A002387 -id:A002387 - cp's OEIS frontend

A115515 a(n) = largest m such that the harmonic number H(m)= Sum_{i=1..m} 1/i is < n.

0, 3, 10, 30, 82, 226, 615, 1673, 4549, 12366, 33616, 91379, 248396, 675213, 1835420, 4989190, 13562026, 36865411, 100210580, 272400599, 740461600, 2012783314, 5471312309, 14872568830, 40427833595, 109894245428

Offset: 1

Artur Jasinski, Jan 23 2006

nonn

H. P. Robinson, Letter to N. J. A. Sloane, Oct 1981

Apart from the initial values, this is simply A002387(n)-1. Cf. A004080.

c:=0: H[0]:=0: for n from 1 to 10^4 do H[n]:=1/n+H[n-1]: if floor(H[n])-floor(H[n-1])=1 then c:=1+c: b[c]:=n-1: else c:=c: fi: od: seq(b[j],j=1..c); # Emeric Deutsch

a[n_] := Ceiling[k /. FindRoot[HarmonicNumber[k] == n, {k, Exp[n]}, WorkingPrecision -> 100]] - 1;
Array[a, 26] (* Jean-François Alcover, Apr 10 2019 *)

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

A081881 Pack bins of size 1 sequentially with items of size 1/1, 1/2, 1/3, 1/4, ... . Sequence gives values of n for which 1/n starts a new bin.

Original entry on oeis.org

1, 2, 4, 10, 26, 69, 186, 504, 1369, 3720, 10111, 27483, 74705, 203068, 551995, 1500477, 4078718, 11087104, 30137872, 81923228, 222690421, 605335323, 1645472007, 4472856655, 12158484965, 33050188741, 89839727480, 244209698681, 663830786257, 1804479163453, 4905082919846

Offset: 1

Wouter Meeussen, Apr 13 2003

For n >= 3, it appears that a(n) = round((a(n-1) - 1/2)*e). Verified through n = 10000 (using the approximation Sum_{j=1..k} 1/j = log(k) + gamma + 1/(2*k) - 1/(12*k^2) + 1/(120*k^4) - 1/(252*k^6) + 1/(240*k^8) - ... + 7709321041217/(16320*k^32), where gamma is the Euler-Mascheroni constant, A001620). - Jon E. Schoenfield, Mar 30 2018

			1/1; 1/2+1/3, 1/4+1/5+1/6+1/7+1/8+1/9 are all just less than or equal to 1; so first four terms are 1, 2, 4, 10.
Lower and upper indices of bin contents are {1,1}, {2,3}, {4,9}, {10,25}, {26,68}, {69,185}, {186,503}, {504,1368}, {1369,3719}, {3720,10110}, {10111,27482}, ...

Jinyuan Wang, Table of n, a(n) for n = 1..1000

Cf. A002387, A136616, A136617, A295572, A300897.

res ={}; FoldList[If[ #1+#2 > 1, AppendTo[res, #2];#2, #1+#2]&, 0, Table[1/k, {k, 1, 1000}]]; 1/res
lst = {1, 2}; n = 2; Do[s = 0; While[s = N[s + 1/n, 64]; s < 1, n++ ]; AppendTo[lst, n]; Print@n, {i, 25}]; lst (* Robert G. Wilson v, Aug 19 2008 *)

default(realprecision, 10^4); e=exp(1);
A136616(k) = floor(e*k + (e-1)/2 + (e-1/e)/(24*k+12));
lista(nn) = {my(k=1); print1(k); for(n=2, nn, k=A136616(k-1)+1; print1(", ", k)); } \\ Jinyuan Wang, Feb 20 2020

a(n) is asymptotic to C*exp(n) where C=0.1688... - Benoit Cloitre, Apr 14 2003
C = 0.16885635666714420373167977550090103410150395689764... (cf. A300897). - Jon E. Schoenfield, Apr 12 2018
a(n) = 1 + (A136616^(n-1))(0), where (f^0)(x)=x, (f^(n+1))(x) = f((f^n)(x)) for any function f. - Rainer Rosenthal, Feb 16 2008, Apr 05 2020

a(13)-a(25) from Robert G. Wilson v, Aug 19 2008

More terms from Jinyuan Wang, Feb 20 2020

A054040 a(n) terms of series {1/sqrt(j)} are >= n.

