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I have these three expressions

Gamma[1/2 + k]*Gamma[7/10 + k]*Gamma[9/10 + k]*Gamma[1 + k]^6*Gamma[11/10 + k]*
  Gamma[13/10 + k]*Gamma[3/2 + k]*Gamma[17/10 + k]*Gamma[19/10 + k]*Gamma[21/10 + k]*
  Gamma[23/10 + k]*Gamma[5/2 + k]*Gamma[13/5 + k]*Gamma[27/10 + k]*Gamma[29/10 + k]*
  Gamma[3 + k]*Gamma[31/10 + k]*Gamma[33/10 + k]*Gamma[7/2 + k]*Gamma[18/5 + k]*
  Gamma[37/10 + k]*Gamma[39/10 + k]*Gamma[41/10 + k]*Gamma[43/10 + k]*Gamma[9/2 + k]*
  Gamma[23/5 + k]*Gamma[47/10 + k]*Gamma[49/10 + k]*Gamma[51/10 + k]*Gamma[53/10 + k]*
  Gamma[28/5 + k]

and

Gamma[n]*Gamma[2/5 + n]^2*Gamma[3/5 + n]^3*Gamma[4/5 + n]^2*Gamma[5/6 + n]*
  Gamma[7/6 + n]*Gamma[6/5 + n]^2*Gamma[7/5 + n]*Gamma[8/5 + n]^2*Gamma[9/5 + n]*
  Gamma[11/6 + n]*Gamma[13/6 + n]*Gamma[11/5 + n]*Gamma[3 + 2*n]

and

Gamma[3/2 + k + n]*Gamma[17/10 + k + n]*Gamma[19/10 + k + n]*Gamma[2 + k + n]*
  Gamma[21/10 + k + n]*Gamma[23/10 + k + n]*Gamma[5/2 + k + n]*Gamma[13/5 + k + n]*
  Gamma[27/10 + k + n]*Gamma[29/10 + k + n]*Gamma[3 + k + n]*Gamma[31/10 + k + n]*
  Gamma[33/10 + k + n]*Gamma[18/5 + k + n]

Also, relatedly

 (1/10)*(23 + 10*n)*(1 + n)!*(2 + n)!*(1 + 2*n)!*(3 + 2*n)!*((1/10)*(7 + 10*n))!*
  ((1/10)*(9 + 10*n))!*((1/10)*(11 + 10*n))!*((1/10)*(13 + 10*n))!*((1/10)*(17 + 10*n))!*
  ((1/10)*(19 + 10*n))!*((1/10)*(21 + 10*n))!

To what extent can these be simplified?

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  • $\begingroup$ I think the answer to your question is "Not much." For example (assuming n and k are integers), Gamma[2/5+n]^2 Gamma[7/5+n] will simplify to (2/5+n) Gamma[2/5+n]^3. $\endgroup$ – JimB Aug 22 '16 at 19:35
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Simplification is many times in the eye of the beholder. But assuming k and n are non-negative integers, use of the following rule might help somewhat (but likely not very much):

rule = {Gamma[x_] /; x > 1 -> (x - 1) Gamma[x - 1], 
   Gamma[x_ + z__] /; x > 1 -> (x + z - 1) Gamma[x - 1 + z]};

Here is that rule applied to your first example:

Gamma[1/2 + k]*Gamma[7/10 + k]*Gamma[9/10 + k]*Gamma[1 + k]^6*
  Gamma[11/10 + k]*Gamma[13/10 + k]*Gamma[3/2 + k]*Gamma[17/10 + k]*
  Gamma[19/10 + k]*Gamma[21/10 + k]*Gamma[23/10 + k]*Gamma[5/2 + k]*
  Gamma[13/5 + k]*Gamma[27/10 + k]*Gamma[29/10 + k]*Gamma[3 + k]*
  Gamma[31/10 + k]*Gamma[33/10 + k]*Gamma[7/2 + k]*Gamma[18/5 + k]*
  Gamma[37/10 + k]*Gamma[39/10 + k]*Gamma[41/10 + k]*Gamma[43/10 + k]*
  Gamma[9/2 + k]*Gamma[23/5 + k]*Gamma[47/10 + k]*Gamma[49/10 + k]*
  Gamma[51/10 + k]*Gamma[53/10 + k]*Gamma[28/5 + k] //. rule

Simplification

Update

If you have some formula that generates the above examples (as I certainly wouldn't want to type each term one at a time), then you should give that formula in your question. That might make it possible to use the Gauss Multiplication Theorem (as suggested by @Lucas) and the duplication formula to simplify things. But once items are multiplied together the order of terms might change or not be so recognizable.

For example in the first equation you have

Gamma[1/2 + k]*Gamma[7/10 + k]*Gamma[9/10 + k]*Gamma[1 + k]^6*
  Gamma[11/10 + k]*Gamma[13/10 + k]*Gamma[3/2 + k]*Gamma[17/10 + k]....

Why would there be Gamma[1 + k]^6 ? None of the other terms are raised to a power. Having such terms that mask the pattern won't make things easier.

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  • $\begingroup$ Aren't all the gammas simplifiable via Gauss' multiplication theorem to something like $\Gamma(10x) / \Gamma(5x)$ $\endgroup$ – Lucas Aug 22 '16 at 21:27
  • $\begingroup$ @Lucas You should post an answer using that theorem. $\endgroup$ – JimB Aug 22 '16 at 21:38
  • $\begingroup$ Yeah, I'm not sure how to do it in generality, or I would have. $\endgroup$ – Lucas Aug 23 '16 at 0:16
  • $\begingroup$ Specifically, how does one match a specific product of gammas $\endgroup$ – Lucas Aug 23 '16 at 0:30
  • $\begingroup$ I've added an update to suggest that you describe how you get the products of gammas. It looks like at least some little bit of simplification has already occurred which might mask any patterns necessary for substantial simplification. $\endgroup$ – JimB Aug 23 '16 at 2:49

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