Ultra magic square of 10-th order

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  • Marzo 1, 2015 alle 9:10 am #303
    Natalia Makarova
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    Scheme of ultra magic square of 10-th 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
x41 x42 x43 x44 x45 x46 x47 x48 x49 x50
k-x50 k-x49 k-x48 k-x47 k-x46 k-x45 k-x44 k-x43 k-x42 k-x41
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 – the associativity constant of square, S = 5k, S – magic constant.

    The general formula

    X(1) = X(43)- X(44)+ X(45)+2*X(46)+2*X(47)+ X(20)- X(21)+3*X(48)- X(22)+ X(2)+ X(15)+ X(13)+ X(27) + X(30)- X(31)+2*X(49)-10*K+ X(16)+ X(36)+2*X(37)+ X(38)+2*X(39)+ X(40) -2*X(8)+2*X(50) 
X(10) = - X(42)-2*X(44)+ X(46)+ X(47)+2*X(48)- X(22)+ X(2)+ X(15)+ X(13)+ X(27)+ X(11)+ X(49)-5*K + X(16)+ X(36)+2*X(37)+ X(38)+2*X(39)-2*X(8) 
X(12) = - X(42)+ X(44)- X(47)+ X(4)+ X(9)- X(20)- X(48)- X(2)- X(15)+ X(26)- X(11)- X(30)+ X(31) + X(34)+ X(35)+ X(36)+ X(8) 
X(14) = 2*X(42)- X(43)+ X(44)-2*X(46)-2*X(47)-2*X(4)-2*X(20)+ X(21)-4*X(48)+ X(22)- X(2) -2*X(15)-2*X(13)- X(26)- X(11)- X(29)+ X(31)-2*X(49)+15*K-2*X(16)- X(34) - X(36)-3*X(37)-2*X(38)-3*X(39)- X(40)+3*X(8)-2*X(50) 
X(17) = - X(44)+ X(48)+ X(2)- X(26)+ X(11)+ X(30)- X(35)+ X(39)- X(8) 
X(18) = X(43)- X(44)+2*X(46)+3*X(47)- X(9)+2*X(20)- X(21)+3*X(48)- X(22)+ X(2)+ X(15)+ X(13) + X(29)+ X(30)- X(31)+2*X(49)-10*K+ X(16)+2*X(37)+2*X(38)+2*X(39)+ X(40) -2*X(8)+2*X(50) 
X(19) = - X(42)+ X(4)+ X(48)+ X(15)+ X(26)- X(30)- X(31)+ X(37)- X(8) 
X(23) = (2*X(42)-2*X(43)+2*X(47)-2*X(4)-2*X(9)-2*X(21)-2*X(22)-2*X(13)-2*X(26) -2*X(27)-2*X(11)+5*K-2*X(16)+2*X(35)+2*X(38)+2*X(40)+2*X(8)) /2 
X(24) = - X(20)+ X(22)- X(15)+ X(27)+ X(11)- X(29)+ X(31)+ X(16)- X(35)+ X(36)- X(40) 
X(25) = -3*X(42)+ X(43)-2*X(44)+2*X(46)+2*X(47)+2*X(4)+ X(9)+3*X(20)+5*X(48)-2*X(22)+ X(2) +3*X(15)+2*X(13)+ X(26)+ X(11)+2*X(29)-2*X(31)+3*X(49)-15*K+ X(16) +4*X(37)+2*X(38)+4*X(39)+2*X(40)-4*X(8)+2*X(50) 
X(28) = -(-4*X(42)-4*X(44)+4*X(46)+6*X(47)+2*X(4)+4*X(20)+10*X(48)-2*X(22)+2*X(2) +4*X(15)+2*X(13)+2*X(26)+2*X(27)+2*X(11)+4*X(29)+2*X(30)-2*X(31) +6*X(49)-35*K+2*X(16)+2*X(36)+8*X(37)+6*X(38)+8*X(39)+4*X(40) -6*X(8)+4*X(50)) /2 
X(3) = X(43)+ X(20)+ X(48)+ X(15)+ X(11)- X(31)+ X(16)- X(35)- X(36)- X(40)- X(8) 
X(32) = - X(42)- X(44)+ X(47)+ X(20)+2*X(48)- X(22)+2*X(15)+ X(11)+ X(29)-2*X(31)+ X(49)- X(34) - X(35)- X(36)+ X(37)+ X(39)-2*X(8) 
X(33) = X(42)+ X(44)- X(47)- X(20)-2*X(48)+ X(22)-2*X(15)- X(11)- X(29)+ X(31)- X(49)+5*K -2*X(37)- X(38)-2*X(39)- X(40)+2*X(8) 
X(41) = - X(42)- X(43)- X(44)- X(45)- X(46)- X(47)- X(48)- X(49)+5*K- X(50) 
X(5) = X(42)- X(43)+ X(44)- X(45)- X(46)- X(47)- X(4)- X(9)- X(20)-3*X(48)+ X(22)- X(2)-2*X(15)- X(13) - X(27)- X(11)+ X(31)-2*X(49)+10*K- X(16)-2*X(37)- X(38)-2*X(39)+2*X(8) - X(50) 
X(6) = -2*X(42)- X(44)+ X(47)+ X(4)+2*X(20)+2*X(48)- X(22)+2*X(15)+ X(13)- X(27)+2*X(29)- X(31) + X(49)-5*K- X(16)+ X(35)- X(36)+2*X(37)+ X(38)+2*X(39)+2*X(40)-2*X(8) + X(50) 
X(7) = 2*X(42)- X(43)+3*X(44)-2*X(46)-3*X(47)- X(4)-3*X(20)+ X(21)-5*X(48)+2*X(22)-2*X(2) -3*X(15)-2*X(13)- X(11)-2*X(29)- X(30)+2*X(31)-2*X(49)+15*K- X(16) -4*X(37)-2*X(38)-4*X(39)-2*X(40)+4*X(8)-2*X(50)

