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A080755
a(n) = ceiling(n*(1+1/sqrt(2))).
3
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
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
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
Equals A003152 + 1. This and its complement A080754 partition the integers >= 2.
Sequence in context: A195127 A292654 A083088 * A372779 A083089 A136617
KEYWORD
nonn,changed
AUTHOR
EXTENSIONS
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007
STATUS
approved