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I performed manually several permutations (layers, rows, columns).
These are some results:diagonals 608 = 409(1,1,1) + 73(2,2,2) + 103(3,3,3) + 23(4,4,4) 926 = 89(4,1,1) + 241(3,2,2) + 313(2,3,3) + 283(1,4,4) 808 = 5(1,4,1) + 389(2,3,2) + 251(3,2,3) + 163(4,1,4) 658 = 67(4,4,1) + 173(3,3,2) + 401(2,2,3) + 17(1,1,4) diagonals 1370 = 409(1,1,1) + 401(2,2,2) + 211(3,3,3) + 349(4,4,4) 420 = 89(4,1,1) + 251(3,2,2) + 37(2,3,3) + 43(1,4,4) 568 = 5(1,4,1) + 313(2,3,2) + 59(3,2,3) + 191(4,1,4) 642 = 67(4,4,1) + 103(3,3,2) + 179(2,2,3) + 293(1,1,4) diagonals 1074 = 197(1,1,1) + 313(2,2,2) + 373(3,3,3) + 191(4,4,4) 724 = 257(4,1,1) + 103(3,2,2) + 71(2,3,3) + 293(1,4,4) 710 = 409(1,4,1) + 7(2,3,2) + 211(3,2,3) + 83(4,1,4) 492 = 89(4,4,1) + 13(3,3,2) + 37(2,2,3) + 353(1,1,4) diagonals 718 = 257(1,1,1) + 281(2,2,2) + 71(3,3,3) + 109(4,4,4) 986 = 199(4,1,1) + 313(3,2,2) + 283(2,3,3) + 191(1,4,4) 786 = 89(1,4,1) + 419(2,3,2) + 37(3,2,3) + 241(4,1,4) 510 = 151(4,4,1) + 7(3,3,2) + 269(2,2,3) + 83(1,1,4)Interesting fact: the sum of all amounts of numbers on the diagonals always equal 3000 = 4*S.
Yes, the probability of convert semimagic cube in magic cube is very small. But we must try the software all the possible permutations.
Besides, one can find many other semimagic cubes and for each of them to perform all the permutations.This is one of the techniques that I use when searching unconventional magic squares of primes and number of Smith. The method worked well and gave solutions.
I found a solution to the model proposed above, but there are 5 errors:
1301,191,1669,389,1249,241, 1609,137,61,1171,1109,953, 577,1429,1367,113,227,1327, 1307,1423,593,1321,199,197, 233,1033,929,463,769,1613, 13,827,421,1583,1487,709, 17,23,1193,1279,1297,1231, 1223,1987,7,863,503,457, 1549,613,2221,107,419,131, 271,239,101,2053,967,1409, 1531,521,1031,337,1471,149, 449,1657,487,401,383,1663, 53,823,631,1571,1021,941, 1019,103,73,1997,1187,661, 997,1847,673,317,523,683, 859,601,461,1009,1289,821, 1373,809,2153,37,361,307, 739,857,1049,109,659,1627, 1061,1597,229,467,1123,563, 181,701,2099,349,211,1499, 947,409,311,919,1721,733, 641,1973,787,269,331,1039, 1093,277,163,1823,1097,587, 1117,83,1451,1213,557,619, 1637,1553,59,1237,157,397, 281,569,1181,151,1459,1399, 617,491,155,2017,697,1063, 479,547,2011,29,773,1201, 743,1753,13,1163,431,937, 1283,127,1621,443,1523,43, 971,853,1259,97,193,1667, 727,1543,1619,509,571,71, 353,251,313,1567,1453,1103, 1483,257,1087,359,1481,373, 67,647,751,1217,911,1447, 1439,1489,11,1291,431,379Checking in the program by Ed Mertensotto showed:
type 1
size 6
361 is not prime
155 is not prime
697 is not prime
13 is not unique
431 is not unique
All Sums = 5040I think can find the right solution for this model.
Here you see a concentric magic cube of order 7 of primes.
Magic constant cube S = 76321.
Inside are magic cube of order 3 (S = 32709) and a concentric magic cube of order 5 (S = 54515).Interesting solution!
Now I will show concentric magic cubes of order 6 of primes.
This is a very interesting solution!This solution is known on the Internet:
http://www.magic-squares.net/c-t-htm/c_prime.htm
Inside is an associative pantriagonal magic cube of order 4 with magic constant S = 19740.
My first solution
Inside is an associative magic cube of order 4 with magic constant S = 13200.
To be continued.
My second solution
Inside is an associative cube of order 4 with a magic constant S = 6720.
My third solution
Attention!
Inside the cube is not associative cube of order 4 with a magic constant S = 4020.You can still try to find concentric magic cubes of order 6 with magic constants S = 5670 and S = 5040.
The border is easy to build, difficult to find not associative magic cube of order 4.
