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Numbers n such that A120292(n) = 1.
4

%I #3 Mar 31 2012 13:20:34

%S 2,3,5,7,8,9,14,15,16,17,23,31,34,35,39,41,43,50,51,56,66,67,70,71,75,

%T 77,92,96,97,98,101,107,112,115,117,123,131,132,153,155,156,160,163,

%U 165,166,170,172,182,185,196,198,200,203,204,207,212,218,231,243,246,249

%N Numbers n such that A120292(n) = 1.

%C A120292(n) = {2, 1, 1, 5, 1, 23, 1, 1, 1, 23, 17, 13, 5, 1, 1, 1, 1, 37, ...} Absolute value of numerator of determinant of n X n matrix with elements M[i,j] = Prime[i]/(1+Prime[i]) if i=j and 1 otherwise.

%t Do[f=Abs[Numerator[Det[DiagonalMatrix[Table[Prime[i]/(Prime[i]+1)-1,{i,1,n}]]+1]]];If[ f == 1, Print[n]],{n,1,256}]

%Y Cf. A120292.

%K hard,nonn

%O 1,1

%A _Alexander Adamchuk_, Feb 02 2007