cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A002387 Least k such that H(k) > n, where H(k) is the harmonic number Sum_{i=1..k} 1/i.

Original entry on oeis.org

1, 2, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422
Offset: 0

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Author

Keywords

Comments

From Dean Hickerson, Apr 19 2003: (Start)
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.
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.
(End)
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)
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

References

  • John H. Conway and R. K. Guy, "The Book of Numbers," Copernicus, an imprint of Springer-Verlag, NY, 1996, pages 258-259.
  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 83, p. 28, Ellipses, Paris 2008.
  • 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.
  • H. P. Robinson, Letter to N. J. A. Sloane, Oct 23 1973.
  • W. Sierpiński, Sur les decompositions de nombres rationnels, Oeuvres Choisies, Académie Polonaise des Sciences, Warsaw, Poland, 1974, p. 181.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane, Illustration for sequence M4299 (=A007340) in The Encyclopedia of Integer Sequences (with Simon Plouffe), Academic Press, 1995.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. Stewart, L'univers des nombres, pp. 54, Belin-Pour La Science, Paris 2000.

Crossrefs

Apart from initial terms, same as A004080.

Programs

  • Haskell
    a002387 n = a002387_list !! n
    a002387_list = f 0 1 where
       f x k = if hs !! k > fromIntegral x
               then k : f (x + 1) (k + 1) else f x (k + 1)
               where hs = scanl (+) 0 $ map recip [1..]
    -- Reinhard Zumkeller, Aug 04 2014
  • Mathematica
    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 *)
    f[n_] := k /. FindRoot[HarmonicNumber[k] == n, {k, Exp[n]}, WorkingPrecision -> 100] // Ceiling; f[0] = 1; Array[f, 28, 0] (* Robert G. Wilson v, Jan 24 2017 after Jean-François Alcover in A014537 *)
  • PARI
    a(n)=if(n,my(k=exp(n-Euler));ceil(solve(x=k-1.5,k+.5,intnum(y=0,1,(1-y^x)/(1-y))-n)),1) \\ Charles R Greathouse IV, Jun 13 2012
    

Formula

Note that the conditionally convergent series Sum_{k>=1} (-1)^(k+1)/k = log 2 (A002162).
Limit_{n->oo} a(n+1)/a(n) = e. - Robert G. Wilson v, Dec 07 2001
It is conjectured that, for n > 1, a(n) = floor(exp(n-gamma) + 1/2). - Benoit Cloitre, Oct 23 2002

Extensions

Terms for n >= 13 computed by Eric W. Weisstein; corrected by James R. Buddenhagen and Eric W. Weisstein, Feb 18 2001
Edited by Dean Hickerson, Apr 19 2003

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

Original entry on oeis.org

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

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Author

Rainer Rosenthal, Jan 13 2008

Keywords

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.
		

Crossrefs

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)) ).

A118051 Denominators of coefficients in a series for the inverse of harmonic number H(x).

Original entry on oeis.org

1, 24, 640, 580608, 199065600, 504627200, 2191186722816000, 44497945755648000, 255806104666112, 15953645581139831685120000, 188420950968830433165312000000, 401521614736326656000000
Offset: 0

Views

Author

David W. Cantrell (DWCantrell(AT)sigmaxi.net), Apr 08 2006

Keywords

Examples

			With InvH(x) being the inverse of H(x), x > 0, an asymptotic series for InvH(x) + 1/2 is u - 1/(24u) + 3/(640u^3) - 1525/(580608u^5) +-... where u = e^(x - g) and g is Euler's gamma constant.
		

Crossrefs

Numerators given in A118050. See also A002387.

Programs

  • Mathematica
    n = 12; coeffs = InverseSeries[Exp[Series[HarmonicNumber[x - 1/2], {x, Infinity, 2n - 1}] - EulerGamma]][[3]]; Table[Denominator[coeffs[[2i - 1]]], {i, 1, n}]
Showing 1-3 of 3 results.