A055980 -id:A055980 - 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

A214075 Triangle read by rows: T(n,k) = floor(A213998(n,k) / A213999(n,k)), 0<=k<=n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 2, 0, 1, 4, 7, 6, 2, 0, 1, 5, 12, 14, 8, 2, 0, 1, 6, 17, 26, 22, 11, 2, 0, 1, 7, 24, 44, 49, 34, 13, 2, 0, 1, 8, 31, 68, 93, 83, 47, 16, 2, 0, 1, 9, 40, 100, 162, 177, 131, 64, 19, 2, 0, 1, 10, 49, 140, 263, 340, 309

Offset: 0

Reinhard Zumkeller, Jul 03 2012

T(n,0) = 1;
T(n,1) = n - 1 for n > 0, cf. A001477;
T(n,2) = A074148(n-2) for n > 2;
T(n,n-2) = A022819(n) for n > 1;
T(n,n-1) = A055980(n) for n > 0;
T(n,n) = A000007(n).

			Start of triangle preceded by triangle "A213998/A213999":
. 0:                      1                                 1
. 1:                   1   1/2                             1  0
. 2:               1    3/2   1/3                         1  1  0
. 3:            1   5/2    11/6   1/4                    1 2  1  0
. 4:        1   7/2   13/3    25/12   1/5               1 3  4  2 0
. 5:     1   9/2   47/6   77/12   137/60   1/6         1 4  7  6 2 0
. 6:  1  11/2   37/3   57/4   87/10   49/20   1/7,    1 5 12 14 8 2 0.

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened

a214075 n k = a214075_tabl !! n !! k
a214075_row n = a214075_tabl !! n
a214075_tabl = zipWith (zipWith div) a213998_tabl a213999_tabl

A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

Original entry on oeis.org

1, 3, 8, 22, 60, 163, 443, 1204, 3273, 8897, 24184, 65739, 178698, 485751, 1320408, 3589241, 9756569, 26521104, 72091835, 195965925, 532690613, 1448003214, 3936080824, 10699376979, 29083922018, 79058296722, 214902731368, 584166189564, 1587928337892, 4316436745787

Offset: 1

Alejandro Argüelles Trujillo and Pablo Hueso Merino, Jan 07 2020

nonn

a(n) is equal to A024581(n) through a(10), and grows very similarly for n > 10.

Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Cf. A331030. - Jon E. Schoenfield, Jan 14 2020

			a(1)=1 because 1 >= 1,
a(2)=3 because 1/2 + 1/3 + 1/4 = 1.0833... >= 1, etc.

Cf. A004080, A024581, A136616, A331030.

Some sequences in the same spirit as this: A002387, A004080, A055980, A115515.

default(realprecision, 10^5); e=exp(1);
lista(nn) = {my(r=1); print1(r); for(n=2, nn, print1(", ", -r+(r=floor(e*r+(e+1)/2+(e-1/e)/(24*(r+1/2)))))); } \\ Jinyuan Wang, Mar 31 2020

x = 0.0
y = 0.0
for i in range(1,100000000000000000000000):
  y += 1
  x = x + 1/i
  if x >= 1:
    print(y)
    y = 0
    x = 0

a(n) = min(p): Sum_{b=r+1..p+r} 1/b >= 1, r = Sum_{k=1..n-1} a(k), a(1) = 1.

a(20)-a(21) from Giovanni Resta, Jan 14 2020

More terms from Jinyuan Wang, Mar 31 2020

A217693 Numbers of distinct integers obtained from summing up subsets of {1, 1/2, 1/3, ..., 1/n}.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Michel Marcus, Oct 11 2012

nonn

a(n) <= A111233(n).
a(n) <= floor(Sum_{k=1..n} 1/k) = A055980(n). - Joerg Arndt, Oct 13 2012
a(n) <= 4 for n <= 94, a(n) <= 5 for n <= 257, a(n) <= 6 for n <= 689. That is because if there is a term 1/a with p dividing a for a prime p, then there must be another term 1/b with p dividing b. Hence, not all terms from 1/1 to 1/n can be summed up. Cf. the "filter" function in my Sage script. - Manfred Scheucher, Aug 17 2015
a(k) = n for all k such that A101877(n) <= k < A101877(n+1). - Jon E. Schoenfield, May 12 2017

			1, 1/2 + 1/3 + 1/6 = 1 and 1 + 1/2 + 1/3 + 1/6 = 2 are integers, but only 2 of them are distinct, so a(6)=2.
a(24)=3 because 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 + 1/24 = 3 and Sum_{k=1..n} 1/k < 4 for all n <= 30.
a(65)=4 because the sum of the reciprocals of the integers in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 27, 28, 30, 33, 35, 36, 40, 42, 45, 48, 52, 54, 56, 60, 63, 65 } is 4 and Sum_{k=1..n} 1/k < 5 for all n <= 82. - _Jon E. Schoenfield_, Apr 30 2018

P. Erdos and R. L. Graham, Old and new problems and results in combinatorial number theory, Université de Genève, 1980.

Manfred Scheucher, Sage Script
H. Yokota, On number of integers representable as sums of unit fractions, Canad. Math. Bull. Vol. 33 (2), 1990.
H. Yokota, On Number of Integers Representable as a Sum of Unit Fractions, II, Journal of Number Theory 67, 162-169, 1997.

