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%I #8 Nov 10 2013 03:09:11
%S 1,2,4,6,10,16,18,22,34,40,56,64,66,74,78,86,130,142,148,160,216,232,
%T 240,256,258,274,282,298,302,318,326,342,514,538,550,574,580,604,616,
%U 640,856,888,904,936,944,976,992,1024,1026,1058,1074,1106,1114,1146,1162,1194,1198
%N Given n numbers n>(n-1)>(n-2)>...>2>1, adding the first and last numbers leads to the identity n+1 = (n-1)+2 = (n-2)+3 = ... In case if some positive x_1, x_2, ... are added to n, (n-1) etc, the strict inequality could be retained. This could be repeated finitely many times till it ends in inequality of form M > N where M-N is minimal. This sequence gives the value of M for different n.
%C Apparently contains 2^(2k+1) and 2^k+2. - _Ralf Stephan_, Nov 10 2013
%H Ramasamy Chandramouli, <a href="/A130320/b130320.txt">Table of n, a(n) for n = 1..17000</a>
%F For n of form 2^k, we have a(n) = 4a(n-1) - 2 with a(1) = 2. For n of form 2^k + 2^(k-1), a(n) = 4a(n-1) with a(1) = 4.
%e a(5) = 10 because we have 5 > 4 > 3 > 2 > 1.
%e To follow a strict inequality we would have 5 + x > 4 + y > 3 > 2 > 1, where x >= 0, y >= 0.
%e The next level of inequality gives 1 + 5 + x > 2 + 4 + y > 3. This implies x > y.
%e Continuing with next level gives 3 + 6 + x > 6 + y. This gives x = 1, y = 0.
%e Hence 10 > 6 giving a(5) = 10.
%K nonn,uned,obsc
%O 1,2
%A _Ramasamy Chandramouli_, May 23 2007
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