    I found a solution to this formula, which has six errors:

    5897 11   4007  2777 2591 2657 4319* 47    4013 3251
53   5153 131   4967 1433 5237 2081  5195* 383  4937
5657 521  4523* 2237 5483 71   5051  263   3221 2543
5351 1787 2357  1913 4091 2153 3917  2063  3491 2447
941  4817 1493  4751 983  4463 1277  4943  1181 4721
1193 4733 971   4637 1451 4931 1163  4421  1097 4973
3467 2423 3851  1997 3761 1823 4001  3557  4127 563
3371 2693 5651  863  5843 431  3677  1391* 5393 257
977  5531 719*  3833 677  4481 947   5783  761  5861
2663 1901 5867 1595* 3257 3323 3137  1907  5903 17

    K=5914, S=29570

    Marzo 2, 2015 alle 6:00 am #305
    Natalia Makarova
    Partecipante

    Progress!
    In this solution there are 4 errors

    5651 761  3671  5273 173  2111 1607* 5231  3011 2081
5351 977  5843  47   5003 2693 5657  1985* 2003 11
653  2591 4217  4007 1223 53   3533  5393  2663 5237
1901 3203 4391  1913 4091 2153 3917  2063  3491 2447
941  4817 1493  4751 983  4463 1277  4943  1181 4721
1193 4733 971   4637 1451 4931 1163  4421  1097 4973
3467 2423 3851  1997 3761 1823 4001  1523  2711 4013
677  3251 521   2381 5861 4691 1907  1697  3323 5261
5903 3911 3929* 257  3221 911  5867  71    4937 563
3833 2903 683  4307* 3803 5741 641   2243  5153 263

    K=5914, S=29570

    Marzo 20, 2015 alle 4:04 am #310
    Natalia Makarova
    Partecipante

    In this solution there are only two non prime numbers – 1661, 1127.

