Home › Forum › Magic Cubes of Prime Numbers › Unsolved Problems “Primes Magic Squares”
- Questo topic ha 11 risposte, 1 partecipante ed è stato aggiornato l'ultima volta 8 anni, 9 mesi fa da Natalia Makarova.
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Maggio 19, 2014 alle 2:33 am #123Natalia MakarovaPartecipante
I tell here about unsolved problems in the field of “Magic Squares”.
Problem # 1
“Pandiagonal Magic Squares of Prime Numbers”The competition was held on this issue (2013):
http://www.azspcs.net/Contest/PandiagonalMagicSquaresLots of good solutions found winner Jarek Wroblewski.
However the problem is not completely solved.For example, the optimal solution found for n = 7:
3,7,173,223,17,197,113, 181,211,11,79,131,23,97, 43,41,149,89,137,191,83, 233,103,107,73,127,31,59, 29,167,101,19,199,67,151, 5,47,139,179,109,61,193, 239,157,53,71,13,163,37
Magic constant S = 733.
The solution by Jarek Wroblewski for n = 8:
5,37,107,157,229,311,271,131, 73,239,397,173,197,13,113,43, 293,313,11,97,181,149,103,101, 223,83,151,71,53,241,233,193, 167,127,179,31,277,317,61,89, 67,59,139,281,269,109,17,307, 137,349,257,211,19,29,199,47, 283,41,7,227,23,79,251,337
Magic constant S = 1248.
It is unknown whether this solution smallest.
And then for all n> 7 we have the same issue.Jarek Wroblewski found solutions also for n> 20:
http://www.math.uni.wroc.pl/~p-k1g4/PMS/The best solution you can send to the website:
http://www.azspcs.net/Contest/PandiagonalMagicSquares/FinalReportand here:
http://www.primepuzzles.net/puzzles/puzz_663.htmAnother similar problem:
Pandiagonal magic squares of consecutive primes
http://www.primepuzzles.net/puzzles/puzz_723.htmI invite everyone to take part in solving these problems.
Links
1. Discussion on scientific forum (in Russian)
http://dxdy.ru/topic12959.html
http://dxdy.ru/topic73817.html
2. Natalia Makarova. Unconventional pandiagonal squares (series of articles). Part 1 here:
http://www.natalimak1.narod.ru/pannetr.htm
3. The algebraic theory of diabolic magic squares. By Barkley Rosser and R. J. Walker
http://yadi.sk/d/tl-_Ab-o5AYhS- Questo topic è stato modificato 9 anni, 10 mesi fa da Natalia Makarova.
Maggio 21, 2014 alle 6:05 am #125Natalia MakarovaPartecipanteProblem # 2
Most Perfect Magic Squares of Prime NumbersDefinitions can be found here:
http://www.primepuzzles.net/puzzles/puzz_671.htmMy solutions for n = 6 and n = 8:
149, 9161, 2309, 6701, 2609, 8861 9067, 1483, 6907, 3943, 6607, 1783 4139, 5171, 6299, 2711, 6599, 4871 3229, 7321, 1069, 9781, 769, 7621 5987, 3323, 8147, 863, 8447, 3023 7219, 3331, 5059, 5791, 4759, 3631
Magic constant S=29790.
19, 5923, 1019, 4423, 4793, 1277, 3793, 2777 4877, 1193, 3877, 2693, 103, 5839, 1103, 4339 499, 5443, 1499, 3943, 5273, 797, 4273, 2297 5297, 773, 4297, 2273, 523, 5419, 1523, 3919 1213, 4729, 2213, 3229, 5987, 83, 4987, 1583 5903, 167, 4903, 1667, 1129, 4813, 2129, 3313 733, 5209, 1733, 3709, 5507, 563, 4507, 2063 5483, 587, 4483, 2087, 709, 5233, 1709, 3733
Magic constant S=24024.
It is not known whether these solutions minimal.
