From: peter.luschny@gmail.com   
      
   This problem has a continuation for hypergeometric functions.   
   Consider   
      
   (*)   
   Table[Table[Sum[2^(k-i)*Binomial[n-i-1,k-i],{i,0,k}],{k,0,n}],{n,0,7}]   
      
   Who does not have Mathematica can enter this formula in Alpha.   
   Alpha provides the answer:   
      
   {{0},   
   {1, 0},   
   {1, 3, 0},   
   {1, 5, 7, 0},   
   {1, 7, 17, 15, 0},   
   {1, 9, 31, 49, 31, 0},   
   {1, 11, 49, 111, 129, 63, 0},   
   {1, 13, 71, 209, 351, 321, 127, 0}}   
      
   As Hans Havermann told me Mathematica 10 yields final terms 1,   
   not 0. So Wolfram's left hand does not know what Wolfram's   
   right hand does.   
      
   But now my question is: what is the right answer for   
   this expression:   
      
   (**)   
   Table[Table[2^k*Binomial[n-1,k]*Hypergeometric2F1[1,-k,-n+1,1/2]   
   {k,0,n}],{n,0,7}]   
      
   Consider only the final terms in the rows.   
      
   (a) 0 (as in the Alpha table above)   
   (b) 1 (as in the Mathematica table above)   
   (c) Indeterminate   
      
   If you think (c) is the right answer then please explain   
   why you think the answer has to be different from (*).   
      
   Peter   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)