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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.

6, 10, 12, 36, 294, 24, 0, 12, 24, 72, 0, 24, 12, 36, 0, 1564, 0, 12, 12, 0, 156, 0, 12, 60, 36, 48, 24, 0, 0, 72, 60, 60, 24, 60, 12, 0, 12, 12, 12, 0, 0, 12, 180, 0, 60, 0, 60, 72, 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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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.

LINKS

Robert Israel, Table of n, a(n) for n = 1..3000

FORMULA

a(n) = sqrt(A117330(A340869(n))).

EXAMPLE

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.

MAPLE

f:= proc(n) local i, t;

t:= LinearAlgebra:-Determinant(Matrix(3, 3, [seq(ithprime(i), i=n..n+8)]));

if issqr(t) then sqrt(t) fi

end proc:

map(f, [$1..10000]);

MATHEMATICA

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 *)

PROG

(PARI) f(n) = matdet(matrix(3, 3, i, j, prime((n+j-1)+3*(i-1)))); \\ A117330

lista(nn) = my(x); for (n=1, nn, if (issquare(f(n), &x), print1(x, ", "))); \\ Michel Marcus, Jan 25 2021

CROSSREFS