For example, for the bordering concentric magic cube with magic constant S = 5040:1301 191 1669 389 1249 241 1609 137 61 1171 1109 953 577 1429 1367 113 227 1327 1307 1423 593 1321 199 197 233 1033 929 463 769 1613 13 827 421 1583 1487 709 17 23 1193 1279 1297 1231 1223 0 0 0 0 457 1549 0 0 0 0 131 271 0 0 0 0 1409 1531 0 0 0 0 149 449 1657 487 401 383 1663 53 823 631 1571 1021 941 1019 0 0 0 0 661 997 0 0 0 0 683 859 0 0 0 0 821 1373 0 0 0 0 307 739 857 1049 109 659 1627 1061 1597 229 467 1123 563 181 0 0 0 0 1499 947 0 0 0 0 733 641 0 0 0 0 1039 1093 0 0 0 0 587 1117 83 1451 1213 557 619 1637 1553 59 1237 157 397 281 0 0 0 0 1399 617 0 0 0 0 1063 479 0 0 0 0 1201 743 0 0 0 0 937 1283 127 1621 443 1523 43 971 853 1259 97 193 1667 727 1543 1619 509 571 71 353 251 313 1567 1453 1103 1483 257 1087 359 1481 373 67 647 751 1217 911 1447 1439 1489 11 1291 431 379Now you have to find a simple magic cube of order 4 with magic constant S = 3360 and insert it into the bordering.
This I can not do.The challenge for all!
I found an approximate solution – concentric magic cube of order 6 with a magic constant S = 5670:
1289 919 599 1789 1031 43 661 677 1831 149 1103 1249 347 937 233 1801 1699 653 1277 797 1579 197 397 1423 607 719 277 827 1429 1811 1489 1621 1151 907 11 491 509 457 563 1627 727 1787 1753 1039 31 887 1823 137 1733 439 2311 383 647 157 19 353 557 2371 499 1871 1553 1949 881 139 811 337 103 1433 1327 263 1163 1381 877 281 1459 761 569 1723 857 97 307 2069 1307 1033 283 2267 523 317 673 1607 1777 967 1451 271 1091 113 1709 449 1499 1123 709 181 167 1609 431 1129 1321 1013 1303 1867 829 131 17 1523 1531 53 2333 751 643 359 773 991 107 373 2309 1117 1667 2423 571 659 127 223 29 313 769 1997 701 1861 367 23 1061 1759 1873 587 293 1877 1481 379 1447 193 227 2591 1109 73 7 1663 1297 83 839 2707 151 593 463 37 1201 479 2063 1427 1693 1069 631 521 1559 197 1697 13 409 1511 443 1597 1399 269 739 983 1879 401 641 1213 59 1741 787 1229 1237 953 1657 89 191 1543 467 1093 311 1693 1493 613 79 1171 1613 1063 461 1283 1847 971 1291 101 859 601Checking in the program by Ed Mertensotto showed:
type 1 size 6 197 is not unique 1693 is not unique All Sums = 5670I think there is a solution without error.
Have the right solution!
971 761 1801 367 157 1613 1447 379 491 1031 569 1753 1033 1667 1709 281 877 103 1117 1777 419 457 1607 293 461 23 787 1723 797 1879 641 1063 463 1811 1663 29 607 409 719 1877 1627 431 1543 1039 31 887 1823 347 599 439 2311 383 647 1291 233 353 557 2371 499 1657 1229 1949 881 139 811 661 1459 1481 1171 13 263 1283 937 149 1303 677 983 1621 1553 97 307 2069 1307 337 727 2267 523 317 673 1163 653 967 1451 271 1091 1237 1531 449 1499 1123 709 359 269 1741 587 1213 907 953 1277 1693 19 1289 829 563 593 53 2333 751 643 1297 1423 991 107 373 2309 467 311 2423 571 659 127 1579 739 313 769 1997 701 1151 1327 197 1871 601 1061 613 17 1831 401 1381 1847 193 397 2591 1109 73 7 1493 101 83 839 2707 151 1789 1759 37 1201 479 2063 131 1699 1069 631 521 1559 191 1697 59 1489 509 43 1873 1861 827 1427 79 227 1249 137 1511 1399 859 1321 443 1787 223 181 1609 1013 857 1597 113 1471 1433 283 773 11 1867 1103 167 1093 1429 277 1129 89 1523 1733 919S=5670
I think that the solution for magic constant S = 5040 also exists, but find it difficult.
Here you see a scheme of concentric magic cube of order 6 with a magic constant
S = 3k
Free variables are highlighted in red.
Inside there is any magic cube of order 4 with magic constant S = 2k.
I found all solutions under this scheme.Interesting to see the classic concentric cube of order 6:
Inside is an unconventional associative magic cube of order 4 with magic constant S = 434.
This cube is built using arithmetic progressions.On this web page
http://www.magic-squares.net/c-t-htm/c_prime.htmyou see concentric magic cube of order 8 with a magic constant S = 39480.
