A004080 -id:A004080 - cp's OEIS frontend

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

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)) ).

A074633 a(n) is the smallest index m such that Sum_{k=2..m} 1/PrimePi(k) >= n, where PrimePi()=A000720().

Original entry on oeis.org

2, 4, 8, 12, 18, 27, 37, 51, 68, 89, 116, 147, 186, 232, 287, 352, 428, 518, 623, 745, 887, 1050, 1240, 1456, 1704, 1987, 2309, 2674, 3090, 3557, 4087, 4684, 5353, 6105, 6949, 7892, 8944, 10121, 11431, 12885, 14502, 16298, 18286, 20485, 22917, 25607

Offset: 1

Labos Elemer, Aug 28 2002

nonn

			a(85) = 927685 because 927686 is the smallest m such that Sum_{k=2..m} 1/PrimePi(k) >= 85.

Cf. A004080, A046024, A074631, A000720.

{s=0, s1=0}; Do[s=s+(1/PrimePi[n]); If[Greater[Floor[s], s1], s1=Floor[s]; Print[{n, Floor[s]}]], {n, 2, 1000000}]

Edited by Jon E. Schoenfield, Apr 04 2023

Name corrected by Sean A. Irvine, Jan 22 2025

A076751 a(n) is the smallest composite k such that Sum_{composites j = 4, ..., k} 1/j exceeds n.

Original entry on oeis.org

16, 63, 216, 715, 2279, 7102, 21722, 65558, 195759, 579465, 1703072, 4975222, 14459492, 41837580, 120585504, 346372172, 991915208, 2832896772, 8071045528, 22944211170

Offset: 1

Jack Brennen, Nov 12 2002

These partial sums, like the harmonic sequence (A004080), can never be integers.

			a(1) = 1 because 1/4 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + 1/14 + 1/15 = 0.97420... < 1 but 1/4 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + 1/14 + 1/15 + 1/16 = 1.03670... > 1.

Cf. A002808, A004080, A016088, A074631.

NextComposite[n_] := Block[{k = n + 1}, While[ PrimeQ[k], k++ ]; k]; k = 4; s = 0; Do[ While[s = s + 1/k; s < n, k = NextComposite[k]]; Print[k]; k = NextComposite[k], {n, 1, 17}]

lista(cmax) = {my(n = 1, s = 0); forcomposite(c = 1, cmax, s += 1/c; if(s > n, print1(c, ", "); n++));} \\ Amiram Eldar, Jul 17 2024

Limit_{n->oo} a(n+1)/a(n) = e.

a(n) = A002808(A074631(n)). - Amiram Eldar, Jul 17 2024

Edited and extended by Robert G. Wilson v, Nov 14 2002
Name edited and a(18) added by Jon E. Schoenfield, Feb 01 2020
a(19)-a(20) from Amiram Eldar, Jul 17 2024

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

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

A306349 Number of terms in the greedy Egyptian fraction representation of n.

Original entry on oeis.org

1, 4, 13, 35, 99

Offset: 1

Pontus von Brömssen, Feb 09 2019

a(n) >= A004080(n).

a(6) > 255 and the denominator of the 255th term in the representation of 6 has 1264021241 digits.

			a(3)=13 is the number of terms of A140335;
a(4)=35 is the number of terms of A281873.

Index entries for sequences related to Egyptian fractions

Cf. A004080, A140335, A281873.

from sympy import egyptian_fraction
def A306349(n): return len(egyptian_fraction((n,1)))

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

A074467 Least k such that Sum_{i=1..k} 1/phi(i) >= n.

Original entry on oeis.org

1, 2, 4, 8, 13, 22, 38, 63, 105, 177, 296, 495, 828, 1386, 2318, 3879, 6489, 10854, 18158, 30375, 50811, 84998, 142187, 237853, 397885, 665589, 1113411, 1862534, 3115683, 5211973, 8718687, 14584783, 24397699, 40812930, 68272636, 114207749, 191048868, 319590137

Offset: 1

Labos Elemer, Aug 29 2002

nonn

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 177, p. 55, Ellipses, Paris 2008.
E. Landau, Uber die Zahlentheoretische Function ϕ(n) und ihre Beziehung zum Goldbachschen satz, Nachrichten der Koniglichten Gesel lschaft der Wissenschaften zu Göttingen mathematisch Physikalische klasse, Jahrgang (1900), pp. 177-186.

H. L. Montgomery, Primes in arithmetic progressions, Michigan Math. J. 17:1 (1970), pp. 33-39. [alternate link]

Cf. A002387, A004080, A046024, A074631, A074633.

{s=0, s1=0}; Do[s=s+(1/EulerPhi[n]); If[Greater[Floor[s], s1], s1=Floor[s]; Print[{n, Floor[s]}]], {n, 1, 1000000}]

a(n)=my(s,k);while(sCharles R Greathouse IV, Jan 29 2013

a(n) ~ k exp(cn) for c = zeta(6)/zeta(2)/zeta(3) = A068468 and k = exp(-gamma + A085609) = 1.0316567993311528...; see Montgomery or Koninck. - Charles R Greathouse IV, Jan 29 2013

More terms from Ryan Propper, Jul 09 2005

a(32)-a(38) from Donovan Johnson, Aug 21 2011

A074468 Least number m such that the Sigma-Harmonic sequence Sum_{k=1..m} 1/sigma(k) >= n.

Original entry on oeis.org

1, 7, 29, 129, 571, 2525, 11167, 49372, 218295, 965177, 4267457, 18868240, 83424514, 368855252, 1630865929, 7210751807, 31881800153

Offset: 1

Labos Elemer, Aug 29 2002

Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 129, p. 44, Ellipses, Paris, 2008.

Cf. A000203 (sigma), A002387, A004080, A046024, A074631, A074633, A074467, A212717, A308039.

{s=0, s1=0}; Do[s=s+(1/DivisorSigma[1, n]); If[Greater[Floor[s], s1], s1=Floor[s]; Print[{n, Floor[s]}]], {n, 1, 1000000}]

Limit_{n->oo} a(n+1)/a(n) = exp(1/c) = 4.42142525588146107878... where c = A308039. - Amiram Eldar, May 05 2024

2 more terms from Lekraj Beedassy, Jul 14 2008
a(11)-a(15) from Donovan Johnson, Aug 22 2011
a(16)-a(17) from Amiram Eldar, May 05 2024

cp's OEIS Frontend

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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A074633 a(n) is the smallest index m such that Sum_{k=2..m} 1/PrimePi(k) >= n, where PrimePi()=A000720().

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A076751 a(n) is the smallest composite k such that Sum_{composites j = 4, ..., k} 1/j exceeds n.

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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).

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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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A306349 Number of terms in the greedy Egyptian fraction representation of n.

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