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Numbers n such that A118679(n) = 1.
2

%I #4 Mar 31 2012 13:20:34

%S 1,3,10,19,24,30,43,51,58,62,73,75,82,94,101,106,115,116,118,128,138,

%T 147,149,159,160,163,167,172,183,186,190,191,195,201,211,214,219,249,

%U 250,252,253,260,266,272,274,277,279,283,290,294,296,306,309,310,318

%N Numbers n such that A118679(n) = 1.

%C A118679[ a(n) ] = 1, where A118679(n) = {1, 2, 1, 13, 19, 13, 17, 43, 53, 1, 19, ...} = Absolute value of numerator of determinant of n X n matrix with M(i,j) = i/(i+1) if i=j otherwise 1. A118679(n) = Numerator[ (n^2+3n-2)/(2(n+1)!) ] = Numerator[ ((2n+3)^2-17)/(4(n+1)!) ].

%F An integer n is in this sequence iff all prime divisors of n^2+3n-2 do not exceed n+1 and n^2+3n-2 is not of the form 2*p^2 for some prime p. [From _Max Alekseyev_, Jun 02 2009]

%t Select[Range[1000],Numerator[(#^2+3#-2)/(2(#+1)!)]==1&]

%Y Cf. A118679, A118680, A127853.

%K nonn

%O 1,2

%A _Alexander Adamchuk_, Feb 03 2007