Inside are:
1. Associative pantriagonal cube of order 4 with a magic constant S = 2k = 19740;
2. Concentric magic cube of order 6 with a magic constant S = 3k = 29610,where k – complementarity constant, k = 9870.
I made a model of a similar cube of order 8 for k = 8400
In my model have:
1. Associative cube of order 4 with a magic constant S = 2k = 16800;
2. Concentric magic cube of order 6 with a magic constant S = 3k = 25200.I have not yet implemented this model.
Perhaps the solution for this model exists.The challenge for all!
Note:
there is an inaccuracy in the illustration.
Variables X135-X141, X143-X147 are not free.I will show new classic concentric cube of order 7
Inside the cube is unconventional associative pantriagonal cube of order 5 with a magic constant S = 860:
296 230 199 133 2 28 322 256 160 94 120 54 283 217 186 212 81 15 309 243 204 173 107 41 335 223 192 126 30 289 315 249 153 87 56 82 276 210 179 113 74 43 302 236 205 166 100 69 328 197 185 119 23 282 251 277 146 80 49 308 269 238 172 106 75 36 295 264 198 67 93 62 321 225 159 147 16 275 244 178 139 108 42 301 270 231 165 134 68 262 288 257 191 95 29 55 314 218 152 121 9 303 237 171 140 101 35 329 263 132 158 127 61 290 224 250 184 88 22 316 342 211 145 114 48This cube is based on the classical associative pantriagonal cube of order 5 using arithmetic progressions:
109 87 70 48 1 11 119 97 55 33 43 21 104 82 65 75 28 6 114 92 77 60 38 16 124 86 69 47 5 108 118 96 54 32 15 25 103 81 64 42 27 10 113 91 74 59 37 20 123 76 68 46 4 107 90 100 53 31 14 117 102 85 63 41 24 9 112 95 73 26 36 19 122 80 58 50 3 106 89 67 52 35 13 116 99 84 62 45 23 101 111 94 72 30 8 18 121 79 57 40 2 110 88 66 49 34 12 120 98 51 61 44 22 105 83 93 71 29 7 115 125 78 56 39 17This cube has sent me Christian Boyer.
So we know that the classic perfect cube of order 4 does not exist.
Maybe there is an unconventional perfect cube of order 4?
This is a very interesting question!
I’m trying to find the answer.Here you can see the system of equations describing the perfect cube of order 4, composed of arbitrary integers, with a magic constant s:
x1+x21+x41+x61=s x2+x22+x42+x62=s x3+x23+x43+x63=s x13+x33+x53+x73=s x7+x27+x47+x67=s x9+x29+x49+x69=s x5+x25+x45+x65=s x1+x27-x41-x42-x43+x45-x47+x62+x63-x65=0 -x1-x2-x3+x29+x65+x41-x49+x42+x43-x45=0 x5+x21-x29+x22+x23-x25-x61-x62-x63+x49=0 x2+x3-x5-x21-x22-x23+x25-x27+x47+x61=0 x1+x22+x43-x61-x62-x63=0 -x1-x2-x3+x23+x42+x61=0 x13+x27+x49-x67-x69-x73=0 -x7-x9-x13+x29+x47+x73=0 -x1-x5-x13+x21+x22+x23-x25-x29-x41-x42-x43+x45-x47+x67+x69+x61+x65+x73=0 x7+x9+x1+x5+x13-x21-x22-x23+x25-x27+x41+x42+x43-x45-x49-x61-x65-x73=0 x5-2*x22+x29+x25-x27-x21-x23+x47+x41+x42-x45-x49+x62+x63-x65=0 x2+x3-x5+x27+x21-x29+x22-x25-2*x42+x49-x41-x43+x45-x47+x65=0 -x1-x2-x3-x27-x29-x33+x47+x49+x41+x45+x53+x62+x63-x65=0 x2+x3-x5+x27+x29+x21+x25+x33-x47-x49-x53-x61-x62-x63=0 x3+x29-x41-x42-x43+x45-x47+x67+x61-x69+x62-x65=0 x7+x1-x9+x2-x5-x21-x22-x23+x25-x27+x49+x63=0 x2+x27+x41-x49+x42+x43-x45-2*x62+x69+x65-x67-x61-x63=0 -2*x2-x1-x3+x5-x7+x9+x21-x29+x22+x23-x25+x47+x62=0 x1+x33-x41-x45-x53+x65=0 x5-x21-x25-x33+x53+x61=0I do not show the cube scheme, first I want to see the solution of this system.
Dear colleagues, someone can help me?Note: errors may occur, the system is rather complicated.
In addition, the system may be incompatible. I do not know whether there is a solution of this system.I checked the system for data from nearly perfect magic cube of order 4 by Fermat, but in this cube are not all the magic line (there are only 64 magic line instead of the required 76 magic lines). So check can not be considered complete.
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