A004080 -id:A004080 - cp's OEIS frontend

A096618 Least k such that H(k) >= 10^n, where H(k) is the harmonic number Sum_{i=0..k} 1/i.

Original entry on oeis.org

1, 12367, 15092688622113788323693563264538101449859497

Offset: 1

Eric W. Weisstein, Jul 01 2004

The next term is approx 1.10*10^434, which too large to include.

Except for the first term, identical to A082912.

R. P. Boas, Jr. and J. W. Wrench, Jr., Partial sums of the harmonic series, Amer. Math. Monthly, 78 (1971), 864-870.
Eric Weisstein's World of Mathematics, High-Water Mark

Cf. A004080, A082912.

A074469 Least m such that Sigma-Composite-Harmonic series Sum_{k=1..m} 1/A000203(A002808(k)) >= n.

Original entry on oeis.org

32, 301, 2123, 13172, 76105, 420007, 2245009, 11719362, 60071831, 303487314, 1515211979

Offset: 1

Labos Elemer, Sep 05 2002

Cf. A004080, A002387, A002808, A000203, A073255, A046024, A074631, A074633, A074468, A074470.

c[x_] := FixedPoint[x+PrimePi[ # ]+1&, x] {s=0, s1=0}; Do[s=s+(1/DivisorSigma[1, c[n]]); If[Greater[Floor[s], s1], s1=Floor[s]; Print[{n, Floor[s]}]], {n, 1, 1000000}]

a(n)=my(m,s=0.);for(c=4,(2*n+2)^(n+2),if(isprime(c),next,m++);s+=1/sigma(c);if(s>=n,return(m))) \\ Charles R Greathouse IV, Feb 19 2013

a(6)-a(11) from Donovan Johnson, Aug 22 2011

A074470 Least m such that Phi-Composite-Harmonic series Sum_{k=1..m} 1/A000010(A002808(k)) >= n.

Original entry on oeis.org

2, 7, 16, 31, 60, 113, 205, 371, 663, 1176, 2069, 3631, 6341, 11039, 19159, 33164, 57287, 98763, 169967, 292061, 501165, 858892, 1470334, 2514423, 4295912, 7333264, 12508213, 21319360, 36312685, 61811287, 105152840, 178787270, 303829041, 516074615, 876190239

Offset: 1

Labos Elemer, Sep 05 2002

Cf. A004080, A002387, A002808, A000203, A073255, A046024, A074631, A074633, A074468, A074469.

c[x_] := FixedPoint[x+PrimePi[ # ]+1&, x] {s=0, s1=0}; Do[s=s+(1/EulerPhi[c[n]]); If[Greater[Floor[s], s1], s1=Floor[s]; Print[{n, Floor[s]}]], {n, 1, 1000000}]

More terms from Lambert Klasen (lambert.klasen(AT)gmx.net), Jul 23 2005

a(30)-a(35) from Donovan Johnson, Aug 21 2011

A094229 Numbers n such that d(n) >= n-th harmonic number H(n)=sum_{i=1..n}1/i.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 64, 66, 68, 70, 72, 75, 76, 78, 80, 81, 84, 88, 90, 92, 96, 98, 99, 100, 102, 104, 105, 108, 110, 112, 114, 116, 117, 120, 124, 126, 128, 130, 132

Offset: 1

Matthew Vandermast, May 29 2004

nonn

A positive integer n belongs to the sequence if and only the number of its divisors (d(n)) is >= the average number of divisors, in the range from 1 through n, for all positive integers (H(n)).

Visible sharp bend on the graph around the 800th term occur where the n-th harmonic number exceeds 8. - Ivan Neretin, Oct 16 2016

			6 is in the sequence because the number of its divisors, 4, is greater than the 6th harmonic number, 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 2.45.

Ivan Neretin, Table of n, a(n) for n = 1..5000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 840. (d(n) is given as sigma_0(n).)

d(n)=A000005(n), H(n)=A001008(n)/A002805(n). See also A004080.

ok[n_] := DivisorSigma[0, n] >= HarmonicNumber[n]; Select[ Range[132], ok] (* Jean-François Alcover, Sep 19 2011 *)

A226187 Least positive integer k such that 1 + 1/2 + ... + 1/k > n/3.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 6, 8, 11, 16, 22, 31, 43, 60, 83, 116, 162, 227, 316, 441, 616, 859, 1199, 1674, 2336, 3260, 4550, 6349, 8861, 12367, 17259, 24088, 33617, 46916, 65477, 91380, 127531, 177984, 248397, 346666, 483812, 675214, 942336, 1315136, 1835421, 2561536, 3574912, 4989191, 6962977, 9717617

Offset: 1

Clark Kimberling, May 30 2013

nonn

Conjecture: a(n+1)/a(n) converges to 1.39...

This constant is probably exp(1/3) = 1.395612425086089528628..., see A004080. - Ralf Stephan, Jun 03 2013

			a(10) = 16 because 1 + 1/2 + ... + 1/15 < 10/3 < 1 + 1/2 + ... + 1/16.

Cf. A226186, A226188, A002387, A004080.

z = 32; f[n_] := 1/n; Do[s = 0; a[n] = NestWhile[# + 1 &, 1, ! (s += f[#]) >= n/3 &], {n, 1, z}]; m = Map[a, Range[z]]

a(n)=local(s,k);s=0;k=1;while(s<=n/3,s=s+1/k;k++);k-1

More terms from Jean-François Alcover, Jun 05 2013

Deleted obsolete b-file. - N. J. A. Sloane, Jan 04 2019

A258255 Least k such that n <= Sum_{i=1..k} 1/A258252(i), where A258252 are the numbers having lowest possible denominators for the sums of reciprocals.

