Home › Forum › Ultra Magic Squares of prime numbers › Ultra magic square 9th order
- Questo topic ha 17 risposte, 1 partecipante ed è stato aggiornato l'ultima volta 8 anni, 11 mesi fa da Natalia Makarova.
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Marzo 7, 2015 alle 10:25 am #306Natalia MakarovaPartecipante
The scheme ultra magic square 9th order:
x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14 x15 x16 x17 x18 x19 x20 x21 x22 x23 x24 x25 x26 x27 x28 x29 x30 x31 x32 x33 x34 x35 x36 x37 x38 x39 x40 k/2 k-x40 k-x39 k-x38 k-x37 k-x36 k-x35 k-x34 k-x33 k-x32 k-x31 k-x30 k-x29 k-x28 k-x27 k-x26 k-x25 k-x24 k-x23 k-x22 k-x21 k-x20 k-x19 k-x18 k-x17 k-x16 k-x15 k-x14 k-x13 k-x12 k-x11 k-x10 k-x9 k-x8 k-x7 k-x6 k-x5 k-x4 k-x3 k-x2 k-x1
Here k is constant associativity of square, S = 9k/2, if S – magic constant square.
The general formula of ultra magic square 9th order
X(1) = (2*X(32)-2*X(34)+2*X(16)-2*X(18)- K+2*X(24)+2*X(8)-2*X(26)+2*X(40))/2 X(11) = - X(32)- X(6)- X(34)+ X(35)-2*X(36)+2*X(10)- X(16)- X(12)-2*X(18)- X(7)+ X(20)-2*X(2) - X(22)+4*K- X(26)- X(27)+ X(28)- X(29)+2*X(30)+2*X(40) X(14) = - X(32)- X(6)- X(16)- X(12)- X(2)- X(22)+5*K- X(24)- X(8)- X(4) X(15) = 3*X(32)+3*X(6)+ X(34)- X(35)+3*X(36)-4*X(10)- X(13)+2*X(16)+2*X(12)+2*X(18)+2*X(7) -2*X(20)+4*X(2)+3*X(22)-9*K+ X(24)+2*X(8)+2*X(26)+2*X(27)- X(28) + X(29)-3*X(30)+ X(4)-3*X(40) X(17) = (-2*X(32)-2*X(6)-2*X(36)+2*X(10)-2*X(16)-2*X(12)-2*X(18)-2*X(7)+2*X(20) -2*X(2)-2*X(22)+9*K-2*X(8)-2*X(26)-2*X(27)+2*X(30)+2*X(40)) /2 X(19) = -(-2*X(6)-4*X(34)+2*X(35)-4*X(36)+6*X(10)-6*X(18)-2*X(7)+4*X(20)+2*X(3) -2*X(2)-2*X(22)-3*K+2*X(24)+2*X(8)-4*X(26)-2*X(27)+4*X(28) -2*X(29)+4*X(30)+2*X(4)+6*X(40)) /2 X(21) = 2*X(32)+2*X(6)- X(35)+ X(36)-2*X(10)- X(13)+2*X(16)+ X(12)+2*X(7)- X(20)+2*X(2)+ X(22) -4*K+ X(24)+2*X(8)-2*X(30)+ X(4)- X(40) X(23) = - X(32)- X(6)+ X(34)+ X(35)+ X(10)- X(16)- X(12)+ X(18)- X(7)- X(3)- X(2)- X(22)+3*K - X(24)- X(8)+ X(29)+ X(30)- X(4) X(25) = - X(32)-2*X(6)-3*X(34)+ X(35)-3*X(36)+4*X(10)+ X(13)- X(16)-4*X(18)-2*X(7)+2*X(20) +2*X(3)-2*X(2)-2*X(22)+4*K-3*X(26)-2*X(27)+2*X(28)-2*X(29) +3*X(30)+ X(4)+4*X(40) X(31) = X(32)+2*X(6)- X(35)+ X(36)-2*X(10)- X(13)+ X(16)+ X(12)+ X(18)+ X(7)- X(20)+2*X(2)+ X(22) -2*K+ X(24)+ X(8)+ X(26)+ X(27)- X(28)-2*X(30)-2*X(40) X(33) = (-4*X(32)-4*X(6)-2*X(34)-4*X(36)+4*X(10)+2*X(13)-2*X(16)-2*X(12)-2*X(18)-2*X(7) +2*X(20)-4*X(2)-2*X(22)+13*K-2*X(24)-2*X(8)-2*X(26)-2*X(27) -2*X(29)+2*X(30)+4*X(40)) /2 X(37) = - X(6)- X(34)+ X(35)- X(36)+ X(10)- X(16)+ X(12)- X(18)- X(7)+ X(20)+ X(3)- X(26)+ X(28)- X(29) + X(30)+ X(4)+ X(40) X(38) = X(34)+ X(36)- X(10)+ X(18)- X(20)+ K+ X(26)- X(28)- X(30)- X(40) X(39) = (-6*X(32)-8*X(6)-4*X(34)+4*X(35)-8*X(36)+12*X(10)+4*X(13)-4*X(16)-4*X(12)-8*X(18) -6*X(7)+6*X(20)+2*X(3)-8*X(2)-6*X(22)+17*K-2*X(24)-4*X(8) -6*X(26)-4*X(27)+4*X(28)-4*X(29)+8*X(30)+10*X(40)) /2 X(5) = -(4*X(32)+2*X(6)-2*X(34)-2*X(10)+2*X(16)+2*X(12)-2*X(18)+2*X(7)-2*X(20) +2*X(3)+4*X(2)+2*X(22)-9*K+2*X(24)+4*X(8)-2*X(26)-2*X(30) +2*X(4)) /2 X(9) = -(-2*X(32)+2*X(10)-2*X(12)+2*X(20)-2*X(2)-2*X(22)- K+2*X(30)+2*X(40))/2
We have 24 free variables (if the constant associative is set) and 16 dependent variables.
According to this formula I found the following solution with 4 errors:
2767 1429 919 3727 883 757 3607 1039 3889 223 3499 1609 1567 3877 2647 853 2239 2503 3457 283 3769 2467 307 3583*451* 1471 3229 3217 1117 1753 1933 1789 3853 1669 2143 1543 3613 2677 643 3259 2113 967 3583 1549 613 2683 2083 2557 373 2437 2293 2473 3109 1009 997 2755 3775*643* 3919 1759 457 3943 769 1723 1987 3373 1579 349 2659 2617 727 4003 337 3187 619 3469 3343 499 3307 2797 1459
K=4226, S=19017
Marzo 15, 2015 alle 9:50 am #307Natalia MakarovaPartecipanteAh! There are still a couple of wrong: (1471, 2755).
It is also a mistake.Marzo 18, 2015 alle 3:41 am #308Natalia MakarovaPartecipanteProgress!
4463 1259 1181 4079 4703 2099 1511 2213 1811 3461 1163 4583 461 2939 4091 3701 1979 941 1559 3989 4799 449 971 929 3041 3209 4373 173 1319 29 5099 3659 3629*2549 2789 4073 4133 5051 2801 3251 2591 1931 2381 131 1049 1109 2393 2633 1553 1523 83 5153 3863 5009 809 1973 2141 4253 4211 4733 383 1193 3623 4241 3203 1481 1091 2243 4721 599 4019 1721 3371 2969 3671 3083 479 1103 4001 3923 719
K=5182, S=23319
In this solution, there is only one element of the wrong – 3629 (not prime number).
I’m not mistaken?Marzo 18, 2015 alle 11:41 am #309Natalia MakarovaPartecipanteThere is a solution with magic constant 23319!
Aprile 29, 2015 alle 6:22 am #327Natalia MakarovaPartecipanteThere is a solution with magic constant 21969!