I have not found solutions for n>8.It is most perfect magic square of order 10 of different natural numbers:
1, 448, 12, 441, 6, 435, 14, 446, 7, 440 418, 33, 407, 40, 413, 46, 405, 35, 412, 41 342, 107, 353, 100, 347, 94, 355, 105, 348, 99 201, 250, 190, 257, 196, 263, 188, 252, 195, 258 156, 293, 167, 286, 161, 280, 169, 291, 162, 285 15, 436, 4, 443, 10, 449, 2, 438, 9, 444 404, 45, 415, 38, 409, 32, 417, 43, 410, 37 356, 95, 345, 102, 351, 108, 343, 97, 350, 103 187, 262, 198, 255, 192, 249, 200, 260, 193, 254 170, 281, 159, 288, 165, 294, 157, 283, 164, 289
See my article:
http://www.natalimak1.narod.ru/sovnetr.htm
http://www.natalimak1.narod.ru/netradic1.htmLuglio 28, 2014 alle 6:53 pm #209Natalia MakarovaPartecipantePublished a new puzzle
http://www.primepuzzles.net/puzzles/puzz_749.htmMaybe this will help us solve this problem.
Dear colleagues, connect, please!Luglio 31, 2014 alle 7:44 am #210Natalia MakarovaPartecipanteThis is the minimal pandiagonal square of order 4 of consecutive primes:
170693941183817 170693941183933 170693941183949 170693941183981 170693941183979 170693941183951 170693941183847 170693941183903 170693941183891 170693941183859 170693941184023 170693941183907 170693941183993 170693941183937 170693941183861 170693941183889
S=2731103058942720
Author Max Alekseyev.
See post on the forum in Russia
http://dxdy.ru/post891839.html#p891839
This problem is solved!I invite all to solve the following problem: find the minimal pandiagonal square of order 5 of consecutive primes.
Agosto 1, 2014 alle 4:01 pm #211Natalia MakarovaPartecipanteS=2731103058942720
Sorry, this is a mistake.
correct:
S=682775764735680Agosto 13, 2014 alle 12:56 am #216Natalia MakarovaPartecipanteHello, dear colleagues!
I bring to your attention an interesting report by my colleague Max Alekseyev on conference MathFest 2014
“An efficient backtracking method for solving a system of
linear equations over a finite set with application for construction of magic squares”http://home.gwu.edu/~maxal/MathFest2014_slides.pdf
- Questa risposta è stata modificata 9 anni, 7 mesi fa da Natalia Makarova.
Agosto 30, 2014 alle 3:59 am #221Natalia MakarovaPartecipanteHello, dear colleagues!
We know two pandiagonal squares of order 4 of consecutive primes.
This is a minimal solution by Max Alekseyev:170693941183817 170693941183933 170693941183949 170693941183981 170693941183979 170693941183951 170693941183847 170693941183903 170693941183891 170693941183859 170693941184023 170693941183907 170693941183993 170693941183937 170693941183861 170693941183889
S = 682775764735680
The second solution is from J. Wroblewski and J. К. Andersen:
320572022166380833 320572022166380921 320572022166380849 320572022166380917 320572022166380909 320572022166380857 320572022166380893 320572022166380861 320572022166380911 320572022166380843 320572022166380927 320572022166380839 320572022166380867 320572022166380899 320572022166380851 320572022166380903
S = 1282288088665523520
See
http://dxdy.ru/post751928.html#p751928
http://www.primepuzzles.net/conjectures/conj_042.htmRequired to find pandiagonal square of order 4 of consecutive primes with a magic constant 682775764735680 < S < 1282288088665523520.
Settembre 6, 2014 alle 9:12 am #222Natalia MakarovaPartecipantePandiagonal magic squares of consecutive primes
n=4 (minimal, author Max Alekseyev)
{170693941183817: 0, 30, 42, 44, 72, 74, 86, 90, 116, 120, 132, 134, 162, 164, 176, 206} 170693941183817 170693941183933 170693941183949 170693941183981 170693941183979 170693941183951 170693941183847 170693941183903 170693941183891 170693941183859 170693941184023 170693941183907 170693941183993 170693941183937 170693941183861 170693941183889 S=682775764735680
n=5, solution is unknown.
This is my solution with 5 errors:
{13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113} 13 47 111* 89 53 79 107 29 43 55* 59 51* 23 97 83 41 73 113 67 19 121* 35* 37 17 103 S=313
n=6 (minimal)
{67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251} 67 193 71 251 109 239 139 233 113 181 157 107 241 97 191 89 163 149 73 167 131 229 151 179 199 103 227 101 127 173 211 137 197 79 223 83 S=930
n=7, solution is unknown.
