%I A136617
%S A136617 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,
%T A136617 43,45,47,48,50,52,54,55,57,59,61,62,64,66,67,69,71,73,74,76,78,79,81,
%U A136617 83,85,86,88,90,91,93,95,97,98,100,102,103,105,107,109,110,112,114,115
%N A136617 a(n) = largest k such that the sum of k consecutive reciprocals 1/n +
... + 1/(n+k-1) does no exceed 1.
%C A136617 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:
%C A136617 .............H........H.....H........***.....H.......
%C A136617 ..............n-1......n.....n+1..............m......
%C A136617 ...........----o-------+------+-----.***.-----+-o----
%C A136617 ................\______________..______________/......
%C A136617 ...............................\/.....................
%C A136617 ............................Length 1..................
%C A136617 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.
%F A136617 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)) )
%e A136617 a(3) = 4 because 1/3+1/4+1/5+1/6 < 1 has 4 summands; adding 1/7 exceeds
1.
%p A136617 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)
%Y A136617 Cf. A136616, A002387, A004080, A079353, A081881, A096618, A118050, A118051.
%Y A136617 Sequence in context: A083088 A080755 A083089 this_sequence A081223 A047293
A076521
%Y A136617 Adjacent sequences: A136614 A136615 A136616 this_sequence A136618 A136619
A136620
%K A136617 easy,nonn
%O A136617 1,2
%A A136617 Rainer Rosenthal (r.rosenthal(AT)web.de), Jan 13 2008
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