Original entry on oeis.org

1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 202, 217, 232, 247, 263, 280, 297, 314, 332, 351, 370, 389, 409, 430, 451, 472, 494, 517, 540, 563, 587, 612, 637, 662, 688, 715, 741, 769, 797, 825

Offset: 1

Asher Auel, Apr 13 2000

nonn

In many cases the first differences have the form {2k, 2k, 2k, 2k+1} (A004524). In such cases the second differences are {0, 0, 1, 1}. See A082915 for the exceptions. In as many as these, the first differences have the form {2k-1, 2k-1, 2k-1, 2k}. - Robert G. Wilson v, Apr 18 2003 [Corrected by Carmine Suriano, Nov 08 2013]
a(100)=2574, a(1000)=250731 & a(10000)=25007302 which differs from Sum{i=4..104}A004524(i)=2625, Sum{i=4..1004}A004524(i)=251250 & Sum{i=4..10004}A004524(i)=25012500. - Robert G. Wilson v, Apr 18 2003
A054040(n) <= A011848(n+2), A054040(10000)=25007302 and A011848(n+2)=25007500. - Robert G. Wilson v, Apr 18 2003

			Let b(k) = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k):
.k.......1....2.....3.....4.....5.....6.....7
-------------------------------------------------
b(k)...1.00..1.71..2.28..2.78..3.23..3.64..4.01
For A019529 we have:
n=0: smallest k is a(0) = 1 since 1.00 > 0
n=1: smallest k is a(1) = 2 since 1.71 > 1
n=2: smallest k is a(2) = 3 since 2.28 > 2
n=3: smallest k is a(3) = 5 since 3.23 > 3
n=4: smallest k is a(4) = 7 since 4.01 > 4
For this sequence we have:
n=1: smallest k is a(1) = 1 since 1.00 >= 1
n=2: smallest k is a(2) = 3 since 2.28 >= 2
n=3: smallest k is a(3) = 5 since 3.23 >= 3
n=4: smallest k is a(4) = 7 since 4.01 >= 4

Sela Fried, On the partial sums of the Zeta function Sum_{n>=1} 1/n^s for 0 < s < 1, 2024.
Hugo Pfoertner, Plot of a(n)-(1/4)*(n^2-2*zeta(1/2)*n), n=1..1000.

Cf. A002387, A004080, A059750, A082915, A054044, A011848, A082915.

See A019529 for a different version.

f[n_] := Block[{k = 0, s = 0}, While[s < n, k++; s = N[s + 1/Sqrt[k], 50]]; k]; Table[f[n], {n, 1, 60}]

a(n)=if(n<0,0,t=1;z=1;while(zBenoit Cloitre, Sep 23 2012

Let f(n) = (1/4)*(n^2-2*zeta(1/2)*n) then we have a(n) = f(n) + O(1). More precisely we claim that for n >= 2 we have a(n) = floor(f(n)+c) where c > Max{a(n)-f(n) : n>=1} = a(153) - f(153) = 1.032880076066608813953... and we believe we can take c = 1.033. - Benoit Cloitre, Sep 23 2012

Definition and offset modified by N. J. A. Sloane, Sep 01 2009

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

Rainer Rosenthal, Jan 13 2008

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.

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

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.

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

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

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 *)

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)) ).

A014537 Number of books required for n book-lengths of overhang in the harmonic book stacking problem. Sum_{i=1..a(n)} 1/i >= 2n and Sum_{i=1..a(n)-1} 1/i < 2n.

Original entry on oeis.org

4, 31, 227, 1674, 12367, 91380, 675214, 4989191, 36865412, 272400600, 2012783315, 14872568831, 109894245429, 812014744422, 6000022499693, 44334502845080, 327590128640500, 2420581837980561, 17885814992891026

Offset: 1

Eric W. Weisstein

Bisection of A002387. - Robert G. Wilson v, Jan 24 2017

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.

Robert G. Wilson v, Table of n, a(n) for n = 1..1000 (terms 1..350 from Alois P. Heinz).
Mike Paterson and Uri Zwick, Overhang, arXiv:0710.2357 [math.HO], 2007.
Eric Weisstein's World of Mathematics, Book Stacking Problem

Cf. A002387.

f[n_] := (k = Floor[ N [ E^(n - EulerGamma) + 1/(2n), 24]] - 2; While[ Floor[ N[ Log[k] + EulerGamma + 1/(2k) - 1/(12k^2) + 1/(120k^4), 24]] < n, k++ ]; k); Table[ f[n], {n, 2, 32, 2} ]
a[n_] := k /. FindRoot[ HarmonicNumber[k] == 2*n, {k, Exp[2*n]}, WorkingPrecision -> 100] // Ceiling; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jun 25 2013 *)

a(n) = A002387(2n), n>=1. Least a(n) with H(a(n)) > 2n with the harmonic numbers H(k):= A001008(k)/A002805(k).