Minimal magic constant for ultra magic square of order 9 is 12249.
Now I’m trying to find a solution to this magic constant.
We have only one array of primes to produce ultra magic square of order 9 with magic constant 1224911 23 29 59 89 101 113 131 173 179 191 263 281 311 383 389 449 479 509 569 593 641 653 659 683 719 743 773 809 821 911 1013 1103 1109 1151 1163 1223 1229 1283 1289 1361 1433 1439 1493 1499 1559 1571 1613 1619 1709 1811 1901 1913 1949 1979 2003 2039 2063 2069 2081 2129 2153 2213 2243 2273 2333 2339 2411 2441 2459 2531 2543 2549 2591 2609 2621 2633 2663 2693 2699 2711
Perhaps there is a solution.
- Questa risposta è stata modificata 9 anni fa da Natalia Makarova.
Maggio 1, 2015 alle 1:28 am #329Natalia MakarovaPartecipanteI found a minimal solution for ultra magic square of order 9 with 3 errors:
59 101 2699 719 1949 2543 2039 911 1229 1619 1061 2339 1613 1499 659 653 113 2693 2531 569 311 2081 1163 2591 1013 11 1979 1913 449 509 1901 173 1289 2333 2243 1439 2579 2459 1541*2129 1361 593 1181 263 143* 1283 479 389 1433 2549 821 2213 2273 809 743 2711 1709 131 1559 641 2411 2153 191 29 2609 2069 2063 1223 1109 383 1661*1103 1493 1811 683 179 773 2003 23 2621 2663
K=2722, S=12249
I found a minimal associative square of order 9 of distinct primes
1283 311 1811 2213 1571 569 2039 1163 1289 773 653 2243 1619 2063 593 2693 383 1229 1979 1499 2699 641 821 89 809 2003 1709 1613 2531 101 131 2333 2441 2663 263 173 113 179 2711 449 1361 2273 11 2543 2609 2549 2459 59 281 389 2591 2621 191 1109 1013 719 1913 2633 1901 2081 23 1223 743 1493 2339 29 2129 659 1103 479 2069 1949 1433 1559 683 2153 1151 509 911 2411 1439
K=2722, S=12249
- Questa risposta è stata modificata 9 anni fa da Natalia Makarova.
Maggio 2, 2015 alle 7:25 pm #331Natalia MakarovaPartecipanteProgress!
In this solution two errors
191 101 1223 2711 773 2411 2243 263 2333 2663 2273 743 593 2543 1163 89 2153 29 2039 653 1151 2549 2339 911 521 23 2063 1013 2081 113 821 1289 1109 1229 2591 2003 1883* 1283 2213 281 1361 2441 509 1439 839 719 131 1493 1613 1433 1901 2609 641 1709 659 2699 2201* 1811 383 173 1571 2069 683 2693 569 2633 1559 179 2129 1979 449 59 389 2459 479 311 1949 11 1499 2621 2531
K=2722, S=12249
Maggio 7, 2015 alle 4:09 am #335Natalia MakarovaPartecipanteI found a solution with magic constant S = 19269.
Now I’m trying to find a solution with magic constant S = 18729.
I have a few solutions with one error.For example
2549 89 3989 383 1193 3323 2543 1409 3251 1499 701 4079 2609 2903 1433 3851 503 1151 4049 2099 569 3803 1949 941 3929 71 1319 1373 3923 3209 1823 29 863 401 3719 3389 1721 971 959* 23 2081 4139 3203 3191 2441 773 443 3761 3299 4133 2339 953 239 2789 2843 4091 233 3221 2213 359 3593 2063 113 3011 3659 311 2729 1259 1553 83 3461 2663 911 2753 1619 839 2969 3779 173 4073 1613
K = 4162, S = 18729
Maggio 7, 2015 alle 3:49 pm #336Natalia MakarovaPartecipanteThere is a solution with magic constant S = 18729 !
