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Now I don't understand why an equivalent thing with an If does not work. If I use the function

C3p[a___, h2_, {n1___, m_, n_, n4___}, b___] := If[m > n,C3p[a, h2, {n1, n, m, n4},  b] + (m - n) C3p[a, h2, {n1, n + m, n4}, b] +  If[m + n == 0, c (m^3 - m)/12 C3p[a, h2, {n1, n4}, b], 0], 0]

This function works indeed as expected, and we have

  In[6]:= C3[1, {}, 1, {1, 1, 1, -4, -3}, 1, {}] // FullSimplify

  Out[6]= 48 (3 C3[1, {}, 1, {-4}, 1, {}] +    5 (C3[1, {}, 1, {-3, -1}, 1, {}] + C3[1, {}, 1, {-2, -2}, 1, {}]))

Now I don't understand why an equivalent thing with an If does not work. If I use the function

C3p[a___, h2_, {n1___, m_, n_, n4___}, b___] := If[m > n,C3p[a, h2, {n1, n, m, n4},  b] + (m - n) C3p[a, h2, {n1, n + m, n4}, b] +  If[m + n == 0, c (m^3 - m)/12 C3p[a, h2, {n1, n4}, b], 0], 0]

This one does not enter in the condition and we obtain

In[8]:= C3p[1, {}, 1, {1, 1, 1, -4, -3}, 1, {}]

Out[9]= 0

In full generality, what I am trying to achive is the following. In a recursive manner, I have this six functions, but I am almost sure that Mathematica is confused since their call is somewhat similar. C3 is a function that take six arguments of the form C3[h1_,{n1___},h2_,{n2___},h3_,{n3___}], three integers and three lists of integers. My code yields the correct result but become incredibly slow. Basically, I am searching a clever way to write

C3[a___, h3_, {n___, m_}] := 0 /; m > 0;
C3[a___, h2_, {n___, m_}, b___] := 0 /; m > 0;
C3[h_, {n___, m_}, a___] := 0 /; m > 0;
C3[a___, h3_, {n1___, m_, n_, n4___}] := C3[a, h3, {n1, n, m, n4}] + (m - n)     C3[a, h3, {n1, n + m, n4}] + 
If[m + n == 0, 24 (m^3 - m)/12 C3[a, h3, {n1, n4}], 0] /; m > n;
C3[a___, h2_, {n1___, m_, n_, n4___}, b___] := 
 C3[a, h2, {n1, n, m, n4}, b] + (m - n) C3[a, h2, {n1, n + m, n4}, b] + 
If[m + n == 0, c (m^3 - m)/12 C3[a, h2, {n1, n4}, b], 0] /; m > n;
C3[h_, {n1___, m_, n_, n4___}, a___] := C3[h, {n1, n, m, n4}, a] + (m - n) C3[h, {n1, n + m, n4}, a] + 
If[m + n == 0, 24 (m^3 - m)/12 C3[h, {n1, n4}, a], 0] /; m > n;

so that mathematica is not confused and know exactly how to commute the thing until the desired order is reached.

I am trying to define a commutator. The way I am working this out is the following. I have a list with integers. I want to put the biggest on the right commuting them. For example, if I have a list like {1,1,1,-4,-3}, then I want to commute the three 1's to the right of the negative number.

I have a recursive function that does the job but I have issues with it since I overcharged this function's definition and the way Mathematica interprets it is not totally clear to me. Therefore, I want to go with an If,else condition. The first recursive function I have is the following

C3[a___, h2_, {n1___, m_, n_, n4___}, b___] := C3[a, h2, {n1, n, m, n4}, b] + (m - n) C3[a, h2, {n1, n + m, n4}, b] + 
If[m + n == 0, c (m^3 - m)/12 C3[a, h2, {n1, n4}, b], 0] /; m > n;

Now I don't understand why an equivalent thing with an If does not work. If I use the function

C3p[a___, h2_, {n1___, m_, n_, n4___}, b___] := If[m > n,C3[a, h2, {n1, n, m, n4},  b] + (m - n) C3[a, h2, {n1, n + m, n4}, b] +  If[m + n == 0, c (m^3 - m)/12 C3[a, h2, {n1, n4}, b], 0], 0]

This function does not even produce an output and never enters the condition. I am also interested in the rapidity of execution since in the end, this function may be called a huge number of time. Any help would be much appreciated.

I am trying to define a commutator. The way I am working this out is the following. I have a list with integers. I want to put the biggest on the right commuting them. For example, if I have a list like {1,1,1,-4,-3}, then I want to commute the three 1's to the right of the negative number.

I have a recursive function that does the job but I have issues with it since I overcharged this function's definition and the way Mathematica interprets it is not totally clear to me. Therefore, I want to go with an If,else condition. The first recursive function I have is the following

C3[a___, h2_, {n1___, m_, n_, n4___}, b___] := C3[a, h2, {n1, n, m, n4}, b] + (m - n) C3[a, h2, {n1, n + m, n4}, b] + 
If[m + n == 0, c (m^3 - m)/12 C3[a, h2, {n1, n4}, b], 0] /; m > n;

Now I don't understand why an equivalent thing with an If does not work. If I use the function

C3p[a___, h2_, {n1___, m_, n_, n4___}, b___] := If[m > n,C3p[a, h2, {n1, n, m, n4},  b] + (m - n) C3p[a, h2, {n1, n + m, n4}, b] +  If[m + n == 0, c (m^3 - m)/12 C3p[a, h2, {n1, n4}, b], 0], 0]

This function does not even produce an output and never enters the condition. I am also interested in the rapidity of execution since in the end, this function may be called a huge number of time. Any help would be much appreciated.