More terms from Robert G. Wilson v, Dec 06 2001

Title corrected by Jeremy Tan, Sep 12 2020

A056054 a(n) = smallest even number 2m such that value of odd harmonic series Sum_{j=0..m} 1/(2j) is > n.

Original entry on oeis.org

8, 62, 454, 3348, 24734, 182760, 1350428, 9978382, 73730824, 544801200, 4025566630, 29745137662, 219788490858, 1624029488844, 12000044999386, 88669005690160, 655180257281000, 4841163675961122, 35771629985782052

Offset: 1

Robert G. Wilson v, Jul 25 2000 and Jan 11 2004

nonn

Numbers 2*m such that floor(f(m)) = floor(f(m-1)) where f(m) = Sum_{j=1..m} ((2*j-1)/(2*j)). Examples:
  floor(f(1))=floor(1/2)=0;
  floor(f(2))=floor(1/2+3/4)=floor(1.25)=1, then 2*2=4 is not in the sequence;
  floor(f(3))=floor(1/2+3/4+5/6)=floor(2.083..)=2, then 2*3=6 is not in the sequence;
  floor(f(4))=floor(1/2+3/4+5/6+7/8)=floor(2.958..)=2, then 2*4=8 is the first term of the sequence. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 15 2007

Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Plenum Press, NY and London, 1996, page 64.

Cf. A002387, A056053, A091463, A091464, A091465.

Cf. A056054.

s = 0; k = 2; Do[ While[s = N[s + 1/k, 24]; s <= n, k += 2]; Print[k]; k += 2, {n, 1, 12}]
(* or assuming that the Mathematica coding in A002387 is correct then *)
b[n_] := Module[{k = Floor[2a[2n]]}, If[ EvenQ[k], k, k + 1]]; Table[ b[n], {n, 19}] (* Robert G. Wilson v, Apr 17 2004 *)

a(n) = 2*A002387(2n).

The next term is approximately the previous term * e^2.

A079526 a(n) = floor( exp(H_n)*log(H_n) ) - sigma(n).

Original entry on oeis.org

-1, -2, -1, -2, 2, -2, 4, 0, 4, 2, 10, -3, 13, 6, 9, 4, 20, 2, 23, 4, 17, 16, 31, -3, 29, 21, 26, 13, 42, 3, 46, 18, 36, 32, 41, 1, 57, 38, 45, 14, 65, 14, 69, 32, 41, 51, 78, 5, 75, 42, 66, 43, 90, 27, 78, 33, 76, 70, 103, -2, 107, 76, 71, 51, 98, 41, 120, 65, 98, 53

Offset: 1

N. J. A. Sloane, Jan 22 2003

sign

M. Kaneko has shown that the Riemann hypothesis is equivalent to the assertion that a(n) > 0 for n > 60.

G. C. Greubel, Table of n, a(n) for n = 1..10000
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, arXiv:math/0008177 [math.NT], 2000-2001; Am. Math. Monthly 109 (#6, 2002), 534-543.

H_n = sum of harmonic series (see A002387), sigma(n) = A000203.

Cf. A057641, A058209, A079527.

[Floor(Exp(HarmonicNumber(n))*Log(HarmonicNumber(n))) - DivisorSigma(1,n): n in [1..80]]; // G. C. Greubel, Jan 15 2019

f[n_] := Floor[Exp[HarmonicNumber[n]]Log[HarmonicNumber[n]]] - DivisorSigma[1, n]; Array[f, 70] (* Robert G. Wilson v, Dec 17 2016 *)

{h(n) = sum(k=1, n, 1/k)};
vector(80, n, floor( exp(h(n))*log(h(n))) - sigma(n,1) ) \\ G. C. Greubel, Jan 15 2019

[floor(exp(harmonic_number(n))*log(harmonic_number(n))) - sigma(n,1) for n in (1..80)] # G. C. Greubel, Jan 15 2019

A281355 a(n) = A092318(n) + 1: Number of terms of the odd harmonic series 1 + 1/3 + 1/5 + 1/7 + ... required to reach a sum >= n.

Original entry on oeis.org

1, 8, 57, 419, 3092, 22846, 168804, 1247298, 9216354, 68100151, 503195829, 3718142208, 27473561358, 203003686106, 1500005624924, 11083625711271, 81897532160125, 605145459495141, 4471453748222757, 33039822589391676, 244133102611731231, 1803913190804074904, 13329215764452299411

Offset: 1

N. J. A. Sloane, Jan 22 2017, following a suggestion from Jerry Polfer

nonn

A different way of listing the sums mentioned in A092318.

