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%I #17 Feb 18 2014 22:42:00
%S 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 43,45,47,48,50,52,54,55,57,59,61,62,64,66,67,69,71,73,74,76,78,79,81,
%U 83,85,86,88,90,91,93,95,97,98,100,102,103,105,107,109,110,112,114,115
%N a(n) = largest k such that the sum of k consecutive reciprocals 1/n + ... + 1/(n+k-1) does not exceed 1.
%C 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 .............H........H.....H........***.....H.......
%C ..............n-1......n.....n+1..............m......
%C ...........----o-------+------+-----.***.-----+-o----
%C ................\______________..______________/......
%C ...............................\/.....................
%C ............................Length 1..................
%C 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.
%H Clark Kimberling, <a href="/A136617/b136617.txt">Table of n, a(n) for n = 1..10000</a>
%H E. R. Bobo, <a href="http://www.jstor.org/stable/2687034">A sequence related to the harmonic series</a>, College Math. J. 26 (1995), 308-310.
%F 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 a(3) = 4 because 1/3+1/4+1/5+1/6 < 1 has 4 summands; adding 1/7 exceeds 1.
%p 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)]; # _Robert Israel_, January 2008
%t Table[Module[{start = Floor[z (E - 1)] - 1},
%t NestWhile[# + 1 &, start, HarmonicNumber[# + z] - HarmonicNumber[z] + 1/z <= 1 &]], {z, 1, 100}] (* _Peter J. C. Moses_, Aug 20 2012 *)
%Y Cf. A136616, A002387, A004080, A079353, A081881, A096618, A103762, A118050, A118051.
%K easy,nonn
%O 1,2
%A _Rainer Rosenthal_, Jan 13 2008
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