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Dear Colleagues!
I’ve decided to hold a contest on the issue pandiagonal squares of consecutive primes.
I invite everyone to take part in solving this complex problem.Here is an article by my colleague Alex Chernov
http://alex-black.ru/article.php?content=121In this article you will find a solution algorithm for n = 6.
You can take this algorithm as a basis and make their changes to solve the problem.Many interesting ideas you can find in forum
http://dxdy.ru/topic73817.htmlYou can ask your questions here, and offer your ideas and algorithms.
For n = 7 we have a theoretical minimum magic constant S = 797.
This square should be composed of the following array:
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 239I tried to make this square, I got a partial solution:
I think that such a solution exists.
Let the scheme pandiagonal square of order 7 is this:
a1 x1 a2 x2 a3 x3 a4 x4 a5 x5 a6 x6 a7 x7 a8 x8 a9 x9 a10 x10 a11 x11 a12 x12 a13 x13 a14 x14 a15 x15 a16 x16 a17 x17 a18 x18 a19 x19 a20 x20 a21 x21 x22 x23 x24 x25 x26 x27 x28Then the general formula pandiagonal square with magic constant S:
x1 = -a11 + a12 + a13 + a15 + a16 + a17 + a20 + a21 - a4 + a5 + a6 + a8 + a9 – 2*S + x16 + x17 + x20 + x25 + x26 + x27, x2 = a12 - a18 - a4 + a5 + a8 + a9 - x16, x3 = a10 + a11 - a12 + a18 - a20 - a21 + a4 - a5 - a8 + S - x17 - x20 - x25 - x26 - x27, x4 = a10 + a12 + a13 + a16 + a19 + a9 - x16 - x17 - x20 - x25 - x27, x5 = a10 + a11 - a21 + a3 + a4 - a5 - x17, x6 = a13 + a14 + a16 + a17 + a18 + a20 + a21 - a3 - S + x16 + x17, x7 = -a10 - a11 - a16 - a18 + a21 - a4 + a5 + x17 + x20 + x25 + x27, x8 = -a10 - a12 - a13 - a16 - a17 - a19 - a20 - a21 - a5 - a8 - a9 + 2*S - x20 - x26, x9 = -a12 - a13 + a18 - a20 + a4 - a5 - a6 - a8 - a9 + S - x25, x10 = -a11 + 2*a12 + 2*a13 + a16 + a17 - a18 + a19 + 2*a20 + a21 - a4 + 2*a5 + a6 + a8 + a9 – 2*S + x20 + x25 + x26, x11 = -a11 + a13 + a16 + a17 + a19 + a20 + a21 - a3 + a5 - S + x16 + x17 + x20, x12 = a17 + a20 + a21 - a4 + a5 + a8 - S + x17 + x20 + x25 + x26, x13 = -a10 - a13 - a14 - a16 – 2*a17 - a18 - a20 - a21 + 2*S - x16 - x17 - x20 - x26, x14 = a10 + a11 - a12 - a13 + a18 - a19 - a20 - a21 + a3 + a4 - 2*a5 - a8 + S - x17 - x20 - x25, x15 = -a15 - a16 - a17 - a18 + S - x16 - x17, x18 = -a10 - a13 - a14 - a16 - a17 - a18 - a19 - a20 - a21 + a3 + S + x25 + x27, x19 = a10 + a12 + a13 + a14 + a15 + a16 + a17 + a18 + a6 + a9 - S - x20 - x25, x21 = -a12 - a15 - a3 - a6 - a9 + S - x27, x22 = a12 + a13 + a14 + a17 + a20 + a21 - a4 + a5 + a6 + a9 - S, x23 = a10 + a11 - a12 + a16 + a17 + a18 + a4 - a5 - a6 - x25 - x27, x24 = -a10 – 2*a12 – 2*a13 - a14 - a15 – 2*a16 – 2*a17 - a19 - 2*a20 - a21 + a4 - a5 - a6 - a8 – 2*a9 + 3*S - x26, x28 = -a11 + 2*a12 + a13 + a15 + a16 - a18 + a19 + a20 - a4 + a5 + a6 + a8 + a9 - S, a1 = a11 – 2*a12 – 2*a13 - a15 - a16 - a17 + a18 - a19 - a20 - a21 + a4 – 2*a5 - a6 - a8 – 2*a9 + 2*S, a2 = -a10 - a11 + a12 + a13 - a18 + a19 + a20 + a21 - a3 - a4 + a5, a7 = -a10 - a12 – 2*a13 - a14 - a16 - a17 - a19 - a20 - a21 - a5 - a6 - a9 + 2*STo construct a square about 7 pandiagonal of consecutive primes with a magic constant S = 797, you can use this pattern of residues modulo 4:
1 3 1 3 3 3 3 3 3 3 3 3 1 1 1 1 1 1 3 3 3 1 3 3 1 3 1 1 3 3 1 1 1 3 1 1 1 1 3 3 3 1 3 3 3 1 1 3 3We have this pattern two groups of prime numbers:
13 17 29 37 41 53 61 73 89 97 101 109 113 137 149 157 173 181 193 197 229 233 7 11 19 23 31 43 47 59 67 71 79 83 103 107 127 131 139 151 163 167 179 191 199 211 223 227 239Let pandiagonal square of order 8 is designated as follows:
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 x51 x52 x53 x54 x55 x56 x57 x58 x59 x60 x61 x62 x63 x64Then the general formula pandiagonal square with magic constant s has the form:
x8=s-x1-x2-x3-x4-x5-x6-x7 x16=s-x10-x11-x12-x13-x14-x15-x9 x24=s-x17-x18-x19-x20-x21-x22-x23 x32=s-x25-x26-x27-x28-x29-x30-x31 x40=s-x33-x34-x35-x36-x37-x38-x39 x42=-s-x1-x10+x12+x13+x14+x15-x17-x18-x2+x20+2x21+2x22+x23-x25-x26+x28+2x29+2x30+x31-x33-x34+x36+x37+x38+x39-x41-x9 x43=4s+x1-x10-2x11-2x12-x13-x14-x15-x18-2x19-2x20-2x21-x22-x26-2x27-2x28-2x29-x3-x30-x34-2x35-2x36-x37-x38-x39+x41 x44=s-x1+x14+x15-x17-x18-x19-x20-x25-x26-x27-x28+x38+x39-x4-x41 x45=4s+x1-x11-2x12-2x13-2x14-x15+x17-x19-2x20-3x21-2x22-x23+x25-x27-2x28-3x29-2x30-x31-x35-2x36-2x37-2x38-x39+x41-x5 x46=-s-x1+x11+x12+x18+x19+x20+x21+x26+x27+x28+x29+x35+x36-x41-x6 x47=x1+x10-x14-x15+x17+x18+x19-x21-x22-x23+x25+x26+x27-x29-x30-x31+x33+x34-x38-x39+x41-x7+x9 x48=-6s+x10+2x11+2x12+2x13+2x14+x15+x18+2x19+x2+3x20+3x21+2x22+x23+x26+2x27+3x28+3x29+x3+2x30+x31+x34+2x35+2x36+2x37+2x38+x39+x4-x41+x5+x6+x7 x49=9s/2-x11-2x12-2x13-2x14-x15-x19-2x20-3x21-2x22-x23-x27-2x28-2x29-x3-2x30-x31-x35-x36-x37-x38-x39-x4-x5-x6-x7 x50=-s/2+x1+x10+x11-x13-x14-x15+x17+x18+x19+x2-x21-2x22-x23+x25+x26+x27-x29+x3-x30-x31+x33+x34+x35+x9 