Multiplet stacking of panmagic squares | ||
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S_{m,k,l} = N_{m,k,l}[P_{v},P_{h},S_{p,k/q,l/q}] |
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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 |
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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 (p^{2})! |
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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 |
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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) |
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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. |
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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. |
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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 |