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A340874 Square root of the determinant of the 3 X 3 matrix [prime(k), prime(k+1), prime(k+2); prime(k+3), prime(k+4), prime(k+5); prime(k+6), prime(k+7), prime(k+8)] when that determinant is a square.

%I A340874 #24 Jan 25 2021 19:03:16
%S A340874 6,10,12,36,294,24,0,12,24,72,0,24,12,36,0,1564,0,12,12,0,156,0,12,60,
%T A340874 36,48,24,0,0,72,60,60,24,60,12,0,12,12,12,0,0,12,180,0,60,0,60,72,
%U A340874 120,0,120,0,2150,0,24,12,0,0,60,0,36,48,120,0,0,0,0,0,0,24,0,0,56,0,24,0,48,0,2266
%N A340874 Square root of the determinant of the 3 X 3 matrix [prime(k), prime(k+1), prime(k+2); prime(k+3), prime(k+4), prime(k+5); prime(k+6), prime(k+7), prime(k+8)] when that determinant is a square.
%C A340874 The prime k-tuples conjecture implies that, for example, there are infinitely many k for which the matrix is of the form [x, x+4, x+10; x+22, x+24, x+30; x+34, x+36, x+42], in which case the determinant is 12^2.
%H A340874 Robert Israel, <a href="/A340874/b340874.txt">Table of n, a(n) for n = 1..3000</a>
%F A340874 a(n) = sqrt(A117330(A340869(n))).
%e A340874 a(3) = 12 because A340869(3) = 14 and the determinant of the 3 X 3 matrix [43, 47, 53; 59, 61, 67; 71, 73, 79] composed of prime(14) to prime(22) in order (by rows or columns) is 144 = 12^2.
%p A340874 f:= proc(n) local i,t;
%p A340874 t:= LinearAlgebra:-Determinant(Matrix(3, 3, [seq(ithprime(i), i=n..n+8)]));
%p A340874 if issqr(t) then sqrt(t) fi
%p A340874 end proc:
%p A340874 map(f, [$1..10000]);
%t A340874 m = 10^4; p = Prime[Range[m + 8]]; Select[Table[Sqrt @ Det @ Partition[p[[n ;; n + 8]], 3], {n, 1, m}], IntegerQ] (* _Amiram Eldar_, Jan 25 2021 *)
%o A340874 (PARI) f(n) = matdet(matrix(3,3,i,j,prime((n+j-1)+3*(i-1)))); \\ A117330
%o A340874 lista(nn) = my(x); for (n=1, nn, if (issquare(f(n), &x), print1(x, ", "))); \\ _Michel Marcus_, Jan 25 2021
%Y A340874 Cf. A117330, A340869.
%K A340874 nonn
%O A340874 1,1
%A A340874 _J. M. Bergot_ and _Robert Israel_, Jan 24 2021