1,1

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}.

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

Sequence in context: A039009 A032793 A083088 this_sequence A083089 A136617 A081223

Adjacent sequences: A080752 A080753 A080754 this_sequence A080756 A080757 A080758

nonn

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

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