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Indices n == 1 (mod 9) such that the 3 X 3 matrix with components (row by row) prime(n+k), 0 <= k <= 8, has zero determinant.
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%I #9 Dec 15 2016 02:32:35

%S 1009,6031,9613,19378,49996,67285,91549,101278,102097,107182,142723,

%T 154792,168562,175006,183718,196345,200530,204031,215407,240292,

%U 263395,264628,277723,289171,299323,307684,313111,369676,372601,376921,425935

%N Indices n == 1 (mod 9) such that the 3 X 3 matrix with components (row by row) prime(n+k), 0 <= k <= 8, has zero determinant.

%C By considering only indices congruent to 1 (mod 9) each prime occurs in exactly one of these matrices. - Subsequence of A117345.

%o (PARI) {m=426000;forstep(n=1,m,9,M=matrix(3,3,i,j,prime(n+3*(i-1)+j-1));if(matdet(M,1)==0,print1(n,",")))}

%Y Cf. A117345.

%K nonn

%O 1,1

%A _Cino Hilliard_, Apr 24 2006

%E Edited by _Klaus Brockhaus_, Apr 28 2006