Cf. A101877, A111233.

ufr(n) = {tab = []; for (i=1, 2^n - 1, vb = binary(i); while(length(vb) < n, vb = concat(0, vb););; val = sum(j=1, length(vb), vb[j]/j); if (denominator(val) == 1, tab = concat(tab, val); ); ); return (length(Set(tab))); }

a(25)-a(46) from Manfred Scheucher, Aug 17 2015

a(47)-a(87) from Jon E. Schoenfield, Apr 30 2018

A096143 a(n) = ceiling(Sum_{i=1..n} 1/i).

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

W. Neville Holmes, Jul 24 2004

			a(4)=3 because the ceiling of 1 + 1/2 + 1/3 + 1/4 is 3.

Cf. A002387, A055980.

Ceiling[HarmonicNumber[Range[110]]] (* Harvey P. Dale, Aug 27 2014 *)

a(n) = ceil(sum(i=1, n, 1/i)) \\ Michel Marcus, Jul 11 2013

a(n) = ceiling(A001008(n)/A002805(n)). - Michel Marcus, Jul 11 2013

A238401 Floor(sum(i/(i+1)),i=1..n).

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63

Offset: 0

Jon Perry, Feb 26 2014

nonn

The first numbers which appear twice in the sequence are 0, 1, 7, 26, 77, 220, 608, 1665, 4540, 12356, 33605, 91367, 248383, 675199,... - Giovanni Resta, Feb 26 2014

These numbers appear at roughly exp(n - gamma). - Charles R Greathouse IV, Feb 26 2014

			a(3) = floor(0/1 + 1/2 + 2/3 + 3/4) = floor(1.91666...) = 1.

Giovanni Resta, Table of n, a(n) for n = 0..1000

Cf. A055980, A001008, A002805.

c=0;
for (i=1;i<50;i++) {
c+=i/(i+1);
document.write(Math.floor(c)+", ");
}

Floor[Accumulate[Table[i/(i+1),{i,0,70}]]] (* Harvey P. Dale, May 18 2023 *)

a(n)=n-ceil(sum(i=2,n,1./i)) \\ Charles R Greathouse IV, Feb 26 2014

a(n) = n - log n + O(1). - Charles R Greathouse IV, Feb 26 2014

a(50)-a(67) from Giovanni Resta, Feb 26 2014

A337904 Numbers k such that the decimal expansion of the k-th harmonic number starts with the digits of k, in the same order.

Original entry on oeis.org

1, 43, 714, 715, 9763, 122968, 122969, 1478366, 17239955, 196746419, 2209316467, 24499118645, 268950072605

Offset: 1

Metin Sariyar, Sep 29 2020

The sequence also includes 196746419, 2209316467, 24499118645, 268950072605, 2928264676792, 31663398162514, 340383084842938, 3640820101879826. - Daniel Suteu, Oct 01 2020

			1/1 + 1/2 + 1/3 + ... + 1/1478366 = 14.78366... .

Cf. A033147, A055980.

s=0;Do[s=s+(1/n);t=IntegerLength[n];m=IntegerLength[Floor[s]];k = Floor[s (10^(t-m))];If[k==n,Print[n]],{n,1,10^11}]

lista(nn) = {my(s=0.); for (n=1, nn, s += 1./n; my(d = digits(floor(10^20*s))); if (fromdigits(vector(#Str(n), j, d[j])) == n, print1(n, ", ")););} \\ Michel Marcus, Sep 30 2020

a(9) from Jinyuan Wang, Sep 30 2020
a(10)-a(12) from Chai Wah Wu, Oct 12 2020
a(13) from Chai Wah Wu, Oct 14 2020

cp's OEIS Frontend

A280520 Triangle read by rows: T(n,k) = number of increasing sequences of n positive integers with reciprocals adding up to k (k=1,2,...,A055980(n)).

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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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A214075 Triangle read by rows: T(n,k) = floor(A213998(n,k) / A213999(n,k)), 0<=k<=n.

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A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

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A217693 Numbers of distinct integers obtained from summing up subsets of {1, 1/2, 1/3, ..., 1/n}.

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PARI

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A096143 a(n) = ceiling(Sum_{i=1..n} 1/i).

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Mathematica

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A238401 Floor(sum(i/(i+1)),i=1..n).

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