A173981 Magic constants of 4 X 4 magic squares which consist of consecutive primes.

258, 276, 5118, 19896, 50478, 13039980, 13297678, 37166532, 96266778, 104100834, 185320518, 383918304, 397075158, 467692578, 683981178, 816166200, 852339780, 874276354, 919926054, 931402662, 1016171040, 1021731906, 1026857286, 1200889680, 1501212942, 1533729354, 1686059670

Offset: 1

Author

Natalia Makarova, Mar 04 2010

Keywords

• nonn

Comments

Necessary conditions for 16 primes from which a magic square of order 4 can be made, are:

1. Their sum S is a multiple of 4

2. Magic constant of possible square K=S/4 is even number.

This is equivalent to the requirement for S to be a multiple of 8.

For a fixed magic constant S, it is easy to obtain the set of n^2 consecutive primes that sum up to n*S, and in particular the smallest one: see the PROGRAM in A260673 which computes the smallest prime for any of the magic sums listed here (for n = 4), and A272386 for the n = 5 analog. The converse is trivial, cf. FORMULA and PROGRAM below. - M. F. Hasler, Oct 28 2018

Examples

			The smallest magic square of order 4 has the constant of 258. See A073520 and A073521.
The following array of 16 consecutive primes:
   37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103
also produces the magic square with the constant of K = 276:
    [ 41 37 97 101]
    [103 83 47  43]
    [ 71 67 79  59]
    [ 61 89 53  73]
But then not every array of 16 consecutive primes produces a magic square. The next magic square can be made from the array (1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321):
    [1229 1249 1321 1319]
    [1301 1303 1231 1283]   (K = 5118)
    [1297 1277 1307 1237]
    [1291 1289 1259 1279]
Two more examples:
    [4943 4933 5011 5009]                   [12553 12583 12689 12653]
    [4999 4973 4967 4957]   (K = 19896),    [12641 12647 12601 12589]   (K = 50478)
    [5003 4969 4987 4937]                   [12671 12611 12619 12577]
    [4951 5021 4931 4993]                   [12613 12637 12569 12659]
		

Links

• Zhao Hui Du, Table of n, a(n) for n = 1..443
• Max Alekseyev, Natalia Makarova and others, Discussion in the Magic square finding thread on scientific forum dxdy.ru (in Russian), February-March 2010.
• Carlos Rivera, Puzzle 3

Crossrefs

Cf. A073520, A073521, A260673 (smallest terms in magic 4 X 4 squares of consecutive primes), A270865 (idem for semimagic squares). Subsequence of A270864 (analog for semimagic squares).

Cf. A270305 (analog for 3 X 3), A177434 (analog for 6 X 6).

Programs

PARI

A173981(n, p=A260673[n], N=4)=sum(i=2, N^2, p=nextprime(p+1), p)/N \\ Illustration of the formula. - M. F. Hasler, Oct 28 2018

Formula

a(n) = Sum_{k=0..15} A000040(A000720(A260673(n))+k)/4. - M. F. Hasler, Oct 28 2018

Extensions

a(24)-a(25) from Arkadiusz Wesolowski, Dec 13 2015

Edited and added a(26)-a(27) (using A260673) by M. F. Hasler, Oct 30 2018