ITEM 1 (Schroeppel): (1/3)! and (2/3)! are interexpressible. (1/4)! and (3/4)! are interexpressible. Thus these two pairs are of dimensionality one. (1/10)! and (2/10)! are sufficient to express (N/10)! for all N. (1/12)! and (2/12)! are sufficient to express (N/12)! for all N. (1/3)! and (1/4)! are sufficient to express (N/12)! for all N. Thus the three cases above are of dimensionality two. PROBLEM: Find some order to this dimensionality business. The reflection and multiplication formulas: pi Z Z! (-Z)! = --------- sin(pi Z) (N-1)/2 -NZ-1/2 (2 pi) N (NZ)! = Z! (Z-1/N)! (Z-2/N)! ... (Z-(N-1)/N)! -------------- I was under the misimpression that Mathematica had finally nailed this, but using LR[L_, N_] := Block[{$MaxExtraPrecision = N}, LatticeReduce[ Transpose[Prepend[IdentityMatrix[Length[L]], Round[10^N*L]]]]] (LR for Lattice Reduce or Linear Relation), which finds multiple solutions, unlike FindIntegerNullVector (PSLQ), to find relations among (k/12)! : In[17]:= Join[{2, 3, 5}, (Range[6]/12)!] Out[17]= {2, 3, 5, (1/12)!, (1/6)!, (1/4)!, (1/3)!, (5/ 12)!, Sqrt[\[Pi]]/2} In[18]:= LR[Log[%], 69][[1 ;; 3]] Out[18]= {{0, 6, -1, -2, 2, 3, -4, -6, 2, 3}, {-6, -2, 7, -4, 4, 0, -8, 0, 4, 0}, {-29420553, 655434586, 1447298892, -1344266042, -1749889224, 1253758671, 366295186, 83356101, -1110973039, -416024934}} In[19]:= Rest /@ %[[1 ;; 2]].Transpose[{Log[%%]}] Out[19]= {{6 Log[2] - Log[3] - 2 Log[5] + 3 Log[Sqrt[\[Pi]]/2] + 2 Log[(1/12)!] + 3 Log[(1/6)!] - 4 Log[(1/4)!] - 6 Log[(1/3)!] + 2 Log[(5/12)!]}, {-2 Log[2] + 7 Log[3] - 4 Log[5] + 4 Log[(1/12)!] - 8 Log[(1/4)!] + 4 Log[(5/12)!]}} In[20]:= First /@ Simplify[E^%] Out[20]= {(8 \[Pi]^(3/2) ((1/12)!)^2 ((1/6)!)^3 ((5/12)!)^2)/( 75 ((1/4)!)^4 ((1/3)!)^6), (2187 ((1/12)!)^4 ((5/12)!)^4)/( 2500 ((1/4)!)^8)} In[21]:= N[%] Out[21]= {1., 1.} In[22]:= MapAt[#^(1/4) &, %%, 2] Out[22]= {(8 \[Pi]^(3/2) ((1/12)!)^2 ((1/6)!)^3 ((5/12)!)^2)/( 75 ((1/4)!)^4 ((1/3)!)^6), (3 3^(3/4) (1/12)! (5/12)!)/( 5 Sqrt[2] ((1/4)!)^2)} In[23]:= Numerator[#] == Denominator[#] & /@ % Out[23]= {8 \[Pi]^(3/2) ((1/12)!)^2 ((1/6)!)^3 ((5/12)!)^2 == 75 ((1/4)!)^4 ((1/3)!)^6, 3 3^(3/4) (1/12)! (5/12)! == 5 Sqrt[2] ((1/4)!)^2} In[24]:= FullSimplify[FunctionExpand[%]] /. Gamma[z_] -> (z - 1)! Out[24]= {75 (1/4)!^4 (1/3)!^6 == 8 π^(3/2) (1/12)!^2 (1/6)!^3 (5/12)!^2, 5 Sqrt[2] (1/4)!^2 == 3 3^(3/4) (1/12)! (5/12)!} In[25]:= N[%] Out[25]= {True, True} Is this really that hard a problem? If we can "sort" these, we should move on to the higher factorials (e.g. BarnesG and Hyperfactorial), which have analogous reflection and tuplication formulæ. gosper.org/facfun.pdf . (Recall Date 2014-03-11 02:53 FunctionExpand also appears to disunderstand the tuplication formula Product[Hyperfactorial[-(i/n) + z], {i, 0, -1 + n}] == (E^(z1/n - n*z1)*Hyperfactorial[n*z]^(1/n))/ n^((1 + 6*n*z + 6*n^2*z^2)/(12*n)) whose need it artificially creates for itself when given the Alternating Sign Matrix formula In[171]:= FunctionExpand[Product[(3 k + 1)!/(n + k)!, {k, 0, n - 1}], n > 0 && n \[Element] Integers] Out[171]= (3^(2 - 5 n + 3/2 n (1 + n)) 8^(-1 + n) (BarnesG[5/3] BarnesG[7/3])^(-2 + n) (BarnesG[8/3] BarnesG[10/3])^(1 - n) BarnesG[2/3 + n] BarnesG[1 + n] BarnesG[4/3 + n] BarnesG[2 + n]^( 2 - n) BarnesG[3 + n]^(-1 + n) Gamma[2 + n]^( 1 - n))/(BarnesG[1 + 2 n] Gamma[1 + n]) ) --rwg Old business: I was unable to reproduce that -2=-1 hypergeometric mess for Wolfram's troubleshooters. Now I get a slightly smaller mess which seems numerically correct.