I tried to solve this problem for the next array of consecutive primes:
{7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239}
This is my solution with 5 errors:
97 167 233 179 11 103 7 59 19 71 163 101 211 173 157 137 89 181 23 83 127 113 131 139 109 121* 123* 61 191 67 53 17 229 47 193 -17* 197 227 107 239 31 13 197* 79 -15* 41 73 199 223 S=797
I think that such a solution exists.
Dear Colleagues!
I ask you to take part in solving this problem.See more information:
http://dxdy.ru/post751921.html#p751921
http://dxdy.ru/topic87170.html
http://www.primepuzzles.net/puzzles/puzz_723.htm
http://www.primepuzzles.net/puzzles/puzz_731.htm
http://www.primepuzzles.net/puzzles/puzz_736.htmGiugno 2, 2015 alle 12:12 pm #363Natalia MakarovaPartecipanteDear colleagues!
Once again I remind interesting problem:
Published sequence in the OEIS
“Smallest magic constant of most-perfect magic squares of order 2n composed of distinct prime numbers”
https://oeis.org/A258082Now we need to find a solution for n > 8.
I invite all to solve this complex problem.
Giugno 2, 2015 alle 1:37 pm #364Natalia MakarovaPartecipanteThe scheme for most perfect square of order 10
Here k – complementarity constant, S = 5k, if S – magic constant of a square.
The general formula of most perfect square of order 10
K = (2*X(36)+ X(20)- X(15)-2*X(45)+2*X(10)+2*X(26))/2 X(1) = (6*X(36)+5*X(20)- X(15)-2*X(45)+2*X(10)-4*X(47)-4*X(48)-4*X(49)+6*X(26))/4 X(11) = (2*X(36)-5*X(20)-3*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(12) = X(20)+ X(45)- X(47) X(13) = 2*X(36)- X(15)-3*X(45)+2*X(10)- X(48)+2*X(26) X(14) = X(20)+ X(45)- X(49) X(16) = (2*X(36)- X(20)-7*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(17) = X(15)+ X(45)- X(47) X(18) = 2*X(36)+ X(20)-2*X(15)-3*X(45)+2*X(10)- X(48)+2*X(26) X(19) = X(15)+ X(45)- X(49) X(2) = - X(45)+ X(10)+ X(47) X(21) = -(-2*X(36)-5*X(20)-3*X(15)-2*X(45)-2*X(10)+4*X(47)+4*X(48)+4*X(49))/2 X(22) = (2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+8*X(47)+4*X(48)+4*X(49)+6*X(26))/4 X(23) = -(2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+4*X(47)+4*X(49)+6*X(26))/4 X(24) = (2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+4*X(47)+4*X(48)+8*X(49)+6*X(26))/4 X(25) = (6*X(36)+5*X(20)- X(15)-2*X(45)+6*X(10)-4*X(47)-4*X(48)-4*X(49)+2*X(26))/4 X(27) = (6*X(36)+5*X(20)- X(15)-6*X(45)+6*X(10)-4*X(48)-4*X(49)+2*X(26))/4 X(28) = -(6*X(36)+5*X(20)- X(15)-6*X(45)+6*X(10)-4*X(47)-8*X(48)-4*X(49)+2*X(26))/4 X(29) = (6*X(36)+5*X(20)- X(15)-6*X(45)+6*X(10)-4*X(47)-4*X(48)+2*X(26))/4 X(3) = X(45)- X(10)+ X(48) X(30) = (2*X(36)-5*X(20)-7*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+6*X(26))/4 X(31) = (-3*X(20)-5*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/2 X(32) = -(-2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+8*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(33) = (6*X(36)- X(20)-11*X(15)-18*X(45)+10*X(10)+4*X(47)+4*X(49)+10*X(26))/4 X(34) = -(-2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+4*X(47)+4*X(48)+8*X(49)+2*X(26))/4 X(35) = (-2*X(36)- X(20)-3*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(37) = (-2*X(36)- X(20)-3*X(15)-2*X(45)+2*X(10)+4*X(48)+4*X(49)+2*X(26))/4 