The Magic Encyclopedia ™

Pan diagonal hypercubes of prime order
{note: investigative article}
(by Aale de Winkel)


(Acknowledgement: Closer analisis I realized that the method described below is a minor variation of <John R Hendricks>'s "digit equations". Analitic coordinates used in stead of regular)

Below a generalisation of the panmagic square investigation is presented, as novel theories (such as here presented) go, it is quite possible that some assumptions made are faulty. The reader is invited to study the below with carefull scruteny and show me the flaws (preferably with some counter example)

The main portion does not concider the magicness of the n-agonals and thus results in {diagonal monagonal} hypercubes, recently a somewhat difficult to formulate, but in practice rather simple condition on some digit changing permutations formulate the {diagonal magic} hypercube. The exact amount is curently not yet computed due to some illusive factors which are yet to be determined. For {perfect} hypercubes see the other article.

NOTE: though presentlyused for the {pandiagonal monagonal [!]} the {pandiagonal magic} and the {perfect} hypercubes of prime order, loosening or restricting conditions on parameters it is quite possible the construction method can be used to find other types of hypercubes. Also it is quite possible to apply different digit changing permutation along each 1-agonal direction, a possibility I'll investigate at some future date.

prime order pan diagonal hypercubes
basic ingredients of pandiagonal hypercubes
Latin hypercube generating formula LH(aj) factor: F = ((m-3)/2n-1)	Latin hypercubes obtained by formula
	LH(aj): LH[ji] = (j=0∑naj ji) % m ; j ε [0,..,n-1]; i ε [0,..,m-1]; aj < aj+1; a0 = 1; aj = 2 .. (m - 1)/2
	The latin hypercubes obtained by the above formula are in normalized position due to the condition aj < aj+1 (can't be equal because that spoils pandiagonality). a0 = 1 because because of digit changing, thus parameters define the LH's structure the range of aj avoids pan-flip variants introduces by parameter range (m+1)/2 .. m-1.
	((m - 3)/2n-1)
	m \ n	2	3	4	5	6	7
	5 7 11 13 17 19	1 2 4 5 7 8	0 1 6 10 21 28	0 0 4 10 35 56	0 0 1 5 35 70	0 0 0 1 21 56	0 0 0 0 7 28
pandiagonal hypercubes
basic pandiagonal hypercubes factor: G = F (F 2(n-1) - 1 n-1) (n-1)!	the basic pandiagonal hypercubes
	H(aj,i) = i=0∑n-1 mn-i-1 LH(aj,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 x-axis as is there are 2n-1 panflips, resulting in 2n-1 F posibilities of which 1 is already chosen, and n-1 LH's need be randomly selected, this explains the listed factor. The factor of (n-1)! is due to reordering of the lower components
	F (F 2(n-1) - 1 n-1) (n-1)! (note: values ?? too large for excell)
	m \ n	2	3	4	5	6	7
	5 7 11 13 17 19	1 6 28 45 91 120	0 6 3.036 14.820 142.926 341.880	0 0 107.880 4.744.740 ?? ??	0 0 32.760 180.300.120 ?? ??	0 0 0 20.389.320 ?? ??	0 0 0 0 ?? ??
pandiagonal hypercubes factor: m!nG/(2m)n= G((m-1)!/2)n	the pandiagonal hypercubes
	{pandiagonal monagonal [!]}
	H(aj,i) = i=0∑n-1 mn-i-1 LH(aj,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: 2n (reflection) ; mn (panrelocation)
pandiagonal magic hypercubes factors (see text): GPm! GN(m-1)! (G - GP + GN)*(?) The various qualities determine possible dividing factors	the pandiagonal magic hypercubes
	{pandiagonal magic}
	The hypercubes n-agonals are given by [00,k0,l(m-1)] <01,k1,l-1> if the parameters (aj) if thus that: (a0 01 + ∑ak k1 - ∑al l1) % m = 0 the digit (∑al (m-1)) % m needs to be changed to (m-1)/2 since that digit is singular on the given n-agonal, the change to (m-1)/2 the n-agonal sums to the magic constant since the sum then also is m(m-1)/2
	The parametervector that combines with a n-agonal pathfinder into a multiple of m gives an n-agonal with a single digit, changing that digit into (m-1)/2 the latin hypercube n-agonal sums to m (m-1) / 2 and thus to its magic sum this pathfinder corresponds with a corner [ci] c1 = m-1 if Pfi = -1; c1 = 0 if Pfi = 1 the digit to change to (m-1)/2 thus is (i=0∑naici) % m the permutation =[(m-1)/2|d,p{{0..m-1}\(m-1)/2] changes the digit 'd' to (m-1)/2 while the remainder p[{0..m-1}\(m-1)/2] applies to all but the digit d of course This only applies to parametervectors that combine multiple m with one of the n-agonal pathfinders the other parmetervectors no restrictions are neccessairy future analysis might reveal a factor which is sligthly less then the (m!)n above though the panrelocation factor can't be used,currenty I hold the figure therefore at for order 7: (4 * 6!3) = 3,977,165,760
	Of the F latin hypercubes FP have all n-agonals showing all digits and the part FN satisfy the 0%m affliction and have single digit n-agonals which gives: GP = FP(FP 2(n-1) - 1 n-1) (n-1)! {pan n-agonal pan diagonal monagonal} hypercubes while: GN = FN(FN 2(n-1) - 1 n-1) (n-1)! {n-agonal pan diagonal monagonal} hypercubes when each component are associated with careful chosen digit changers G - GP - GN {n-agonal pan diagonal monagonal} hypercubes have both types of components. (calculating the total of these is thus more difficult) (FP and FN are yet to be determined)
	FP factors. Unknown counting argument (numbers manually obtained)
	m \ n	2	3	4	Cube numbers where obtained in the {perfect} hypercube article. The tesseract numbers I'll have to correct for the non-pantriagonal tesseracts also in future upload the non-panquadragonal probably will get investigated
	5 7 11 13 17 19	1 2 4 5 7 8	0 0 3 6 15 21	0 0 0? 0? 2? 5?

