The Magic Encyclopedia ™

Multiplet stacking of panmagic squares
{note: investigative article}
(by Aale de Winkel)

Once again <George Chen> sent me an order 15 panmagic square which was build using a permutation of the 15 digits 1..15 where each pentlet was summing up to 40. One such permutation was to permute the rows and one to permute the columns of the "Natural square" M_i,j = m * i + j + 1 i,j = 0 .. m-1. This permutation rearanges the set of number in the order m natural square in q by q blocks of numbers (m = p * q). I supposed those q² numbers could be used to substitute the numbers of a regular order q panmagic square, the excell order 15 triplet stacking spreadsheet showed me the validity of this idea and is currently my evidence for the following generalisation of this sample.
NOTE: currently the reader must hold these as preliminairy remarks, due to other things to do I might need some time to check these notes against mantiond spreadsheet and aquire further evidence on other orders. The reader is invited to create simular spreadsheet to the above or improve upon it

Multiplet stacking of panmagic squares
S_m,k,l = N_m,k,l[P_v,P_h,S_p,k/q,l/q]

The above formula formulates the substitution of the order p panmagic square numbers by the
numbers in the p by p blocks of the square obtained from the natural square permuted by the
multiplet permutations, a square which can be formulated as N_i,j = m * P_v[i] + P_h[j] + 1; i,j = 0.m-1

Stacker S_p
factor: (p!)²
(preliminairy) The order p square which stacks the order q square
The order p square which is needed to stack panmagic square to create panmagic square needs
to be of "skew magic" quality (ie: every (broken) 2-agonal sums to the same sum)
curently I hold these as row/column permutations of the "natural squares"
(N_i,j = i * p + j ; i,j = 0 .. p-1
suggesting the factor of (p!)²
NOTE: currently not yet determined wether the quality reallly need to be skew magic
it might well be that the stackers quality doesn't mattter, raising the factor to (p²)!

Stackey S_q
factor: (see order q) The order q square which stacked by the order p square
The panmagic square to be stacked adheres of course to the order q theory, current evidence
the numbers of the stackey merely needs to be replaced by the numbers of each q by q square
of the permuted natural square in order for the order m square

Multiplet permutation
factor:
[^pq_p|q;
∑ q(pq-1)/2]
symbolically a permutation of the numbers [0..pq], where each set of q numbers sum to q(pq-1)/2]
The Multiplet permutation is hereby defined as a permutation of the m = pq numbers which
split into p sets of q numbers each set summing to the same sum, since
_i=0∑^m-1 i = m(m-1)/2 = pq(pq-1)/2 devided over the p sets each sum
therefore must be q(pq-1)/2, as for the counting argument I sybolically denote by [^pq_p|q; ∑ s]
the amount of p-plet of q numbers where each set of q numbers are summing to the sum s,
a number I currently have no way to calculate for lack of a formula
(it ought to be somewhat simular to the formula for the binomial factor, if there is one)

Note: The numbers [^pq_p|q; ∑ s] is curently under investigation, the only numbers I
currently know of are [¹⁵_5|3; ∑ 24] = 10 and [¹⁵_3|5; ∑ 40] = 305.
Data uploaded onto the database.

Note: For all involve 1-agonal directions one needs a multiplet permutation the corresponding
factor thus multiply. George suggest 6 alternates for using the natural square splitup of the
order 15 squares in bloks 3 by 3, 3 by 5, 5 by 3 and 5 by 5 blocks (need to investigate how to
incorporate these into this formalism. since order 3 squares are not pandiagonal, and 3 by 5
pandiagoannlity is somewhat foggy this will take some time.

At first glance the above seems to suggest that with q = 2 and odd p doubly odd order squares might be reachably
however p(pq-1)/2 is a half integer, henche there are no duet permutations in the doubly odd order case. This
safeguard the non existeance of doubly odd order panmagic square
Currently I hold the method can be used for any compound order, save of course the doubly odd orders

Note: The above notes are preliminairy at best, much needs to be investigated by programmatic means

Multiplet stacking of panmagic squares
S_m,k,l = N_m,k,l[P_v,P_h,S_p,k/q,l/q]
The above formula formulates the substitution of the order p panmagic square numbers by the numbers in the p by p blocks of the square obtained from the natural square permuted by the multiplet permutations, a square which can be formulated as N_i,j = m * P_v[i] + P_h[j] + 1; i,j = 0.m-1
Stacker S_p factor: (p!)² (preliminairy)	The order p square which stacks the order q square
Stacker S_p factor: (p!)² (preliminairy)	The order p square which is needed to stack panmagic square to create panmagic square needs to be of "skew magic" quality (ie: every (broken) 2-agonal sums to the same sum) curently I hold these as row/column permutations of the "natural squares" (N_i,j = i * p + j ; i,j = 0 .. p-1 suggesting the factor of (p!)² NOTE: currently not yet determined wether the quality reallly need to be skew magic it might well be that the stackers quality doesn't mattter, raising the factor to (p²)!
Stackey S_q factor: (see order q)	The order q square which stacked by the order p square
Stackey S_q factor: (see order q)	The panmagic square to be stacked adheres of course to the order q theory, current evidence the numbers of the stackey merely needs to be replaced by the numbers of each q by q square of the permuted natural square in order for the order m square
Multiplet permutation factor: [^pq_p\|q; ∑ q(pq-1)/2] symbolically	a permutation of the numbers [0..pq], where each set of q numbers sum to q(pq-1)/2]
	The Multiplet permutation is hereby defined as a permutation of the m = pq numbers which split into p sets of q numbers each set summing to the same sum, since _i=0∑^m-1 i = m(m-1)/2 = pq(pq-1)/2 devided over the p sets each sum therefore must be q(pq-1)/2, as for the counting argument I sybolically denote by [^pq_p\|q; ∑ s] the amount of p-plet of q numbers where each set of q numbers are summing to the sum s, a number I currently have no way to calculate for lack of a formula (it ought to be somewhat simular to the formula for the binomial factor, if there is one)
Note: The numbers [^pq_p\|q; ∑ s] is curently under investigation, the only numbers I currently know of are [¹⁵_5\|3; ∑ 24] = 10 and [¹⁵_3\|5; ∑ 40] = 305. Data uploaded onto the database.
Note: For all involve 1-agonal directions one needs a multiplet permutation the corresponding factor thus multiply. George suggest 6 alternates for using the natural square splitup of the order 15 squares in bloks 3 by 3, 3 by 5, 5 by 3 and 5 by 5 blocks (need to investigate how to incorporate these into this formalism. since order 3 squares are not pandiagonal, and 3 by 5 pandiagoannlity is somewhat foggy this will take some time.
At first glance the above seems to suggest that with q = 2 and odd p doubly odd order squares might be reachably however p(pq-1)/2 is a half integer, henche there are no duet permutations in the doubly odd order case. This safeguard the non existeance of doubly odd order panmagic square Currently I hold the method can be used for any compound order, save of course the doubly odd orders