%I #21 Sep 08 2022 08:45:09
%S 2,4,6,7,9,11,12,14,16,18,19,21,23,24,26,28,30,31,33,35,36,38,40,41,
%T 43,45,47,48,50,52,53,55,57,59,60,62,64,65,67,69,70,72,74,76,77,79,81,
%U 82,84,86,88,89,91,93,94,96,98,100,101,103,105,106,108,110,111,113,115,117
%N a(n) = ceiling(n*(1+1/sqrt(2))).
%C Equivalently, numbers m such that {rm} > {r}, where r=2^(1/2) and { } denotes fractional part - see comments below.
%C _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_.
%C First, suppose m satisfies {rm} > {r}. Define n := 2m - [rm] - 1 = m (2-r) + {rm} - 1
%C Then n is integer and n (1 + 1/r) = m-1 + {rm}(1+1/r) - 1/r.
%C 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.
%C 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.
%C 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}.
%H G. C. Greubel, <a href="/A080755/b080755.txt">Table of n, a(n) for n = 1..10000</a>
%t With[{c=1+1/Sqrt[2]},Ceiling[c #]&/@Range[80]] (* _Harvey P. Dale_, Apr 26 2011 *)
%o (PARI) vector(100, n, ceil(n*(1+1/sqrt(2)))) \\ _G. C. Greubel_, Aug 16 2018
%o (Magma) [Ceiling(n*(1+1/Sqrt(2))): n in [1..100]]; // _G. C. Greubel_, Aug 16 2018
%Y Equals A003152 + 1. This and its complement A080754 partition the integers >= 2.
%K nonn
%O 1,1
%A _Benoit Cloitre_ and _N. J. A. Sloane_, Mar 09 2003
%E Edited by _N. J. A. Sloane_ at the suggestion of _Andrew S. Plewe_, Jun 08 2007