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A145265
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A positive integer n is included if there exists a positive integer m such that sum{k>=0} floor(n/(m+k)) = n.
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2
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1, 4, 5, 8, 15, 18, 19, 22, 23, 26, 33, 36, 37, 40, 41, 44, 51, 54, 55, 58, 59, 62, 69, 72, 73, 76, 77, 80, 87, 90, 91, 94, 95, 98, 105, 108, 109, 112, 113, 116, 123, 126, 127, 130, 131, 134, 141, 144, 145, 148, 149, 152, 159, 162, 163, 166, 167, 170, 177, 180, 181, 184
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| This sequence is the complement of sequence A145266. A145264(a(n)) >= 1.
Does this sequence contain all of those and only those, positive integers that are congruent to 0, 1, 4, 5, 8, 15 (mod 18)? [From 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. [From Bryce Herdt (mathidentity(AT)yahoo.com), Nov 02 2008]
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LINKS
| Leroy Quet, sum{k>=m} floor(n/k) = n, rec.puzzles [From Bryce Herdt (mathidentity(AT)yahoo.com), Nov 02 2008]
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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.
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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 [From Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Oct 17 2008]
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CROSSREFS
| A145264, A145266
Sequence in context: A019526 A145488 A050892 * A171938 A072808 A104884
Adjacent sequences: A145262 A145263 A145264 * A145266 A145267 A145268
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KEYWORD
| nonn
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AUTHOR
| Leroy Quet Oct 05 2008
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EXTENSIONS
| More terms from Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Oct 17 2008
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