A014537 -id:A014537 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

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.

Robert G. Wilson v, Table of n, a(n) for n = 0..2303 (first 101 terms from T. D. Noe)
John V. Baxley, Euler's constant, Taylor's formula, and slowly converging series, Math. Mag. 65 (1992), 302-313. (Gives terms up to n = 24.)
R. P. Boas, Partial sums of the harmonic series, II, Mimeographed manuscript, no date.
R. P. Boas, Jr. and J. W. Wrench, Jr., Partial sums of the harmonic series, Amer. Math. Monthly, 78 (1971), 864-870. (Gives terms up to n = 20.)
Nick Hobson, Solution to puzzle 34: Harmonic sum 2.
H. P. Robinson, Letter to N. J. A. Sloane, Oct 28 1973.
H. P. Robinson, Letter to N. J. A. Sloane, Oct 1981
H. P. Robinson, Letter to N. J. A. Sloane, Dec 20 1983
John A. Rochowicz, Jr., Harmonic Numbers: Insights, Approximations and Applications, Spreadsheets in Education (eJSiE) (2015), Vol. 8: Iss. 2, Article 4.
R. G. Wilson, Letter to N. J. A. Sloane with attachment, Jan 1989 (A006509 is mentioned in the attachment)
R. G. Wilson v, Letter to N. J. A. Sloane, Oct 12 1993
J. W. Wrench, Jr., Selected Partial Sums of the Harmonic Series, Manuscript, no date [Annotated scanned copy]

Apart from initial terms, same as A004080.

Cf. A006509, A055980, A115515, A242654.

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

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 *)

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

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

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

A136616 a(n) = largest m with H(m) - H(n) <= 1, where H(i) = Sum_{j=1 to i} 1/j, the i-th harmonic number, H(0) = 0.

Original entry on oeis.org

1, 3, 6, 9, 11, 14, 17, 19, 22, 25, 28, 30, 33, 36, 38, 41, 44, 47, 49, 52, 55, 57, 60, 63, 66, 68, 71, 74, 76, 79, 82, 85, 87, 90, 93, 96, 98, 101, 104, 106, 109, 112, 115, 117, 120, 123, 125, 128, 131, 134, 136, 139, 142, 144, 147, 150, 153, 155, 158, 161, 163, 166

Offset: 0

Rainer Rosenthal, Jan 13 2008

			a(3) = 9 because H(9) - H(3) = 1/4 + ... + 1/9 < 1 < 1/4 + ... + 1/10 = H(10) - H(3).

E. R. Bobo, A sequence related to the harmonic series, College Math. J. 26 (1995), 308-310.

Cf. A001008, A002805, A002387, A004080, A079353, A096618, A115515, A014537, A055980, A103762, A136617.

e:= exp(1):
A136616 := n -> floor( e*n + (e-1)/2 + (e - 1/e)/(24*(n + 1/2))):
seq(A136616(n), n=0..50);

default(realprecision, 10^5); e=exp(1);
a(n) = floor(e*n+(e-1)/2+(e-1/e)/(24*n+12)); \\ Jinyuan Wang, Mar 06 2020

a(n) = floor(e*n + (e-1)/2 + (e - 1/e)/(24*(n + 1/2))), after a suggestion by David Cantrell.

a(n) = A103762(n+1) - 1 = A136617(n+1) + n for n > 0. - Jinyuan Wang, Mar 06 2020

Definition corrected by David W. Cantrell, Apr 14 2008

A065071 Minimum number of identical bricks of length 1 which, when stacked without mortar in the naive way, form a stack of length >=n.