x51=-7s/2+2x10+2x11+2x12+x13+x17+2x18+2x19+x2+2x20+x21-x23+x25+2x26+2x27+2x28+x29+x3+x34+x35+x36+x4+x9 x52=-9s/2+x10+2x11+2x12+2x13+x14+x17+2x18+3x19+3x20+3x21+2x22+x23+x26+2x27+2x28+2x29+x3+x30+x35+x36+x37+x4+x5 x53=-7s/2+x11+2x12+2x13+2x14+x15-x17+x19+2x20+2x21+2x22+x23+x27+2x28+2x29+2x30+x31+x36+x37+x38+x4+x5+x6 x54=-s/2-x10-x11+x13+x14+x15-x17-2x18-x19+x21+x22+x23-x25-x26-x27+x29+x30+x31+x37+x38+x39+x5+x6+x7-x9 x55=9s/2-x1-2x10-2x11-2x12-x13-x17-2x18-3x19-x2-2x20-x21-x25-2x26-2x27-2x28-x29-x3-x33-x34-x35-x36-x37-x4-x5-x9 x56=9s/2-x10-2x11-2x12-2x13-x14-x18-2x19-x2-3x20-2x21-x22-x26-2x27-2x28-2x29-x3-x30-x34-x35-x36-x37-x38-x4-x5-x6 x57=-7s/2-x1+x11+2x12+2x13+2x14+x15-x17+x19+2x20+3x21+2x22+x23-x25+x27+2x28+2x29+x3+2x30+x31-x33+x35+x36+x37+x38+x39+x4-x41+x5+x6+x7-x9 x58=5s/2-x10-x11-x12-x18-x19-x2-x20-x21-x26-x27-x28-x29-x3-x30-x34-x35-x36-x37-x38-x39+x41 x59=s/2-x1-x10-x11+x14+x15-x17-x18-x19-x2+x21+x22+x23-x25-x26-x27+x29-x3+x30+x36+x37+x38+x39-x4-x41-x9 x60=9s/2+x1-x10-2x11-3x12-2x13-2x14-x15-x18-2x19-3x20-3x21-2x22-x23+x25-x27-2x28-2x29-x3-x30-x35-2x36-x37-x38-x39-x4+x41-x5 x61=s/2-x1-x13-x25+x35+x36+x38+x39-x4-x41-x5-x6 x62=5s/2+x1+x10-x12-x13-2x14-x15+x17+x18-x20-2x21-2x22-x23+x25-x28-2x29-2x30-x31-x35-x36-x37-2x38-x39+x41-x5-x6-x7+x9 x63=-7s/2+x10+2x11+2x12+x13+x14+x18+2x19+x2+2x20+2x21+x22+x26+x27+2x28+2x29+x3+x30+x35+x36+x37+x38+x4-x41+x5 x64=-5s/2+x1+x10+x11+x12+x13+x17+x18+x19+x2+x20+x25+x26+x27+x3+x33+x34+x4+x41+x5+x6+x9See
http://dxdy.ru/post906522.html#p906522This is a potential array of 64 consecutive prime numbers with a minimal of magic constant S = 2016:
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 439You can use the pattern of residues modulo 4:
3 1 3 1 1 3 1 3 3 1 3 3 3 3 3 1 1 1 1 3 1 1 1 3 3 1 1 3 3 1 1 3 1 1 1 3 1 3 3 3 1 1 3 1 1 1 1 3 3 3 1 1 3 1 3 1 1 3 3 1 3 3 3 3My solution with 14 errors:
79 173 283 313 421 107 389 251 167 277 311 419 103 383 199 157 409 89 373 227 229 233 109 347 331 349 197 139 211 293 137 359 97 317 101 239 337 439 223 263 401 181 127 353 269 281 241 163 191 415* 473* 333* 191* 213* 127* 73* 341* 215* 151 -7* 255* 67* 591* 403* S=2016This is a bad solution.
Suggest your ideas, colleagues!
Denote pandiagonal square of order 6 as follows:
a1 a2 a3 b1 b2 b3 a4 a5 a6 b4 b5 b6 a7 a8 a9 b7 b8 b9 b1’ b2’ b3’ a1’ a2’ a3’ b4’ b5’ b6’ a4’ a5’ a6’ b7’ b8’ b9’ a7’ a8’ a9’Let the magic constant of this square is S.
