A268790 Magic sums of 3 X 3 magic squares composed of primes.

177, 213, 219, 267, 309, 327, 381, 393, 411, 417, 447, 453, 471, 501, 519, 537, 573, 579, 633, 681, 717, 723, 753, 771, 789, 807, 813, 843, 849, 879, 921, 933, 1011, 1041, 1047, 1059, 1077, 1101, 1119, 1137, 1149, 1167, 1191, 1203, 1227, 1257, 1263, 1293

Offset: 1

Author

Arkadiusz Wesolowski, Feb 13 2016

Keywords

• nonn

Comments

From Robert Israel, Feb 16 2016: (Start)

All terms are 3 times odd primes.

3*p is a term if and only if p is a prime not in A073350.

Conjecture: 3*p is a term for every prime > 859.

I verified this for all primes < 100000.

The Green-Tao theorem implies the sequence is infinite: given one magic square with entries a(i,j), there are infinitely many pairs of positive integers x,y such that b(i,j) = x + y*a(i,j) are all prime. Then b(i,j) form another magic square. (End)

Every number of the form 3*(A227284(n) + 840) is in this sequence. - Arkadiusz Wesolowski, Feb 22 2016

The terms equal three times the central elements (and equivalently, one third of the sum of all elements) of the 3 X 3 magic squares made of primes, which are listed in A320872. - M. F. Hasler, Oct 28 2018

Examples

			Examples of 3 X 3 magic squares composed of primes.
.
+---+---+---+
| 17| 89| 71|
+---+---+---+
|113| 59| 5 |
+---+---+---+
| 47| 29|101|
+---+---+---+
The magic constant is 177 = a(1).
.
+---+---+---+
| 41| 89| 83|
+---+---+---+
|113| 71| 29|
+---+---+---+
| 59| 53|101|
+---+---+---+
The magic constant is 213 = a(2).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..9552
• G. L. Honaker, Jr. and Chris Caldwell, Prime Curios!: 859
• Wikipedia, Magic square
• Index entries for sequences related to magic squares

Crossrefs

Cf. A000040, A024351, A073350, A164843, A227284.

Cf. A320872, A320871, A320873.

Programs

Maple

N:= 10000: # to get all terms <= N P:= select(isprime,{seq(p,p=3..2*N/3,2)}):
count:= 0:
for ic from 1 while P[ic] <= N/3 do
   c:= P[ic];
   V:= map(`-`,P[ic+1..-1],c) intersect map(t -> c-t, P[1..ic-1]);
   nv:= nops(V);
   VV:= {seq(seq(V[j]-V[i],j=i+1..nv),i=1..nv-1)} intersect V;
   nvv:= nops(VV);
   found:= false;
   for ia from 1 to nvv while not found do
     a:= VV[ia];
     for ib from ia+1 to nvv while VV[ib] < c - a do
       b:= VV[ib];
       if b <> 2*a and {c-a-b,c-a+b,c-b+a,c+a+b} subset P then
          found:= true;
          count:= count+1;
          A[count]:= 3*c;
          break
       fi
     od
   od:
od:
seq(A[i],i=1..count); # Robert Israel, Feb 16 2016

PARI

c=3;A268790_vec=3*vector(50,i,c=A320872_row(1,0,c+1)[2,2]) \\ Illustrates formula & comment. - M. F. Hasler, Oct 28 2018

PARI

is_A268790(c)={denominator(c/=3)==1&& isprime(c)&& forstep(a=2,c\2-1,2, isprime(c-a)&& isprime(c+a)&& forstep(b=2,c-2*a-2,2, isprime(c-2*a-b)&& isprime(c-a-b)&& isprime(c-b)&& isprime(c+b)&& isprime(c+a+b)&& isprime(c+2*a+b)&& return(1)))} \\ M. F. Hasler, Oct 28 2018

Formula

If conjecture is true, a(n) = 3*prime(n+40) for n >= 110. - Robert Israel, Feb 16 2016

A268790 = 3*{column 5 of A320872} as a set, i.e., with duplicates removed. - M. F. Hasler, Oct 28 2018