Original entry on oeis.org

1, 5, 32, 228, 1675, 12368, 91381, 675215, 4989192, 36865413, 272400601, 2012783316, 14872568832, 109894245430, 812014744423, 6000022499694, 44334502845081, 327590128640501, 2420581837980562, 17885814992891027

Offset: 1

John W. Layman, Nov 08 2001

Note that one can do "better" in terms of projections if one groups the bricks asymmetrically into lozenges with holes. See the Ainsley and Drummond references. Ainsley considers only the case of four bricks, but achieves an overhang of (15 - 4*sqrt(2))/8, compared with 25/24 for the harmonic pile. - D. G. Rogers, Aug 31 2005

Lim_{n -> inf} a(n)/a(n-1) = exp(2). - Robert G. Wilson v, Jan 26 2017

			Obviously a(1)=1. If the center of gravity of one brick is placed at the end of a second brick, the length of the stack of 2 bricks is 1.5. If the c.g. of that stack is placed at the end of a third brick, the length of the stack is 1.75. Continuing, we get a stack of length 1.916666... for 4 bricks and a stack of length 2.0416666... for 5 bricks. Thus a(2)=5.

N. J. A. Sloane, Illustration for sequence M4299 (=A007340) in The Encyclopedia of Integer Sequences (with Simon Plouffe), Academic Press, 1995.

Robert G. Wilson v, Table of n, a(n) for n = 1..1000
S. Ainley, Finely Balanced, Math. Gaz., 63 (1979), 272.
R. Dickau, Harmonic numbers and the book-stacking problem
J. E. Drummond, On stacking bricks to achieve a large overhang, Math. Gaz., 65 (1981), 40-42.
Eric Weisstein's World of Mathematics, Book Stacking Problem

Cf. harmonic numbers H(n) = A001008/A002805, A002387, A004080.

A002387[n_] := Floor[ Exp[n - EulerGamma] + 1/2]; a[n_] := A002387[2n - 2] + 1; a[1] = 1; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Dec 13 2011, after Charles R Greathouse IV *)
f[n_] := k /. FindRoot[HarmonicNumber[k -1] == 2n, {k, Exp[ 2n]}, WorkingPrecision -> 100] // Ceiling; Array[f, 21, 0] (* Robert G. Wilson v, Jan 26 2017 after Jean-François Alcover in A014537 *) (* note that the index is off by one *)

a(n) = A002387(2n) + 1 = A014537(n) + 1.

More terms from Vladeta Jovovic, Nov 14 2001

A335094 Decimal expansion of (15 - 4*sqrt(2))/8.

Original entry on oeis.org

1, 1, 6, 7, 8, 9, 3, 2, 1, 8, 8, 1, 3, 4, 5, 2, 4, 7, 5, 5, 9, 9, 1, 5, 5, 6, 3, 7, 8, 9, 5, 1, 5, 0, 9, 6, 0, 7, 1, 5, 1, 6, 4, 0, 6, 2, 3, 1, 1, 5, 2, 5, 9, 6, 3, 4, 1, 1, 6, 6, 0, 1, 3, 1, 0, 0, 4, 6, 3, 3, 7, 6, 0, 7, 6, 8, 9, 4, 6, 4, 8, 0, 5, 7, 4, 8, 0, 6, 2, 3, 2, 8, 3, 6, 1, 7, 9, 2, 1, 3, 6, 3

Offset: 1

Jeremy Tan, Sep 12 2020

Largest overhang off an edge achievable with four unit-length bricks.

Corresponding values for one, two and three bricks are 1/2, 3/4 and 1 respectively.

			1.1678932188134524755991556378951...

S. Ainley, Finely Balanced, The Mathematical Gazette, 63 (1979), p. 272.
M. Paterson and U. Zwick, Overhang, arXiv:0710.2357 [math.HO], 2007.
Eric Weisstein's World of Mathematics, Book Stacking Problem

Cf. A020761 (1/2), A152627 (3/4).

Cf. A014537, A157215, A268682.

First@ RealDigits@ N[(15 - 4 Sqrt[2])/8, 102]

PARI
```
(15-4*sqrt(2))/8
```

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

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A136616 a(n) = largest m with H(m) - H(n) <= 1, where H(i) = Sum_{j=1 to i} 1/j, the i-th harmonic number, H(0) = 0.

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A065071 Minimum number of identical bricks of length 1 which, when stacked without mortar in the naive way, form a stack of length >=n.

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A335094 Decimal expansion of (15 - 4*sqrt(2))/8.

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