Denote: q = S/3.Theorem (author S. Belyaev)
ai + ai’ – q = pi
bi + bi’ – q = -pi(i = 1, 2, 3, …, 9)
where
p1 + p5 + p9 = 0
p3 + p5 + p7 = 0
p1 – p6 – p8 = 0
p9 – p2 – p4 = 0
p3 – p4 – p8 = 0
p7 – p2 – p6 = 0so that the square:
p1 p2 p3 -p1 -p2 -p3 p4 p5 p6 -p4 -p5 -p6 p7 p8 p9 -p7 -p8 -p9 -p1 -p2 -p3 p1 p2 p3 -p4 -p5 -p6 p4 p5 p6 -p7 -p8 -p9 p7 p8 p9is pandiagonal square with magic constant S = 0.
Example
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 83We have in this pandiagonal square:
S = 930, q = 310
p1 = -14
p2 = 34
p3 = -60
p4 = -70
p5 = 50
p6 = -24
p7 = 10
p8 = 10
p9 = -36More see here
http://www.natalimak1.narod.ru/pannetr1.htmSuppose we are given an array of 49 numbers:
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 239Denote
S1 – the sum of all the numbers array
S = S1/7Definition 1
Exact-square representation of an array as a series of magical rows is the following matrix 7×7, composed of 7 magic strings so that:
1. is contained in a matrix each number of array only once;
2. any two rows of the matrix does not contain the same numbers.Example 1
7 29 89 107 157 181 227 11 67 79 97 139 193 211 13 41 53 149 163 179 199 17 37 73 113 127 191 239 19 47 59 131 151 167 223 23 43 61 103 137 197 233 31 71 83 101 109 173 229Definition 2
Two exact-square representations of an array are called orthogonal if each row of the first representation has exactly one element in common with each row in the second representation.
Example 2
7 29 89 107 157 181 227 11 67 79 97 139 193 211 13 41 53 149 163 179 199 17 37 73 113 127 191 239 19 47 59 131 151 167 223 23 43 61 103 137 197 233 31 71 83 101 109 173 229 7 43 97 109 151 191 199 11 23 71 107 167 179 239 13 59 73 103 139 181 229 17 53 67 83 157 197 223 19 79 89 101 127 149 233 29 37 47 137 163 173 211 31 41 61 113 131 193 227Theorem 1
Two orthogonal exact-square representations of an array give semi-magic square with magic constant S, consisting of the numbers of the array.
Example 3
using two orthogonal exact-square representations of an array of Example 27 29 89 107 157 181 227 43 137 233 23 197 103 61 97 211 79 11 67 139 193 109 173 101 71 83 229 31 151 47 19 167 223 59 131 191 37 127 239 17 73 113 199 163 149 179 53 13 41To be continued.
