{perfect} hypercubes of odd order  

basic ingredients of perfect hypercubes denoted by: LP(....{a_{0},...,a_{n1}}=[perm(0,..,m1)],....) 

Component parameters {a_{0},...,a_{n1}} 
parameters to create a component by formula  
LP({a_{1},...,a_{n}}) [_{j}i] = (_{j=0}∑^{n1}a_{j} _{j}i) % m  
Note: to explain the above for the square this means: LP({a_{1},a_{2}}) [x,y] = (a_{0} x + a_{1} y) % m 

digit permutation =[perm(0,..,m1)] 
permutation to change digits  
LP[_{j}i] < perm[LP[_{j}i]]  
Note: simple replacement of value given by the permutation  
further ingredients  
ragonal pathfinder <v_{0},...,v_{n1}> 
nVector denoting the direction of the ragonal  
v_{i} = 1,0,1  
Note: ragonal directions are given by nvectors with only values 1,0,1 formula's use general 'v' since they are also valid for other vectors 

inproduct <v1,v2> 
mathematical notion  
<v1,v2> = _{j=0}∑^{n1} (v1_{j} v2_{j})  
Note: v1 and v2 are two nVectors, the in(ner)product denotes a number in our case an integer whis should be taken %m. 

component conciderations given a pathfinder  
LP({a_{0},...,a_{n1}}) <v_{0},...,v_{n1}> = (LP[k==0] + k<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>>) % m ; k = 0.. m1  
the numbers on an ragonal turn out to be a constant plus a sequence of numbers given by the inproduct of the parameters and the ragonals pathfinder 

<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>> relatively prime to m 
all digits present on these ragonals  
these ragonals naturally sum to S_{m} = m (m1) / 2  
<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>> NOT relatively prime to m 
the ragonals show q times the same s digits (q s = m)  
q = lcm(m,<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>>) /
<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>> s = m / q 

these ragonals sum to: m LP[k==0] + q _{k=0}∑^{s1} {(k<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>>) % m} = m LP[k==0] + q min{(k<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>>) % m; k=1..s1} S_{s} = m LP[k==0] + q gcd(m,q) S_{s} = so the lines that do sum have: LP[k==0] = (S_{m}  q gcd(m,q) S_{s}) / m 

one can probably always panrelocate to posititions where the above values LP[k==0] fall on the main ragonals making the hypercube {magic} for these ragonals 

in order to make all ragonals sum one needs to digitchange such that: _{k=0}∑^{m} perm[LP({a_{0},...,a_{n1}}) <v_{0},...,v_{n1}>] = _{k=0}∑^{m} perm[(LP[k==0] + k<<a_{0},...,a_{n1}>,<v_{0},...,v_{n1}>>) % m)] = S_{m} the diagonals of s by q {magic} rectangles provide such permutations 
counting {pandiagonal} hypercubes of odd order  

Basic Components LP({1,a_{1},...,a_{n1}}) factor: B_{n,m} = (^{(m3)/2}_{n1}) quality: {pandiagonal} 
concidering the digit change option one can set a_{0} = 1, to put the component in normalized position one set a_{k} > a_{k1}, to avoid reflected variants we limit a_{n1} <= (m1) / 2 

giving the above conditions it can be deduced that there exist B_{n,m} = (^{(m3)/2}_{n1}) such basic components 

Note: The quality of these "Basic Hypercubes" is {pandiagonal}, filtering out the {perfect} is subject to investigation 

B_{n,m} = (^{(m3)/2}_{n1}) 

m\n  1  2  3  4  5  6  7  
3 5 7 9 11 13 15 
1 1 1 1 1 1 1 
1 2 3 4 5 6 
1 3 6 10 15 
1 4 10 20 
1 5 15 
1 6 
1 

combining components  
In order to get a magic hypercube n components needs to be combined in thus a manner as to avoid double appearing numbers. Component with different parameters can be combined as well as different aspects of the same components. Below I keep the hypercube in normalized position and obtain the aomount of {pandiagonal} hypercubes. Numbers D_{n,m} yet to be determined 

Basic Hypercubes LP(...,{1,a_{1},...,a_{n1}},..) factor: B_{n,m} [ 2^{n1} B_{n,m}  1 ] ! / [ 2^{n1} B_{n,m}  n ] !) quality: {pandiagonal} 
Combining Basic Components give the basic hypercubes  
Choosing the highest component out of the B_{n,m} possibil1ties selects one of the possible 2^{n1} B_{n,m} that exist for the lower components, the factor 2 ^{n1} is caused by reflecting the parameters a_{i} > ma_{i} i=1..n1 ; thus gives a factor B_{n,m} (^{2n1Bn,m1} _{n1}) the lower components can of course permute there position giving an extra (n1)! resulting in: B_{n,m} [ 2^{n1} B_{n,m}  1 ] ! / [ 2^{n1} B_{n,m}  n ] !) 