X(38) = (10*X(36)+5*X(20)- X(15)-6*X(45)+6*X(10)-4*X(47)-8*X(48)-4*X(49)+6*X(26))/4 X(39) = (-2*X(36)- X(20)-3*X(15)-2*X(45)+2*X(10)+4*X(47)+4*X(48)+2*X(26))/4 X(4) = - X(45)+ X(10)+ X(49) X(40) = -(-2*X(36)-5*X(20)-7*X(15)-6*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(41) = -(2*X(36)-5*X(20)-7*X(15)-10*X(45)+2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(42) = 2*X(36)-2*X(15)-4*X(45)+2*X(10)+ X(47)+2*X(26) X(43) = -2*X(36)+2*X(15)+4*X(45)-2*X(10)+ X(48)-2*X(26) X(44) = 2*X(36)-2*X(15)-4*X(45)+2*X(10)+ X(49)+2*X(26) X(46) = (6*X(36)+5*X(20)- X(15)-6*X(45)+6*X(10)-4*X(47)-4*X(48)-4*X(49)+6*X(26))/4 X(5) = 2*X(36)-2*X(15)-2*X(45)+ X(10)+2*X(26) X(50) = 2*X(36)-2*X(15)-3*X(45)+2*X(10)+2*X(26) X(6) = -(2*X(36)-5*X(20)-7*X(15)-6*X(45)-2*X(10)+4*X(47)+4*X(48)+4*X(49)+2*X(26))/4 X(7) = 2*X(36)-2*X(15)-3*X(45)+ X(10)+ X(47)+2*X(26) X(8) = -2*X(36)+2*X(15)+3*X(45)- X(10)+ X(48)-2*X(26) X(9) = 2*X(36)-2*X(15)-3*X(45)+ X(10)+ X(49)+2*X(26)
We have 9 free variables and 42 dependent variables.
Giugno 4, 2015 alle 11:34 am #365Natalia MakarovaPartecipanteI tried to create a most perfect square of order 10 with magic constant S = 240240
(K = 48048), using the general formula.
We have a large array of primes, which consists of 972 complementary pairs:19 31 67 71 79 97 101 109 131 137 167 179 191 211 229 239 241 251 257 269 271 307 311 331 337 347 349 367 389 409 419 439 449 457 467 479 521 . . . . . . . . . . . . . . . . . . . . . . . . . 47527 47569 47581 47591 47599 47609 47629 47639 47659 47681 47699 47701 47711 47717 47737 47741 47777 47779 47791 47797 47807 47809 47819 47837 47857 47869 47881 47911 47917 47939 47947 47951 47969 47977 47981 48017 48029
However, the result was bad:
9712 28661 13627 45341 10477 34316 4057 38231 20737 35081 20744 36979 16829 20299 19979 31324 26399 27409 9719 30559 33851 4522 37766 21202 34616 10177 28196 14092 44876 10942 37589 20134 33674 3454 36824 14479 43244 10564 26564 13714 16214 22159 20129 38839 16979 27814 10559 31729 27239 28579 13732 43991 9817 27311 12967 38336 19387 34421 2707 37571 16724 21649 20639 38329 17489 27304 11069 31219 27749 28069 37871 19852 33956 3172 37106 14197 43526 10282 26846 13432 33569 4804 37484 21484 34334 10459 27914 14374 44594 11224 20234 37489 16319 20809 19469 31834 25889 27919 9209 31069
K = 48048, S = 240240
We have in this solution 53 primes and 47 are not prime numbers.
- Questa risposta è stata modificata 8 anni, 10 mesi fa da Natalia Makarova.
Giugno 5, 2015 alle 6:20 pm #367Natalia MakarovaPartecipanteProgress!
8755 28661 13627 46298 10477 33359 4057 38231 21694 35081 21701 36979 16829 19342 19979 32281 26399 27409 8762 30559 31937 5479 36809 23116 33659 10177 27239 15049 44876 11899 39503 19177 34631 1540 37781 14479 44201 9607 26564 12757 15257 22159 20129 39796 16979 26857 10559 31729 28196 28579 14689 43991 9817 26354 12967 39293 19387 34421 1750 37571 15767 21649 20639 39286 17489 26347 11069 31219 28706 28069 37871 20809 32999 3172 36149 16111 42569 11239 24932 14389 33569 3847 38441 21484 35291 8545 28871 13417 46508 10267 21191 37489 16319 19852 19469 32791 25889 27919 8252 31069
K = 48048, S = 240240
This solution has 65 primes and 35 are not prime numbers.
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