

A130320


Given n numbers n>(n1)>(n2)>...>2>1, adding the first and last numbers leads to the identity n+1 = (n1)+2 = (n2)+3 = ... In case if some positive x_1, x_2, ... are added to n, (n1) etc, the strict inequality could be retained. This could be repeated finitely many times till it ends in inequality of form M > N where MN is minimal. This sequence gives the value of M for different n.


1



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
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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(n1)  2 with a(1) = 2. For n of form 2^k + 2^(k1), a(n) = 4a(n1) 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



