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Least m>0 such that n-m divides 2*n-1+m.
1

%I #10 Aug 02 2012 14:58:13

%S 1,1,3,3,5,2,7,7,9,3,5,11,13,4,15,7,17,5,19,19,9,6,23,23,15,7,27,27,

%T 29,8,13,19,33,9,35,15,37,10,23,39,17,11,43,43,45,12,35,47,49,13,21,

%U 51,31,14,55,23,57,15,59,35,25,16,63,63,65,17,39,67,51,18,29,71

%N Least m>0 such that n-m divides 2*n-1+m.

%H Clark Kimberling, <a href="/A214745/b214745.txt">Table of n, a(n) for n = 2..1000</a>

%e Write x#y if x|y is false; then 10#22, 9#23, 8|24, so a(11) = 3.

%t Table[m = 1; While[! Divisible[2n-1+m,n-m], m++]; m, {n, 2, 100}]

%K nonn,easy

%O 2,3

%A _Clark Kimberling_, Jul 29 2012