Note: The quality of these "Basic Hypercubes" is {pandiagonal} this reproduces the numbers in the {perfect} square article as well as predict the 6 order 7 {pandiagonal} cubes order 7 uploaded onto the database 

B_{n,m} [ 2^{n1} B_{n,m}  1 ] ! / [ 2^{n1} B_{n,m}  n ] !)  
m\n  1  2  3  4  5  6  7  
3 5 7 9 11 13 15 
1 1 1 1 1 1 1 
1 6 15 28 45 66 
6 330 3,036 14,820 51.330 
210 107,880 4,744,740 78,883,080 
32,760 180,300,120 47,722,860,360 
20,389,320 1,446,769,946,160 
48,920,775,120 

Basic Hypercubes LP(...,{1,a_{1},...,a_{n1}},..) compond orders factor: unknown 
Combining Basic Components compund order  
For compound order the situaton is a bitmore difficult. The general combinations calculted above give multipe appearing numbers when the differance or factors is an integral multiple of the factors of the order, suppose m = qr then (a + kq) r = (ar + kqr) = ar + km = ar (mod m) which confirms the fact stated. Yet I don't know a solid counting argument which give me the numbers mentioned below. I'm looking for a solid counting argument though. NOTE These numbers not yet incorporated into the following tables, nor have I not figure out wheter the multiple numbers can be controlled by some mixture of digit changing permutation 

experimental estimate  
m\n  2  factors  
9 15 21 25 27 33 35 39 45 49 
12 39 93 193 192 273 351 388 471 906 
3*3 3*5 3*7 5*5 3*3*3 3*11 5*7 3*13 3*3*5 7*7 

{pandiagonal} hypercubes in normalized position amount: F_{n,m} * (m1)!^{n1} (m2n1)! * (^{m1}_{2n}) (2n1)! * B_{n,m} [ 2^{n1} B_{n,m}  1 ] ! / [ 2^{n1} B_{n,m}  n ] !) 
lower component digit permutations completes the proces  
To keep the highet component in normalized position the used digit changing permutation are limited to those where one randomlyselect 2n numbers to be placed on the i'th monagonal through [_{j}0] at the first (P_{i}) and last (R_{i}) thus that: P_{0} < P_{i} < R_{i} < P_{i+1} < R_{i+1} < R_{0} ; i = 1..n2 except for the lowest these can permute, thus explains (^{m1}_{2n}) (2n1)! the orher (m12n) numbers can be randomly permuted: (m12n)! the lower components all permutations [0,perm(1,m1)] can be used: (m1)^{n1} 

(m1)!^{n1} (m2n1)! (^{m1}_{2n}) (2n1)!
B_{n,m} [ 2^{n1} B_{n,m}  1 ] ! / [ 2^{n1} B_{n,m}  n ] !) note; numbers are huge so numbers not displayed 

m\n  2  3  
3 5 7 9 11 13 15 
144 777,600 6,096,384,000 92,177,326,080,000 2,581,228,494,028,800,000 125,400,898,533,107,957,760,000 
373,248,000 3,605,157,642,240,000 24,179,071,274,975,232,000,000 271,461,250,591,582,448,517,120,000,000 5,668,198,751,497,959,264,452,912,087,040,000,000 

(BE AWARE: this reasoning is generalized from known data for prime order squares (so highly speculative)) the reader is invited to check the reasoning carefully and point out a better one. The not explained factor F_{n,m} is added to denote the fraction of working permutation yet to be determined (see 1st table above) several factors are entering here at various places I do think this amounts to a global (integral) fraction of the amount given above. 

note: the numbers above give the amount of {pandiagonal} hypercubes in normalized position I haven't factored in the "not working" permutations which might effect the numbers in the case of compound order, so the above numbers can be taken as an "upperbound". Currently don't see a flaw in the reaoning but is subject to scruteny upon the factor 2^{n1} / (2n1)!! numbers will be corrected to this factor next week. 