Search: id:A002387
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%I A002387 M1249 N1385
%S A002387 1,2,4,11,31,83,227,616,1674,4550,12367,33617,91380,248397,675214,
%T A002387 1835421,4989191,13562027,36865412,100210581,272400600,740461601,
%U A002387 2012783315,5471312310,14872568831,40427833596,109894245429,298723530401,
               812014744422
%N A002387 Least k such that H(k) > n, where H(k) is the harmonic number sum_{i=1..k} 
               1/i.
%C A002387 For k>=1, log(k+1/2) + gamma < H(k) < log(k+1/2) + gamma + 1/(24k^2), 
               where gamma is Euler's constant (A001620). It is likely that the 
               upper and lower bounds have the same floor for all k>=2, in which 
               case a(n) = floor(exp(n-gamma)+1/2) for all n>=0. - Dean Hickerson 
               (dean.hickerson(AT)yahoo.com), Apr 19 2003
%C A002387 This remark is based on a simple heuristic argument. The lower and upper 
               bounds differ by 1/(24k^2), so the probability that there's an integer 
               between the two bounds is 1/(24k^2). Summing that over all k >= 2 
               gives the expected number of values of k for which there's an integer 
               between the bounds. That sum equals pi^2/144 - 1/24 ~ 0.02687. That's 
               much less than 1, so it is unlikely that there are any such values 
               of k. - Dean Hickerson (dean.hickerson(AT)yahoo.com), Apr 19 2003
%C A002387 Referring to A118050 and A118051, using a few terms of the asymptotic 
               series for the inverse of H(x), we can get an expression which, with 
               greater likelihood than mentioned above, should give a(n) for all 
               n >= 0. For example, using the same type of heuristic argument given 
               by Dean Hickerson, it can be shown that, with probability > 99.995%, 
               we should have, for all n >= 0, a(n) = floor(u + 1/2 - 1/(24u) + 
               3/(640u^3)) where u = e^(n - gamma). - David W. Cantrell (DWCantrell(AT)sigmaxi.net)
%C A002387 For k > 1, H(k) is never an integer. Hence apart from the first two terms 
               this sequence coincides with A004080. - Nick Hobson Nov 25 2006
%D A002387 J. V. Baxley, Euler's constant, Taylor's formula and slowly converging 
               series, Math. Mag., 65 (1992), 302-313.
%D A002387 R. P. Boas, Jr. and J. W. Wrench, Jr., Partial sums of the harmonic series, 
               Amer. Math. Monthly, 78 (1971), 864-870.
%D A002387 John H. Conway and R. K. Guy, "The Book of Numbers," Copernicus, an imprint 
               of Springer-Verlag, NY, 1996, page 258-259.
%D A002387 J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 83, p. 28, Ellipses, 
               Paris 2008.
%D A002387 Ronald Lewis Graham, Donald Ervin Knuth and Oren Patashnik, "Concrete 
               Mathematics, a Foundation for Computer Science," Addison-Wesley Publishing 
               Co., Reading, MA, 1989, Page 258-264, 438.
%D A002387 W. Sierpi\'{n}ski, Sur les decompositions de nombres rationnels, Oeuvres 
               Choisies, Acad\'{e}mie Polonaise des Sciences, Warsaw, Poland, 1974, 
               p. 181.
%D A002387 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A002387 N. J. A. Sloane, Illustration for sequence M4299 (=A007340) in The Encyclopedia 
               of Integer Sequences (with S. Plouffe), Academic Press, 1995.
%D A002387 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A002387 I. Stewart, L'univers des nombres, pp. 54, Belin-Pour La Science, Paris 
               2000.
%H A002387 T. D. Noe, <a href="b002387.txt">Table of n, a(n) for n=0..100</a> (using 
               Hickerson's formula)
%H A002387 N. Hobson, <a href="http://www.qbyte.org/puzzles/p034s.html">Harmonic 
               Sum</a>.
%F A002387 Note that the conditionally convergent series Sum_{ k >= 1 } (-1)^(k+1)/
               k = log 2 (A002162).
%F A002387 Lim as n -> inf. a(n+1)/a(n) = e. - Robert G. Wilson v (rgwv(AT)rgwv.com), 
               Dec 07 2001
%F A002387 It is conjectured that, for n>1, a(n) = floor(exp(n-gamma)+1/2). - Benoit 
               Cloitre (benoit7848c(AT)orange.fr), Oct 23 2002
%t A002387 fh[0]=0; fh[1]=1; fh[k_] := Module[{tmp}, If[Floor[tmp=Log[k+1/2]+EulerGamma]==Floor[tmp+1/
               (24k^2)], Floor[tmp], UNKNOWN]]; a[0]=1; a[1]=2; a[n_] := Module[{val}, 
               val=Round[Exp[n-EulerGamma]]; If[fh[val]==n&&fh[val-1]==n-1, val, 
               UNKNOWN]]; (* fh[k] is either floor(H(k)) or UNKNOWN *)
%Y A002387 Apart from initial terms, same as A004080.
%Y A002387 Cf. A055980.
%Y A002387 Sequence in context: A148159 A102814 A034770 this_sequence A148160 A148161 
               A148162
%Y A002387 Adjacent sequences: A002384 A002385 A002386 this_sequence A002388 A002389 
               A002390
%K A002387 nonn,nice
%O A002387 0,2
%A A002387 N. J. A. Sloane (njas(AT)research.att.com).
%E A002387 Terms for n >= 13 computed by Eric Weisstein (eric(AT)weisstein.com). 
               Corrected by Jim Buddenhagen (jbuddenh(AT)gmail.com) and Eric W. 
               Weisstein (eric(AT)weisstein.com), Feb 18 2001.
%E A002387 Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Apr 19 2003
%E A002387 a(27) from Thomas Gettys (tpgettys(AT)comcast.net), Dec 05 2006
%E A002387 a(28) from Thomas Gettys (tpgettys(AT)comcast.net), Jan 21 2007

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