Theorem 2
If
a11 a12 a13 a14 a15 a16 a17 a21 a22 a23 a24 a25 a26 a27 a31 a32 a33 a34 a35 a36 a37 a41 a42 a43 a44 a45 a46 a47 a51 a52 a53 a54 a55 a56 a57 a61 a62 a63 a64 a65 a66 a67 a71 a72 a73 a74 a75 a76 a77is pandiagonal square of order 7 of different numbers, then there are four mutually orthogonal exact-square representations of an array of numbers that make up this square:
#1 a11 a12 a13 a14 a15 a16 a17 a21 a22 a23 a24 a25 a26 a27 a31 a32 a33 a34 a35 a36 a37 a41 a42 a43 a44 a45 a46 a47 a51 a52 a53 a54 a55 a56 a57 a61 a62 a63 a64 a65 a66 a67 a71 a72 a73 a74 a75 a76 a77 #2 a11 a21 a31 a41 a51 a61 a71 a12 a22 a32 a42 a52 a62 a72 a13 a23 a33 a43 a53 a63 a73 a14 a23 a34 a44 a54 a64 a74 a15 a25 a35 a45 a55 a65 a73 a16 a26 a36 a46 a56 a66 a76 a17 a27 a37 a47 a57 a67 a77 #3 a11 a22 a33 a44 a55 a66 a77 a12 a23 a34 a45 a56 a67 a71 a13 a24 a35 a46 a57 a61 a72 a14 a25 a36 a47 a51 a62 a73 a15 a26 a37 a41 a52 a63 a74 a16 a27 a31 a42 a53 a64 a75 a17 a21 a32 a43 a54 a65 a76 #4 a17 a26 a35 a44 a53 a62 a71 a16 a25 a34 a43 a52 a61 a77 a15 a24 a33 a42 a51 a67 a76 a14 a23 a32 a41 a57 a66 a75 a13 a22 a31 a47 a56 a65 a74 a12 a21 a37 a46 a55 a64 a73 a11 a27 a36 a45 a54 a63 a72The converse theorem is not true.
Example
this pandiagonal square 7th order of prime numbers with the magic constant S = 1597:191 89 397 409 43 157 311 379 103 101 491 17 313 193 317 241 109 163 439 47 281 223 383 227 107 541 37 79 331 337 7 139 167 563 53 83 347 389 277 127 307 67 73 97 367 11 263 173 613This square is made up of numbers of the following array:
7 11 17 37 43 47 53 67 73 79 83 89 97 101 103 107 109 127 139 157 163 167 173 191 193 223 227 241 263 277 281 307 311 313 317 331 337 347 367 379 383 389 397 409 439 491 541 563 613We have four mutually orthogonal exact-square representations of an array:
#1 191 89 397 409 43 157 311 379 103 101 491 17 313 193 317 241 109 163 439 47 281 223 383 227 107 541 37 79 331 337 7 139 167 563 53 83 347 389 277 127 307 67 73 97 367 11 263 173 613 #2 191 379 317 223 331 83 73 89 103 241 383 337 347 97 397 101 109 227 7 389 367 409 491 163 107 139 277 11 43 17 439 541 167 127 263 157 313 47 37 563 307 173 311 193 281 79 53 67 613 #3 191 103 109 107 167 307 613 89 101 163 541 563 67 73 397 491 439 37 53 83 97 409 17 47 79 331 347 367 43 313 281 223 337 389 11 157 193 317 383 7 277 263 311 379 241 227 139 127 173 #4 311 313 439 107 7 347 73 157 17 163 227 337 83 613 43 491 109 383 331 67 173 409 101 241 223 53 307 263 397 103 317 79 563 127 11 89 379 281 37 167 277 367 191 193 47 541 139 389 97My program has found a solution n = 7, S = 797 with 5 errors:
131 137 223 149 11 139 7 67 13 97 173 57* 233 157 163 167 83 179 17 37 151 73 109 117* 127 181* 101 89 191 71 43 31 227 53 181 41 239 199 79 197 23 19 131* 61 35* 59 107 211 193I think there is a right solution.
My program has found a solution with 4 errors:
151 139 227 181 13 79 7 43 11 73 157 49* 253* 211 167 137 97 193 23 31 149 107 109 131 113 199 97* 41 173 101 59 17 197 71 179 53 239 127 89 233 37 19 103 61 83* 47 83 229 191This I do a better approximation to the pandiagonal square of order 7 of consecutive primes with a magic constant S = 797.
Does anybody have a better approximation?
Interesting conversion pandiagonal squares of order 7
Let there pandiagonal square with magic constant S.
also have pandiagonal square with magic constant S.
