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Extending addictive factorial?

  1. Dec 30, 2009 #1
    If we define an addictive factorial for any integer n:

    f(n) = n + (n-1) + (n-2) ..... 0

    1!+ = 1
    2!+ = 2+1 = 3
    3!+ = 3+2+1 = 6
    4!+ = 4+3+2+1 = 10
    5!+ = 15

    is it possible to extend it to real or possibly complex numbers by analytic continuation?

    just like the gamma function extends the factorial.
     
    Last edited: Dec 30, 2009
  2. jcsd
  3. Dec 30, 2009 #2
    Observe that
    [tex]\sum_{k=1}^n k=\frac{n(n+1)}{2}[/tex]
    is your function.
     
  4. Dec 30, 2009 #3
    never mind, forgot about the binomial
     
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