To toandhsp! I solved your problem by Classic Worksheet Maple 13  (time about 6 min).

To Markiyan Hirnyk! Your solving found only 98 solutions (and accurate of them will be 50). Therefore, I believe that the application of DirectSearch package for these tasks is inefficient. It is interesting your comment on this.

Dear Markiyan! Thank you for finding error in my code. I did not notice the coefficients 2, 3 and so on before x_2, x_3, and so on. The corrected code:

P:=proc(k, i)

local L, s, M, K, m;

if not (k>=2*i and 2*i>=4) then

error `should be k>=2*i>=4` fi;

if not type(k, posint) or not type(i, posint) then

error `should be k is integer and i is integer` fi;

L:=[];

for s from k-2*i to k-i do

M:=combinat[composition](s+i, i);

K:=[seq([seq(M[j,l]-1, l=1..i)], j=1..nops(M))];

for m from 1 to nops(M) do

if convert([seq(type(K[m,t]/t, integer), t=2..i)], `and`) then

L:=[op(L), [K[m,1],seq(K[m,t]/t, t=2..i)]]; fi; od;

od;

L;

end proc;

Dear Markiyan! Thank you for finding error in my code. I did not notice the coefficients 2, 3 and so on before x_2, x_3, and so on. The corrected code:

P:=proc(k, i)

local L, s, M, K, m;

if not (k>=2*i and 2*i>=4) then

error `should be k>=2*i>=4` fi;

if not type(k, posint) or not type(i, posint) then

error `should be k is integer and i is integer` fi;

L:=[];

for s from k-2*i to k-i do

M:=combinat[composition](s+i, i);

K:=[seq([seq(M[j,l]-1, l=1..i)], j=1..nops(M))];

for m from 1 to nops(M) do

if convert([seq(type(K[m,t]/t, integer), t=2..i)], `and`) then

L:=[op(L), [K[m,1],seq(K[m,t]/t, t=2..i)]]; fi; od;

od;

L;

end proc;

@Markiyan Hirnyk 

Unfortunately, there is a limit in Maple on the length of the list. Run the following code (all permutations of 10 elements):

combinat[permute](10);

There are ways around this restriction, for example, dividing the list into several parts.

@Markiyan Hirnyk 

Unfortunately, there is a limit in Maple on the length of the list. Run the following code (all permutations of 10 elements):

combinat[permute](10);

There are ways around this restriction, for example, dividing the list into several parts.

That's right!

That's right!

And where is your Maple  code and PDF?

@Markiyan Hirnyk 

1) Brief comments can be found in the attached file.

 2) The reason for the error in the Maple's calculation of the second integral G i don't understand.

@Markiyan Hirnyk 

1) Brief comments can be found in the attached file.

 2) The reason for the error in the Maple's calculation of the second integral G i don't understand.

Of cause, it's way more difficult, but i don't know different way.

Of cause, it's way more difficult, but i don't know different way.

@Markiyan Hirnyk  You have given a definition by reference to Wikipedia: "A Latin square is said to be reduced (also, normalized or in standard form) if both its first row and its first column are in their natural order. " That is the definition I had in mind. If the condition is imposed only on the increase in the first row, the number of Latin squares is 8:

n:=0: NList:=[]:

for i from 1 to 960 do

if [seq(List[i][1,k], k=1..5)]=[1,2,3,4,5] then n:=n+1: NList:=[op(NList), List[i]]: fi:

od:

n;

op(NList[1..4]);

op(NList[5..8]);

@Markiyan Hirnyk  You have given a definition by reference to Wikipedia: "A Latin square is said to be reduced (also, normalized or in standard form) if both its first row and its first column are in their natural order. " That is the definition I had in mind. If the condition is imposed only on the increase in the first row, the number of Latin squares is 8:

n:=0: NList:=[]:

for i from 1 to 960 do

if [seq(List[i][1,k], k=1..5)]=[1,2,3,4,5] then n:=n+1: NList:=[op(NList), List[i]]: fi:

od:

n;

op(NList[1..4]);

op(NList[5..8]);

@Markiyan Hirnyk  The number of normalized Latin squares with this property is 0, which is obvious without any calculations.