Maggio 7, 2015 alle 7:43 pm #337Natalia MakarovaPartecipanteNow I’m trying to find a solution with magic constant S = 16407.
We have the following array of 101 primes to produce this ultra magic square:23 29 53 89 107 113 179 197 233 239 257 317 347 389 443 479 509 557 563 647 677 683 719 743 809 827 857 947 953 983 1013 1097 1103 1187 1223 1229 1289 1307 1373 1409 1433 1439 1493 1559 1583 1607 1619 1667 1697 1733 1823 1913 1949 1979 2027 2039 2063 2087 2153 2207 2213 2237 2273 2339 2357 2417 2423 2459 2543 2549 2633 2663 2693 2699 2789 2819 2837 2903 2927 2963 2969 2999 3083 3089 3137 3167 3203 3257 3299 3329 3389 3407 3413 3449 3467 3533 3539 3557 3593 3617 3623
I suggest using this pattern of residues modulo 5
3 3 3 3 2 4 4 2 3 2 2 3 3 4 2 4 4 3 2 2 2 4 2 4 4 3 4 2 3 3 3 4 2 4 2 4 3 4 3 2 3 4 3 2 3 2 4 2 4 2 3 3 3 4 2 3 2 2 4 2 4 4 4 3 2 2 4 2 3 3 4 4 3 4 2 2 4 3 3 3 3
I will write a program, using the general formula and the pattern.
- Questa risposta è stata modificata 8 anni, 11 mesi fa da Natalia Makarova.
Maggio 8, 2015 alle 5:57 pm #339Natalia MakarovaPartecipanteI found several solutions with one error.
For example:1013 113 743 2543 1697 3449 3299 857 2693 2837 1187 3203 2423 3329 557 509 2339 23 3557 2207 2087 2549 647 1289 1979 53 2039 827 3083 233 2963 389 3467 239 2237 2969 1313* 719 1583 107 1823 3539 2063 2927 2333 677 1409 3407 179 3257 683 3413 563 2819 1607 3593 1667 2357 2999 1097 1559 1439 89 3623 1307 3137 3089 317 1223 443 2459 809 953 2789 347 197 1949 1103 2903 3533 2633
K=3646, S=16407
Maggio 10, 2015 alle 10:32 am #340Natalia MakarovaPartecipanteI tried to find a solution with magic constant S = 13059.
In this solution is the one error839 1931 59 1319 1709 2633 2459 1871 239 659 2621 2309 1973 1949 1889 503 863 293 2381 1619 761 1601 2333 41 1493 101 2729 773 149 2789 71 353 1289 2423 2393 2819 2879 359 2411 2339 1451 563 491 2543 23 83 509 479 1613 2549 2831* 113 2753 2129 173 2801 1409 2861 569 1301 2141 1283 521 2609 2039 2399 1013 953 929 593 281 2243 2663 1031 443 269 1193 1583 2843 971 2063
K=2902, S=13059
Maggio 11, 2015 alle 8:44 am #341Natalia MakarovaPartecipanteThere is a solution with magic constant S = 14967 !
Now I’m trying to find a solution with magic constants 13997, 13059 and 12249.Maggio 12, 2015 alle 9:08 pm #343Natalia MakarovaPartecipanteI found a solution S = 13977 with one error:
263 353 443 2003 2957 2909 2969 797 1283 2837 1667 2543 1583 1889 167 809 2459 23 1787 2657 2267 2789 557 1019 1709 83 1109 977 1733 773 1613 419 2447 1229 1697 3089 1193 179 2243 107 1553 2999 863 2927 1913 17 1409 1877 659 2687 1493 2333 1373 2129 1997 3023 1397*2087 2549 317 839 449 1319 3083 647 2297 2939 1217 1523 563 1439 269 1823 2309 137 197 149 1103 2663 2753 2843
K = 3106, S = 13977
Maggio 24, 2015 alle 7:32 am #355Natalia MakarovaPartecipanteThere is a solution with magic constant S = 13977!
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