binomial no longer uses Float64 internally.

[jiahao]

Replaces the old accumulation step, using x=*(nn::Float64, rr::Float64),
with x::T=div(widemul(x,nn), rr) as suggested by @JeffBezanson.

Previously, binomial(54,23) computed the wrong answer. The new thresholds are:

type	arguments	failure mode
Int32	(34,16)	integer overflow (throws OverflowError())
Int64	(67,29)	integer overflow (throws OverflowError())
Int128	(131,63)	integer overflow (throws OverflowError())

Fix #6154.

[JeffBezanson]

Are you sure x will always be negative if there's an overflow? That doesn't seem plausible.

[jiahao]

This is sufficient but not necessary. Maybe this should be in the inner loop? x always starts positive and is multiplied and divided by positive numbers, so the only way to make it negative is due to overflow. I don't think it is possible in the inner loop to overflow and remain positive because the factors aren't that large.

[JeffBezanson]

I still don't think x<0 is a correct test for overflow. The only thing I can think of is

temp = div(widemul(x, nn), rr)
x = temp
x != temp && throw(OverflowError())

[lbenet]

Using your code above I still get InexactError for binomial(67,30).

Forcing x::T as you suggested #6154, so the code becomes:

temp = div(widemul(x,nn), rr)
x::T = temp
x != temp && throw(OverflowError())

renders the OverflowError as wanted.

[lbenet]

The code seems ok, though I don't think you need/use oldx

[jiahao]

Thanks for catching that.