    4787 263  641  3323 2357 1277 1661*5393 4007 5861
947  5087 5741 5807 3137 1523 197  1433 3251 2447
563  2333 1913 2003 5903 5897 3851 2711 2153 2243
4127 3803 3917 131  617  1193 4691 2543 4931 3617
4421 5867 3701 113  2423 4547 977  3833 3011 677
5237 2903 2081 4937 1367 3491 5801 2213 47   1493
2297 983  3371 1223 4721 5297 5783 1997 2111 1787
3671 3761 3203 2063 17   11   3911 4001 3581 5351
3467 2663 4481 5717 4391 2777 107  173  827  4967
53   1907 521  4253 4637 3557 2591 5273 5651 1127*

    K=5914, S=29570

    Marzo 20, 2015 alle 4:24 am #311
    Natalia Makarova
    Partecipante

    According to the solution I made pattern of residues modulo 4:

    3  3  1  3  1  1  1  1  3  1 
3  3  1  3  1  3  1  1  3  3 
3  1  1  3  3  1  3  3  1  3 
3  3  1  3  1  1  3  3  3  1 
1  3  1  1  3  3  1  1  3  1 
1  3  1  1  3  3  1  1  3  1 
1  3  3  3  1  1  3  1  3  3 
3  1  3  3  1  3  3  1  1  3 
3  3  1  1  3  1  3  1  3  3 
1  3  1  1  1  1  3  1  3  3

    This pattern can be used to find solutions, if K = 2 (mod 4), S = 2 (mod 4).

    Marzo 21, 2015 alle 4:22 am #312
    Natalia Makarova
    Partecipante

    In these solutions there is only one not a prime number.

    7561 601  4129 97   1279 6043 8713 9007 8689 31
5557 3163 8599 6793 5521 8353 1063 1861 337  4903
6709 8707 7927 2593 8377 1051 1237 3793 3823 1933
73   6151 6841 4243 4663 5869 1741 6553 1723 8293
2833 4051 709  1021 7177 5701 1993 6991 8101 7573
1657 1129 2239 7237 3529 2053 8209 8521 5179 6397
937  7507 2677 7489 3361 4567 4987 2389 3079 9157
7297 5407 5437 7993 8179 853  6637 1303 523  2521
4327 8893 7369 8167 877  3709 2437 631  6067 3673
9199 541  223  517* 3187 7951 9133 5101 8629 1669
    8947* 601 1549 127  3079 3373 9151 9007 6709 3607 
5557 4639 8353 6481 6547 9199 1063 2467 1303 541 
4447 9091 8821 5647 3727 1051 3571 103  7759 1933 
163  7213 5689 4243 4663 5869 1741 6553 1723 8293 
2833 4051 709  1021 7177 5701 1993 6991 8101 7573 
1657 1129 2239 7237 3529 2053 8209 8521 5179 6397 
937  7507 2677 7489 3361 4567 4987 3541 2017 9067 
7297 1471 9127 5659 8179 5503 3583 409  139  4783 
8689 7927 6763 8167 31   2683 2749 877  4591 3673 
5623 2521 223  79   5857 6151 9103 7681 8629 283
    8233 601  2389 307  1069 6679 7069 9007 7687 3109 
5557 4447 8821 5719*5923 8689 1063 2557 523  2851 
6637 7213 9103 2437 9157 1051 1861 1237 5521 1933 
499  7243 5323 4243 4663 5869 1741 6553 1723 8293 
2833 4051 709  1021 7177 5701 1993 6991 8101 7573 
1657 1129 2239 7237 3529 2053 8209 8521 5179 6397 
937  7507 2677 7489 3361 4567 4987 3907 1987 8731 
7297 3709 7993 7369 8179 73   6793 127  2017 2593 
6379 8707 6673 8167 541  3307 3511 409  4783 3673 
6121 1543 223  2161 2551 8161 8923 6841 8629 997

    K=9230, S=46150

    Marzo 22, 2015 alle 5:43 am #313
    Natalia Makarova
    Partecipante

    There is a solution with magic constant S = 46150 !

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