A136617 a(n) = largest k such that the sum of k consecutive reciprocals 1/n + ... + 1/(n+k-1) does not exceed 1.

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

Offset: 1

Author

Rainer Rosenthal, Jan 13 2008

Keywords

• easy
• nonn

Comments

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.

Examples

			a(3) = 4 because 1/3+1/4+1/5+1/6 < 1 has 4 summands; adding 1/7 exceeds 1.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• E. R. Bobo, A sequence related to the harmonic series, College Math. J. 26 (1995), 308-310.

Crossrefs

Cf. A136616, A002387, A004080, A079353, A081881, A096618, A103762, A118050, A118051.

Programs

Maple

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, Jan 2008

Mathematica

Table[Module[{start = Floor[z (E - 1)] - 1},
  NestWhile[# + 1 &, start, HarmonicNumber[# + z] - HarmonicNumber[z] + 1/z <= 1 &]], {z, 1, 100}] (* Peter J. C. Moses, Aug 20 2012 *)

Formula

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)) ).