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A130320
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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.
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1
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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
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1,2
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COMMENTS
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Apparently contains 2^(2k+1) and 2^k+2. - Ralf Stephan, Nov 10 2013
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn,uned,obsc
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AUTHOR
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STATUS
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approved
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