Re: [math-fun] When (-1)-choose-(-1) is 1/2
Rich's proposal is the b=a/2 case of In[722]:= TableForm[Table[ Limit[Binomial[a*t + n, b*t + k], t -> 0], {n, -3, 3}, {k, -3, 3}]] Out[722]//TableForm= b b 3 b 6 b 10 b - 1 - - -3 + --- 6 - --- -10 + ---- a 0 0 a a a a 2 b b b 2 b 3 b 4 b -(---) - 1 - - -2 + --- 3 - --- -4 + --- a a 0 a a a a b b b b b b b - -(-) - 1 - - -1 + - 1 - - -1 + - a a a a a a a 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 2 1 0 0 0 0 1 3 3 1 In[723]:= % == Table[Limit[Binomial[a*t + n, a*t + n - b*t - k], t -> 0], {n, -3, 3}, {k, -3, 3}] == Table[Limit[(a*t + n)!/(b*t + k)!/(a*t + n - b*t - k)!, t -> 0], {n, -3, 3}, {k, -3, 3}] Out[723]= True GKP says b=0*a. Kronenburg and Mma say that 97% of mathematicians agree that the science is settled. --rwg
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Bill Gosper