OFFSET
1,2
COMMENTS
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
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
LINKS
Leroy Quet, sum{k>=m} floor(n/k) = n, rec.puzzles [From Bryce Herdt (mathidentity(AT)yahoo.com), Nov 02 2008]
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Oct 05 2008
EXTENSIONS
More terms from Stefan Steinerberger, Oct 17 2008
STATUS
approved