compound order panmagic squares (orders: m = _{i=0}∏^{n}p_{i})  

Latin component squares LSC(P_{i}) 
a m by m square of squares with digits radix p_{i} (ie.: {0 .. p_{i}1})  
when p_{i} is a prime > 3 the seperate squares are easily made panmagic by the formulae given in my prime order page and so easily put together in a resulting order m panmagic component square. The order 15 evidence suggest it also possible when p_{i} = 3 horewver evidence shows that components need not be panmagic themselves to make up a panmagic square, off sums in one component can be compensated by off sums in an other component. 

Latin squares LC(....) parametrizable perhaps (suggested be square orders) 
LC(....) = _{i=0}∑^{n} _{j=i+1}∏^{n} p_{j} LSC(P_{i})  
The somewhat strange generalisation of regular decomposition with a single number as a base The product factor is thus that enough space is given to the lower factors just as in the regular number decomposition. So in principle given the right LSC one ought to be able to obtain a regular panmagic square. 

prime digital equations  S_{i,j} = (a i + b j + c) mod p; a,b,c,i,j ε [0 .. m1]; p prime factor of m  
The current evidence I have suggest it possible to have a kind of pdigital equations to obtain squares with radix p digits, given these squares it is then a matter of combining these squares, just like in all other cases (note: this is a new field of investigation!) 

Colorisation patterns 
an m/p by m/p square of digit changing permutations to be applied on the p by p subsquares having radix p digits 

C_{i,j} = (=[perm(p)])_{i,j}; i,j ε [0 .. m/p  1]  
Colorisation patterns are needed to be applied on the squares obtained by means of forementioned prime digital equatations, depending on future investigation the above might be quantized further. 

LSC_{i,j} = S_{i,j}(C_{i,j}) i,j ε [0 .. m1]  
Carpet Colorisation 
Carpet colorisation (see order 9 investigation) might also here be applied to obtain LSC's not obtainable by proposed prime digital scheme 

The method discussed above notwithstanding, notice the prime order investigation, according to the obtained numbers, which matches amounts obtained by others, the outlined parameter method results in all possible LSC's for prime order p, thus obtained LSC's pasted onto a m/p by m/p square of p by p cells, thus the needed panmagic m by m LSC for order m is obtained. Of course carpet colorisation is applicable independently on each p by p subblock. When p = 3 things are a bit trickier, however the obtained order 9 LSC's (see database) which can be bordered to obtain order m LSC's for primefactor 3. Thus obtained LSC's need simply be combined and checked for regularity. 

Pandiagonal bordering 
In case the trivial border (border with only '1') is aplicable (as in case of the primefactor 3 lSC's) one can place in 1/8 of the border a random selection of patterns, and distribute these patterns over the rest of the border compatible with the "pandiagonal addon", thus obtained a pandiagonal LSC. 

note: 
Aside from using pandiagonal LSC's it is possible to have nonpandiagonal LSC's as components for latin squares, one prime factor LSC's off sums can be compensated by higher prime factors LSC's multiplied by this prime factor. Evidence for this I have however the obtained Latin Square is also obtainable by regular pandiagonal LSC's, further investigation is needed to study this peculiar fact probably one need to decompose latin squares by highest prime factor to the lowest, to avoid nonpandiagonal LSC's altogether (as said this needs further study to be certain) 