Example
S = 1597, z = 107 139 167 563 53 331 337 389 277 127 307 67 83 347 367 11 263 173 613 73 97 397 409 43 157 311 191 89 101 491 17 313 193 379 103 109 163 439 47 281 317 241 227 107 541 37 79 223 383After the conversion
7 129 177 563 53 331 337 399 277 117 307 67 83 347 357 21 263 173 613 73 97 397 409 43 157 311 191 89 101 491 17 313 193 389 93 109 163 439 47 271 317 251 227 107 541 37 89 213 383I’m trying to make pandiagonal square of order 8 of the following consecutive primes:
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 439Magic constant S = 2016.
I use a template from the residues modulo 4:
1 3 3 3 3 1 3 3 1 3 3 1 3 3 3 3 3 1 1 3 1 3 3 1 3 1 3 3 1 1 1 3 1 1 3 1 3 3 3 1 3 1 1 3 1 1 1 1 1 1 1 1 3 3 3 3 3 1 1 1 1 1 3 1My program has found a solution with 11 errors:
89 103 139 179 263 373 431 439 313 271 359 389 151 131 191 211 383 337 349 307 197 167 163 113 227 269 107 127 353 409 241 283 229 193 83 149 379 419 331 233 367 433 293 347 137 181 101 157 257 273* 281 285* 195* 239 263* 223 151* 137* 405* 233* 341* 97 295* 357*The bad elements are marked *
I think that can find the best solution.
My computer runs very slowly.This my pandiagonal square of order 8 of primes with magic constant
S = 2000:11 13 683 293 647 263 59 31 311 557 61 191 53 149 229 449 29 233 193 443 197 353 163 389 653 173 83 73 107 211 569 131 23 67 641 269 659 317 17 7 277 521 71 251 19 113 239 509 103 347 167 281 271 467 137 227 593 89 101 199 47 127 587 257Enlarge all elements of the square by 2, we obtain pandiagonal square of arbitrary natural numbers with magic constant S = 2016:
13 15 685 295 649 265 61 33 313 559 63 193 55 151 231 451 31 235 195 445 199 355 165 391 655 175 85 75 109 213 571 133 25 69 643 271 661 319 19 9 279 523 73 253 21 115 241 511 105 349 169 283 273 469 139 229 595 91 103 201 49 129 589 259On this square, I got a template (see above).
We have a very similar pandiagonal square – a prototype of the desired solution.I have another prototype:
63 117 171 225 279 333 387 441 339 297 423 381 123 81 207 165 405 351 321 267 237 183 153 99 201 243 93 135 369 411 261 303 219 177 111 69 435 393 327 285 375 429 291 345 159 213 75 129 273 315 357 399 105 147 189 231 141 87 249 195 309 255 417 363S=2016
This pandiagonal square of arbitrary natural numbers, but it gives us an idea about the structure of the desired solution.
I take into account in its program structure representation – a sample. It helped me to find the best approximation to the solution.My program has found a solution with 10 errors:
89 83 151 223 227 373 431 439 337 283 419 397 107 127 179 167 367 353 313 251 257 199 163 113 191 269 139 131 349 389 241 307 233 193 103 137 379 383 311 277 359 421 197 331 157 281 97 173 229 305* 389* 397* 167* 83* 223* 223* 211 109 305* 149 373* 181 371* 317Bad elements are marked *
This pandiagonal square of order 8 with magic constant S = 2016, in there are only three non-prime numbers (305, 305, 371).
But there is the same prime numbers (389, 397, 167, 83, 223, 373).I changed the algorithm.
Now my program has found a solution with 6 errors:79 401 107 89 383 397 367 193 307 157 239 433 227 173 131 349 313 167 389 151 181 263 241 311 409 211 233 347 317 103 229 167* 127 101 179 269 431 97 439 373 271 337 331 197 191 353 223 113 281 283 257 199 149 379 109 359 229* 359* 281* 331* 137 251 277 151*This pandiagonal square of order 8 of consecutive primes with magic constant
S = 2016.
This solution is made only of primes, but there are the same numbers (167, 229, 359, 281, 331, 151).
I got a good approximation to the solution.
I think that you can find the best solution.Engaged someone with this problem? Please inform your preliminary results.
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