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A183867
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a(n) = n + floor(2*sqrt(n)); complement of A184676.
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4
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3, 4, 6, 8, 9, 10, 12, 13, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 80, 81, 82, 83
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OFFSET
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1,1
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COMMENTS
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Also equals n + floor(sqrt(n) + sqrt(n+1/2)). Proof: floor(2*sqrt(n)) is the largest k such that k^2/4 <= n, while floor(sqrt(n) + sqrt(n+1/2)) is the largest k such that (k^2 - 1)/4 + 1/(16*k^2) <= n. All perfect squares are 0 or 1 (mod 4). In either case, it is easily verified that one of the inequalities is satisfied if and only if the other inequality is satisfied. - Nathaniel Johnston, Jun 26 2011
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LINKS
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Nathaniel Johnston, Table of n, a(n) for n = 1..5000
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MAPLE
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seq(n+floor(2*sqrt(n)), n=1..67); # Nathaniel Johnston, Jun 26 2011
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MATHEMATICA
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a=4; b=0;
Table[n+Floor[(a*n+b)^(1/2)], {n, 100}]
Table[n-1+Ceiling[(n*n-b)/a], {n, 70}]
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PROG
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(PARI) a(n) = n+floor(2*sqrt(n)); \\ Michel Marcus, Dec 08 2015
(Magma) [n+Floor(2*Sqrt(n)): n in [1..100]]; // Vincenzo Librandi, Dec 09 2015
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CROSSREFS
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Cf. A179272.
Sequence in context: A263098 A317391 A134745 * A182829 A112800 A294456
Adjacent sequences: A183864 A183865 A183866 * A183868 A183869 A183870
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KEYWORD
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nonn,easy
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AUTHOR
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Clark Kimberling, Jan 07 2011
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STATUS
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approved
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