OFFSET
1,1
COMMENTS
Equivalently, numbers m such that {rm} > {r}, where r=2^(1/2) and { } denotes fractional part - see comments below.
Andrew S. 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, 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 an 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 ceiling(n (1+1/r)) = m-1+ceiling({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 = ceiling(n(1 + 1/r)) = ceiling(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}.
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..10000
MATHEMATICA
With[{c=1+1/Sqrt[2]}, Ceiling[c #]&/@Range[80]] (* Harvey P. Dale, Apr 26 2011 *)
PROG
(PARI) vector(100, n, ceil(n*(1+1/sqrt(2)))) \\ G. C. Greubel, Aug 16 2018
(Magma) [Ceiling(n*(1+1/Sqrt(2))): n in [1..100]]; // G. C. Greubel, Aug 16 2018
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Benoit Cloitre and N. J. A. Sloane, Mar 09 2003
EXTENSIONS
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007
STATUS
approved