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Integers k such that f(k) - f(k-1) equals 1, where f(n) = floor(n/exp(sqrt(log(n)))).
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%I #14 May 02 2020 02:44:14

%S 3,9,16,24,33,42,51,61,71,82,93,104,115,127,139,151,163,175,188,200,

%T 213,226,239,253,266,279,293,307,321,335,349,363,377,392,406,421,436,

%U 451,465,480,495,511,526,541,557,572,588,603,619,635,650,666,682,698,714,730

%N Integers k such that f(k) - f(k-1) equals 1, where f(n) = floor(n/exp(sqrt(log(n)))).

%H J. M. De Koninck and V. Ouellet, <a href="http://ac.inf.elte.hu/Vol_044_2015/153_44.pdf">On the n-th element of a set of positive integers</a>, Annales Univ. Sci. Budapest Sect. Comput. 44 (2015), 153-164. See 3. on p. 162.

%t f[n_]:=Floor[n/Exp[Sqrt[Log[n]]]]; Select[Range[2, 1000], f[#] - f[# - 1]==1 &] (* _Indranil Ghosh_, Jul 28 2017 *)

%o (PARI) f(n) = floor(n/exp(sqrt(log(n))));

%o isok(n) = f(n) - f(n-1) == 1;

%o (Python)

%o from sympy import floor, exp, sqrt, log

%o def f(n): return floor(n/exp(sqrt(log(n))))

%o print([n for n in range(2, 1001) if f(n) - f(n - 1) == 1]) # _Indranil Ghosh_, Jul 28 2017

%K nonn

%O 1,1

%A _Michel Marcus_, Jul 28 2017