The studies above correspond with the prime order panmagic square investigation, since for squares pandiagonal and perfect are the same no further investigation is needed there. The perfect treshold of 2ⁿ suggest none of the three order 7 cubes is perfect, a close look at the 1.518 order 11 cubes might reveal a dividing factor to derive the number of perfect cubes (supposing there is one)

SAMPLE: the 6 basic order 7 pandiagonal cubes
{pandiagonal monagonal}
The 4 possible LH's
LH(1,2,3)	LH(1,2,4)	LH(1,5,3)	LH(1,5,4)
<LH(1,2,3),LH(1,2,4),LH(1,5,3)> <LH(1,2,3),LH(1,5,3),LH(1,2,4)>	<LH(1,2,3),LH(1,2,4),LH(1,5,4)> <LH(1,2,3),LH(1,5,4),LH(1,2,4)>	<LH(1,2,3),LH(1,5,3),LH(1,5,4)> <LH(1,2,3),LH(1,5,4),LH(1,5,3)>	note that '5' = '2' panflipped and '4' = '3' panflipped and '6' would be '1' panflipped.
The upload onto the database of the above mentioned show clearly the correlations LH(1,2,4) = LH(1,2,3)~2 LH(1,5,3) = LH(1,2,3)~4 and LH(1,5,4) = LH(1,2,3)~6 Thus showing that all order 7 cubes are based on the single latin cube LH(1,2,3)
{pandiagonal magic}
The 4 possible LH's (and digit chaging permutations)
LH(1,2,3)=[3|4,P[0,1,2,4,5,6]]	LH(1,2,4)=[3|0,P[0,1,2,4,5,6]]	LH(1,5,3)=[3|6,P[0,1,2,4,5,6]]	LH(1,5,4)=[3|2,P[0,1,2,4,5,6]]
LH(1,2,3) has single digit along the <1,1,-1> direction (1+2-3)%7 = 0, the corner position is [6,6,0] thus the digit to change to 3 is (1*6+2*6+3*0)%7 = 18%7 = 4 LH(1,2,4) has single digit along the <1,1,1> direction (1+2+3)%7 = 0, the corner position is [0,0,0] thus the digit to change to 3 is (1*0+2*0+4*0)%7 = 0 LH(1,5,3) has single digit along the <-1,1,1> direction (-1+5+3)%7 = 7%7 = 0, the corner position is [6,0,0] thus the digit to change to 3 is (1*6+5*0+3*0)%7 = 6 LH(1,5,4) has single digit along the <1,-1,1> direction (1-5+4)%7 = 0, the corner position is [0,6,0] thus the digit to change to 3 is (1*0+5*6+4*0)%7 = 30%7 = 2

Note: the counting arguments above did not taken into account that in case used parameters are lineair dependent in all the used latin hypercubes in a directional pair, doubly numbers appear. So the actual number are a bit lower then the Ca's result, by what factor I currrently don't know but will be investigated some future date.