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Progress!
My program has found a solution with 5 errors.79 101 107 373 379 401 383 193 151 233 211 349 307 229 103 433 353 283 257 199 149 127 317 331 409 223 337 367 157 83 109 331* 131 97 163 269 431 397 439 89 191 281 359 113 347 277 251 197 313 419 181 179 109* 263 241 311 389 379* 401* 167 137 239 173 131*There are only 5 identical prime numbers:
331, 109, 379, 401, 131.And more progress!
My program has found a solution with 4 errors.79 101 439 109 227 401 379 281 359 233 263 409 191 197 251 113 89 419 257 271 269 383 149 179 389 311 173 131 229 83 353 347 283 97 167 181 431 397 107 353* 307 313 211 433 139 277 199 137 193 163 349 331 373 127 241 239 317 379* 157 151 157* 151* 337 367S=2016
In this solution there are only four identical prime numbers –
353, 379, 157, 151.Conversion for pandiagonal square of order 8
Suppose we have pandiagonal square 8th order with a magic constant S:
a11 a12 a13 a14 a15 a16 a17 a18 a21 a22 a23 a24 a25 a26 a27 a28 a31 a32 a33 a34 a35 a36 a37 a38 a41 a42 a43 a44 a45 a46 a47 a48 a51 a52 a53 a54 a55 a56 a57 a58 a61 a62 a63 a64 a65 a66 a67 a68 a71 a72 a73 a74 a75 a76 a77 a78 a81 a82 a83 a84 a85 a86 a87 a88Then for any integer z next square
a11 a12 a13 a14+z a15-z a16 a17 a18 a21 a22 a23-z a24 a25 a26+z a27 a28 a31 a32 a33+z a34 a35 a36-z a37 a38 a41 a42 a43 a44-z a45+z a46 a47 a48 a51 a52 a53 a54 a55 a56 a57 a58 a61 a62 a63 a64 a65 a66 a67 a68 a71 a72 a73 a74 a75 a76 a77 a78 a81 a82 a83 a84 a85 a86 a87 a88also have pandiagonal square with magic constant S.
Example
Given pandiagonal square:
359 433 379 179 101 157 211 197 229 79 311 439 109 217 349 283 269 191 263 331 409 137 193 223 163 89 167 151 401 281 367 397 307 389 251 313 139 113 83 421 353 383 149 263 233 431 97 107 137 271 317 239 277 307 337 131 199 181 179 101 347 373 379 257S=2016
Square after conversion, z = -90:
359 433 379 89 191 157 211 197 229 79 401 439 109 127 349 283 269 191 173 331 409 227 193 223 163 89 167 241 311 281 367 397 307 389 251 313 139 113 83 421 353 383 149 263 233 431 97 107 137 271 317 239 277 307 337 131 199 181 179 101 347 373 379 257S=2016
I tried my algorithm for an array of primes:
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499My program found pandiagonal square 8th order of the numbers of the array with a magic constant S = 2016 in 5 minutes!
79 389 167 229 239 241 263 409 7 61 443 457 191 173 251 433 421 163 37 463 149 487 97 199 349 419 317 31 277 131 353 139 271 257 283 53 431 109 379 233 307 337 211 113 491 449 19 89 313 59 401 311 41 383 461 47 269 331 157 359 197 43 193 467The algorithm works well.
This algorithm is described in my article
http://www.natalimak1.narod.ru/pannetr2.htmNow you have to make pandiagonal square 8th order with a magic constant
S = 2016 of the numbers of the next array:79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439This is a difficult task!
Is there a solution?In the program by colleague S. Belyaev I found a solution for n = 6 with 2 errors:
151+ 16 120 118 48 196 90 76 142 160 82 22 106 162 94* 6 88 198 40 126 46 72 202 30 112 78 186 66* 156 42 60 130 0 166 12 100 180S=1494
Dmitry Petukhov found two new solutions – pandiagonal squares of order 4 of consecutive primes.
#1
11796223202765101+0 148 232 336 268 300 36 112 126 22 358 210 322 246 90 58S=47184892811061120
http://dxdy.ru/post974234.html#p974234
#2
17537780902038437+0 192 150 330 210 270 60 132 186 6 336 144 276 204 126 66S=70151123608154420
The new solution by D. Petukhov
12548708437706431+ 0 58 222 256 240 238 18 40 46 12 268 210 250 228 28 30S=50194833750826260
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