OFFSET
1,1
COMMENTS
For a 9 X 9 magic square composed of 81 consecutive primes, it is necessary that the sum of these primes is a multiple of 9.
This sequence consists of integers equal the sum of 81 consecutive primes divides by 9. It is not known whether each such set of consecutive primes can be arranged into 9 X 9 magic square but it looks plausible.
LINKS
Stefano Tognon, Squares from 37 (in Italian).
Natalia Makarova, Sequence of Magic Numbers MK 9th Order (in Russian).
EXAMPLE
a(1)=2211 for a square containing prime(12)..prime(92):
[37 127 163 179 229 233 379 421 443
41 431 463 457 59 139 433 109 79
409 311 389 71 307 347 281 53 43
373 137 181 251 401 239 317 89 223
173 419 101 103 113 353 313 277 359
97 383 397 479 47 197 107 263 241
349 131 193 149 367 199 73 467 283
439 61 257 191 227 167 151 449 269
293 211 67 331 461 337 157 83 271]
a(2)=2261 for a square containing prime(13)..prime(93):
[41 379 281 467 349 257 229 199 59
313 223 127 337 131 101 479 107 443
409 71 331 79 137 263 347 271 353
211 307 487 149 251 293 181 113 269
191 419 109 439 173 233 103 397 197
97 283 193 317 433 457 241 157 83
461 139 239 359 373 179 67 401 43
89 277 73 53 367 167 463 389 383
449 163 421 61 47 311 151 227 431]
MAPLE
s:= proc(n) option remember;
`if` (n=1, add (ithprime(i), i=1..81),
ithprime(n+80) -ithprime(n-1) +s(n-1))
end:
a:= proc(n) option remember; local k, m;
a(n-1);
for k from 1+b(n-1) while irem (s(k), 9, 'm')<>0 do od;
b(n):= k; m
end:
a(0):=0: b(0):=0:
seq (a(n), n=1..50);
MATHEMATICA
Total[#]/9&/@Select[Partition[Prime[Range[500]], 81, 1], Divisible[ Total[ #], 9]&] (* Harvey P. Dale, Jan 08 2014 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Natalia Makarova, Jun 11 2011
EXTENSIONS
Edited by Max Alekseyev, Jun 18 2011
STATUS
approved