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

1,2

Comments

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

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.

Example

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.

Crossrefs

Sequence in context: A073805 A352587 A301374 * A339574 A258599 A101176

Adjacent sequences: A130317 A130318 A130319 * A130321 A130322 A130323

Keyword

nonn,uned,obsc

Author

Ramasamy Chandramouli, May 23 2007

Status

approved