prime order perfect hypercubes  

basic ingredients of pandiagonal hypercubes  
Latin hypercube generating formula LH(a_{j}) 
Latin hypercubes obtained by formula 
LH(a_{j}): LH[_{j}i] = (_{j=0}∑^{n}a_{j} _{j}i) % m ;
j ε [0,..,n1]; i ε [0,..,m1]; a_{j} < a_{j+1}; a_{0} = 1; a_{j} = 2 .. (m  1)/2 

The latin hypercubes obtained by the above formula are in normalized position due to the condition a_{j} < a_{j+1} (can't be equal because that spoils pandiagonality). a_{0} = 1 because because of digit changing, thus parameters define the LH's structure the range of a_{j} avoids panflip variants introduces by parameter range (m+1)/2 .. m1. 

pan ragonal hypercubes  
ragonal pathfinder  P_{r} = <_{j}i; j = 0..n1; _{j}i ε {1,0,1}; _{j=0}∑^{n1}_{j}i = r> 
The formula merely captures all ragonal direction the rdimensional subhypercube posesses 

pan ragonal condition C_{r} 
C_{r} : _{j=0}∑^{n1} P_{r}^{j} a_{j} relatively prime to m. 
since m is prime it means that the sum is no integral multiple of m this condition posed on the latin hypercubes parameter ensures that also on the ragonal lines all digits are present 

perfect condition P_{n} 
P_{n} : C_{r} for all r = 1 .. n 
In order for the hypercube to be perfect all ragonal conditions must be satisfied 
prime order perfect hypercubes  

basic ingredients of perfect hypercubes  
Latin hypercube generating formula LH(a_{j}) factor: Counting argument unknown F 
Latin hypercubes obtained by formula  
LH(a_{j}): LH[_{j}i] = (_{j=0}∑^{n}a_{j} _{j}i) % m ;
j ε [0,..,n1]; i ε [0,..,m1]; a_{j} < a_{j+1}; a_{0} = 1; a_{j} = 2 .. (m  1)/2 (_{j=0}∑^{n}a_{j} Pf_{p}[j]) % m != 0; p ε [0,..,3^{n}1]; Pf_{p}[j] ε {1,0,1}; 

The latin hypercubes obtained by the above formula are in normalized position due to the condition a_{j} < a_{j+1} (can't be equal because that spoils pandiagonality). a_{0} = 1 because because of digit changing, thus parameters define the LH's structure the range of a_{j} avoids panflip variants introduces by parameter range (m+1)/2 .. m1. 

Unknown counting argument (numbers manually obtained)  
m \ n  2  3  4  5  6  7  
5 7 11 13 17 19 
1 2 4 5 7 8 
0 0 3 6 15 21 
0 0 0 0 1 3 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 

As with the pandiagonal hypercube the same arguments give numbers of possible {perfect} hypercubes, so we have G = F (^{F 2(n1)  1} _{n1}) (n1)! basic {perfect} hypercubes and thus eventually G((m1)!/2)^{n} {perfect} hypercubes. (stated numbers preliminairy) 

{perfect} hypercubes  
basic {perfect} hypercubes factor: G = F (^{F 2(n1)  1} _{n1}) (n1)! 
the basic {perfect} hypercubes  
H(a_{j,i}) = _{i=0}∑^{n1} m^{ni1} LH(a_{j,i})  
In order to retain the normalized position the highest component (denoted with i=0) need to remain in normalized position, the other component are added in posible panflip and transpositional variants, leaving the xaxis as is there are 2^{n1} panflips, resulting in 2^{n1} F posibilities of which 1 is already chosen, and n1 LH's need be randomly selected, this explains the listed factor. The factor of (n1)! is due to reordering of the lower components 

F (^{F 2(n1)  1} _{n1}) (n1)!  
m \ n  2  3  4  5  6  7  
5 7 11 13 17 19 
1 6 28 45 91 120 
0 0 330 3,036 51,330 142,926 
0 0 0 0 1,632 22,230 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 

{perfect} hypercubes factor: m!^{n}G/(2m)^{n}= G((m1)!/2)^{n} 
the {perfect} hypercubes  
{perfect}  
H(a_{j,i}) = _{i=0}∑^{n1} m^{ni1} LH(a_{j,i})_{=[perm(i)]}  
Applying independently digits changers to the various components generate all(?) possible pandiagonal hypercubes, which introduces a factor (m!)^{n} (if not mistaken). This however also introduces all panvariants which gives a deviding factor of (2m)^{n} with: 2^{n} (reflection) ; m^{n} (panrelocation) 