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A080755 a(n) = ceiling(n*(1+1/sqrt(2))). 2
2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 24, 26, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 45, 47, 48, 50, 52, 53, 55, 57, 59, 60, 62, 64, 65, 67, 69, 70, 72, 74, 76, 77, 79, 81, 82, 84, 86, 88, 89, 91, 93, 94, 96, 98, 100, 101, 103, 105, 106, 108, 110, 111, 113, 115, 117 (list; graph; refs; listen; history; internal format)
OFFSET

1,1

COMMENTS

Equivalently, numbers m such that {rm} > {r}, where r=2^(1/2) and { } denotes fractional part - see comments below.

Andrew Plewe, May 18 2007, observed that the sequence defined by a(n) = ceiling(n*(1+1/sqrt(2))) appeared to give the same numbers as the sequence, originally due to Clark Kimberling (ck6(AT)evansville.edu), Jul 01 2006, defined by: numbers m such that {rm} > {r}, where r=2^(1/2). The following proof that these sequences are indeed the same is due to David Applegate.

First, suppose m satisfies {rm} > {r}. Define n := 2m - [rm] - 1 = m (2-r) + {rm} - 1

Then n is integer and n (1 + 1/r) = m-1 + {rm}(1+1/r) - 1/r.

Now {rm} < 1 so {rm}(1+1/r)-1/r < 1. And {rm}>{r}=r-1, so {rm}(1+1/r)-1/r > (r-1)(1+1/r)-1/r = 0. Thus ceil(n (1+1/r)) = m-1+ceil({rm}(1+1/r) - 1/r) = m. So m is in the sequence.

Conversely, let m be in the sequence, that is, there exists n such that m = ceil(n(1 + 1/r)) = ceil(n + [n/r] + {n/r}) = n + [n/r] + 1.

Then mr = rn + [n/r]r + r = r(r-1)(n/r-[n/r]) + r + n + 2[n/r] = r(r-1){n/r} + r + n + 2[n/r] and, since 0 < {n/r} < 1, r < r(r-1){n/r} + r < r^2=2, which implies {mr} = {r(r-1){n/r}+r} > {r}.

MATHEMATICA

With[{c=1+1/Sqrt[2]}, Ceiling[c #]&/@Range[80]] (* From Harvey P. Dale, Apr 26 2011 *)

CROSSREFS

Equals A003152 + 1. This and its complement A080754 partition the integers >= 2.

Sequence in context: A032793 A195127 A083088 * A083089 A136617 A198081

Adjacent sequences:  A080752 A080753 A080754 * A080756 A080757 A080758

KEYWORD

nonn

AUTHOR

Benoit Cloitre and N. J. A. Sloane (njas(AT)research.att.com), Mar 09 2003

EXTENSIONS

Edited by N. J. A. Sloane (njas(AT)research.att.com) at the suggestion of Andrew Plewe, Jun 08 2007

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Last modified February 17 18:41 EST 2012. Contains 206074 sequences.