%I #11 Mar 08 2015 18:21:48
%S 1,4,5,8,15,18,19,22,23,26,33,36,37,40,41,44,51,54,55,58,59,62,69,72,
%T 73,76,77,80,87,90,91,94,95,98,105,108,109,112,113,116,123,126,127,
%U 130,131,134,141,144,145,148,149,152,159,162,163,166,167,170,177,180,181,184
%N A positive integer n is included if there exists a positive integer m such that Sum_{k>=0} floor(n/(m+k)) = n.
%C This sequence is the complement of sequence A145266. A145264(a(n)) >= 1.
%C Does this sequence contain all of those and only those, positive integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)? - _Leroy Quet_, Oct 31 2008
%C This sequence and its cross-referents may be calculated more easily by checking whether a partial sum of sum{k>=0} floor(n/(n-k)) ever equals n; that is, calculating from the top down. It appears that the terms are precisely those congruent to 0, 1, 4, 5, 8, or 15 modulo 18. - Bryce Herdt (mathidentity(AT)yahoo.com), Nov 02 2008
%H Leroy Quet, <a href="http://groups.google.com/group/rec.puzzles/browse_thread/thread/458f20b5f40a6493#">sum{k>=m} floor(n/k) = n, rec.puzzles</a> [From Bryce Herdt (mathidentity(AT)yahoo.com), Nov 02 2008]
%e Checking n = 8: floor(8/3) + floor(8/4) + floor(8/5) + floor(8/6) + floor(8/7) + floor(8/8) = 2 + 2 + 1 + 1 + 1 + 1 = 8. So 8 is included in the sequence. Checking n = 6: floor(6/2) + floor(6/3) + floor(6/4) + floor(6/5) + floor(6/6) = 3 + 2 + 1 + 1 + 1 = 8, which is > 6. But floor(6/3) + floor(6/4) + floor(6/5) + floor(6/6) = 2 + 1 + 1 + 1 = 5, which is < 6. So 6 is not included in the sequence.
%t a = {}; For[n = 1, n < 200, n++, c = 0; For[m = 1, m < n + 1, m++,If[Sum[Floor[n/(m + k)], {k, 0, n}] == n, c = 1; Break]]; If[c == 1, AppendTo[a, n]]]; a (* _Stefan Steinerberger_, Oct 17 2008 *)
%Y Cf. A145264, A145266.
%K nonn
%O 1,2
%A _Leroy Quet_, Oct 05 2008
%E More terms from _Stefan Steinerberger_, Oct 17 2008
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