binomial theorem. Step through rows, n, and entries, k, and compute the Pascal's triangle value as

n!/(k!*(n-k)!)

You'll actually have better luck if you use the natural log of a factorial, then you can use laws of exponents to get:

exp(log(n!/k!/(n-k)!))

= exp(log(n!)-log(k!)-log((n-k)!))

= exp(logfact(n)-logfact(k)-logfact(n-k))

which won't run into the integer overflow problems that a plain factorial function would have.

To fill up a logfact array, something like this might work:

while(i<maxn)

logfact(i)=logfact(i-1)+log(i)

i=i+1

Wend

Be careful to initialize correctly, and watch your conversion between integers and doubles (probably have to do some rounding to your final answers).