A092318 is the main entry for this problem.

Gerhard Kirchner, Table of n, a(n) for n = 1..1000 [replacing an older version by Vincenzo Librandi]
Jerry V Polfer, found a new sequence, not sure if interesting, and how to properly submit, Seqfan list, Jan. 22, 2017 (and follow-up messages).
Albert Säfström, n describes the expected number of loops created by tieing together two random loose ends of a(n) ropes until there is none left.

Cf. A002387, A092318.

a[n_]:=Floor[Exp[2*n-EulerGamma]/4+1]-Boole[n==1]; Array[a,23] (* Stefano Spezia, Jun 25 2024 *)

Lim_{n -> oo} a(n)/exp(2*n) = 1/4e^gamma ~ 0.140364870891721292456...;

a(n) = floor(exp(2*n-gamma)/4+1), for all given values a(n) > 1. - M. F. Hasler and Robert G. Wilson v, Jan 23 2017 [corrected by Gerhard Kirchner, Jul 25 2020]

More explicit, self-contained definition by M. F. Hasler, Jan 22 2017
More terms (computed using A056053) from M. F. Hasler, Jan 23 2017
a(17) corrected in data and 127 terms in the b-file, according to the corrections in A092315, Gerhard Kirchner, Jul 27 2020

A079353 Numbers n such that the best rational approximation to H(n) with denominator <=n is an integer, where H(n) denotes the n-th harmonic number (A001008/A002805).

Original entry on oeis.org

1, 3, 4, 10, 11, 30, 31, 82, 83, 226, 227, 615, 616, 1673, 1674, 4549, 4550, 12366, 12367, 33616, 33617

Offset: 1

Benoit Cloitre, Feb 14 2003

From Robert Israel, May 19 2014: The definition is unclear. For example, how does 10 fit in? H(10) = 7381/2520, and the best approximation with denominator <= 10 is 29/10, which is not an integer. Similarly, I don't see how 31, 82, 227, 616, or 1674 fit the definition, as according to my computations the best approximations in these cases are 125/31, 409/82, 1363/227, 4313/616, 13393/1674.
Response from David Applegate, May 20 2014: I suspect, without deep investigation, that what was meant by "best rational approximation to" is "continued fraction convergent". The continued fraction convergents to H(10)=7381/2520 are 2, 3, 41/14, 495/169, ... The continued fraction convergents to H(31) are 4, 145/36, 149/37, 443/110, ... The continued fraction convergents to H(82) are 4, 5, 499/100, 2001/401, ... I haven't verified that the rest of the terms match this definition.
Response from Ray Chandler, May 20 2014: I confirm that definition matches the listed terms and continues with 4549, 4550 and no others less than 10000.
Added by Ray Chandler, May 29 2014: Except for the beginning terms A079353 appears to be the union of A115515 and A002387 (compare A242654).

			H(11)=83711/27720 and the best approximation to H(11) among the fractions of form k/11, k>=0, is 33/11=3, an integer. Hence 11 is in the sequence.

Cf. A001008/A002805, A002387, A004080, A115515.

See A242654 for the most likely continuation.

okQ[n_] := Select[Convergents[N[HarmonicNumber[n], 30], 10], Denominator[#] <= n &][[-1]] // IntegerQ;
Reap[For[n = 1, n <= 40000, n++, If[okQ[n], Print[n]; Sow[n]]]][[2, 1]] // Quiet (* Jean-François Alcover, Apr 10 2019 *)

a(16)-a(17) from Ray Chandler, May 20 2014
Edited by N. J. A. Sloane, May 29 2014
a(18)-a(21) from Jean-François Alcover, Apr 10 2019

cp's OEIS Frontend

A115515 a(n) = largest m such that the harmonic number H(m)= Sum_{i=1..m} 1/i is < n.

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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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A081881 Pack bins of size 1 sequentially with items of size 1/1, 1/2, 1/3, 1/4, ... . Sequence gives values of n for which 1/n starts a new bin.

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A054040 a(n) terms of series {1/sqrt(j)} are >= n.

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

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A014537 Number of books required for n book-lengths of overhang in the harmonic book stacking problem. Sum_{i=1..a(n)} 1/i >= 2n and Sum_{i=1..a(n)-1} 1/i < 2n.

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A056054 a(n) = smallest even number 2m such that value of odd harmonic series Sum_{j=0..m} 1/(2j) is > n.

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