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Search: id:A136617
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| A136617 |
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a(n) = largest k such that the sum of k consecutive reciprocals 1/n + ... + 1/(n+k-1) does no exceed 1. |
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+0
3
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| 1, 2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 24, 26, 28, 30, 31, 33, 35, 36, 38, 40, 42, 43, 45, 47, 48, 50, 52, 54, 55, 57, 59, 61, 62, 64, 66, 67, 69, 71, 73, 74, 76, 78, 79, 81, 83, 85, 86, 88, 90, 91, 93, 95, 97, 98, 100, 102, 103, 105, 107, 109, 110, 112, 114, 115
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Heuristic formula from David Cantrell (SeqFan mailing list January 2008). Think of a ruler with harmonic numbers H(n) as marks. Then A136617(n) gives the number of marks m-n+1 = A136616(n)-n+1:
.............H........H.....H........***.....H.......
..............n-1......n.....n+1..............m......
...........----o-------+------+-----.***.-----+-o----
................\______________..______________/......
...............................\/.....................
............................Length 1..................
The first 23 terms of A083088 are identical to those of A136617 but the limits of A083088(n)/n and A136617(n)/n for n->oo are different.
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FORMULA
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a(n) = A136616(n-1) - n + 1 with David Cantrell's heuristics: a(n) = floor( (e - 1)*(n - 1/2) + (e - 1/e)/(24*(n - 1/2)) )
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EXAMPLE
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a(3) = 4 because 1/3+1/4+1/5+1/6 < 1 has 4 summands; adding 1/7 exceeds 1.
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MAPLE
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A136617 := proc(n) local t, m; t:= 0; for m from n do t:= t+1/m; if t > 1 then return m-n; fi; od; end proc; [seq(A136617(n), n=1..100)]; (Courtesy Robert Israel January 2008)
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CROSSREFS
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Cf. A136616, A002387, A004080, A079353, A081881, A096618, A118050, A118051.
Sequence in context: A083088 A080755 A083089 this_sequence A081223 A047293 A076521
Adjacent sequences: A136614 A136615 A136616 this_sequence A136618 A136619 A136620
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KEYWORD
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easy,nonn
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AUTHOR
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Rainer Rosenthal (r.rosenthal(AT)web.de), Jan 13 2008
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