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Numbers n such that A118680(n) = 1.
2

%I #2 Mar 31 2012 13:20:34

%S 10,17,26,36,37,45,50,59,61,65,67,78,82,90,91,94,101,102,105,108,110,

%T 122,136,138,145,147,149,153,155,165,170,173,181,183,188,189,193,197,

%U 210,213,220,224,226,231,232,239,249,250,257,262,263,266,268,276,279

%N Numbers n such that A118680(n) = 1.

%C Also a(n) are the numbers n such that 1 + Sum[ k, {k,1,n} ] = 1 + n(n+1)/2 divides Product[ k, {k,1,n} ] = n!. A118680[ a(n) ] = 1, where A118680(n) = {2, 2, 7, 11, 2, 11, 29, 37, 23, 1, 67, 79, 23, 53, 11, 137, 1, ...} = Absolute value of numerator of determinant of n X n matrix with M(i,j) = (i+1)/i if i=j otherwise 1. A118680(n) = Numerator[ (1 + n(n+1)/2) / n! ].

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

%Y Cf. A118680, A118679, A127852.

%K nonn

%O 1,1

%A _Alexander Adamchuk_, Feb 03 2007