A130320 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.

1, 2, 4, 6, 10, 16, 18, 22, 34, 40, 56, 64, 66, 74, 78, 86, 130, 142, 148, 160, 216, 232, 240, 256, 258, 274, 282, 298, 302, 318, 326, 342, 514, 538, 550, 574, 580, 604, 616, 640, 856, 888, 904, 936, 944, 976, 992, 1024, 1026, 1058, 1074, 1106, 1114, 1146, 1162, 1194, 1198

Offset: 1

Author

Ramasamy Chandramouli, May 23 2007

Keywords

• nonn
• uned
• obsc

Comments

Apparently contains 2^(2k+1) and 2^k+2. - Ralf Stephan, Nov 10 2013

Examples

			a(5) = 10 because we have 5 > 4 > 3 > 2 > 1.
To follow a strict inequality we would have 5 + x > 4 + y > 3 > 2 > 1, where x >= 0, y >= 0.
The next level of inequality gives 1 + 5 + x > 2 + 4 + y > 3. This implies x > y.
Continuing with next level gives 3 + 6 + x > 6 + y. This gives x = 1, y = 0.
Hence 10 > 6 giving a(5) = 10.
		

Links

• Ramasamy Chandramouli, Table of n, a(n) for n = 1..17000

Formula

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.