Original entry on oeis.org

1, 4, 14, 46, 153, 535, 1855, 6449, 22460, 81237

Offset: 1

Ivan Neretin, May 24 2015

Presumably, every natural number is reached at some step exactly, rather than "stepped over" (as is the case with harmonic series).

			For the first few terms of A258252, the sums of their reciprocal are: 1, 3/2, 5/3, 2, 9/4, 7/3, 12/5, 5/2, 18/7, 13/5, 14/5, 17/6, 20/7, 3, ... that are equal to 1, 2, 3 for n=1, 4, 14. So a(1)=1, a(2)=4, a(3)=14.

Cf. A258252, A004080 (analog for harmonic series), A002387.

A078141 a(n) = floor(exp(n - gamma)), where gamma is Euler's constant.

Original entry on oeis.org

0, 1, 4, 11, 30, 83, 226, 615, 1673, 4549, 12366, 33616, 91380, 248397, 675213, 1835420, 4989191, 13562027, 36865412, 100210580, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245428

Offset: 0

Michael Somos, Nov 20 2002

For any fixed real value x > 0 let u(1) = 1 and u(n) = u(n-1) - x*sign(u(n-1))/n. Then the series S(x) = Sum_{k>=1} u(k) converges and has value S(x) = x*floor(exp(1/x + gamma - 1)) - x - 1 where gamma = 0.5772... is the Euler gamma constant. Thus for n >= 0 a(n+1) = n*S(1/n) + n + 1.
From Gerhard Kirchner, May 02 2020: (Start)
It is interesting to compare T(x)=exp(x-gamma) and G(x)=[T(x)] with F(x)= index k of the harmonic sequence H(k) such that H(k) is closest to x. More exactly: For H(k-1)=x1, where x1=H(k)-1/(2k) is the point of dicontinuity. Harmonic sequence: H(0)=0 and H(k)=H(k-1)+1/k for k>0 with the asymptotic formula H(k)=log(k)+gamma+1/(2k)-1/(12k^2)+O(1/k^4).
G(x) has a point of dicontinuity x2=x1+1/(12k^2), which can be derived from T(x2)=k and the asymptotic formula. Within the "critical" section x1What about n>200? Let x1Regarding the difference as a random number between 0 and 1 and considering the exponential increase of T(n), it is unlikely that any n is located in a critical section.
(End)

Cf. A002387, A004080.

PARI
```
a(n)=if(n<0,0,floor(exp(n-Euler)))
```

A168214 Least k such that Sum_{i=n..k} 1/i >= n.

Original entry on oeis.org

1, 11, 51, 192, 669, 2222, 7135, 22374, 68916, 209348, 628916, 1872269, 5531641, 16238866, 47410139, 137758585, 398617683, 1149205715, 3302324374, 9461757569, 27038402095, 77082571383, 219276117983, 622541323482, 1764242459656

Offset: 1

Zhining Yang, Nov 20 2009

			1/2 + 1/3 + ... + 1/10 < 2, but 1/2 + 1/3 + ... + 1/11 >= 2, so a(2) = 11.

Northwolves, Harmonic number sum

Cf. A004080, A002387

a[n_] := k /. FindRoot[Sum[1/i, {i, n, k}] == n, {k, n*E^n}, WorkingPrecision -> 32] // Ceiling; Table[a[n], {n, 1, 25}] (* Jean-François Alcover, Jun 08 2013 *)

a(n)=my(k=n,s);while((s+=1./k)Charles R Greathouse IV, Jun 17 2013

Sub Harmonic_number_sum()
Dim s As Double, i As Long, j As Long, n As Long
For n = 1 To 15
  s = 0
  For i = n To 1000000000
    s = s + 1 / i
    If s >= n Then Exit For
  Next
Debug.Print "a(" & n & ")=" & i:
Next
End Sub

a(18)-a(25) from Donovan Johnson, Jun 19 2010

Example edited by Jon E. Schoenfield, Dec 20 2014

A231405 Least integer j such that Sum_{i=1..j} 1/i^(1/3) >= n.

Original entry on oeis.org

1, 1, 3, 4, 6, 8, 10, 12, 15, 17, 20, 23, 25, 28, 32, 35, 38, 41, 45, 49, 52, 56, 60, 64, 68, 72, 76, 81, 85, 89, 94, 98, 103, 108, 113, 117, 122, 127, 132, 138, 143, 148, 153, 159, 164, 170, 175, 181, 187, 192, 198, 204, 210, 216, 222, 228, 234, 240, 247, 253

Offset: 0

Carmine Suriano, Nov 08 2013

nonn

			a(7)=12 since Sum_{i=1..12} 1/i^(1/3) = 7.106248... and Sum_{i=1..11} 1/i^(1/3) = 6.669458... .

Sela Fried, On the partial sums of the Zeta function Sum_{n>=1} 1/n^s for 0 < s < 1, 2024.

Cf. A004080, A054040, A054041 (condition >n).

Cf. A067086.

s=0;n=1;
for (i=1;i<30;i++) {
s+=1/Math.pow(i,1/3);
if (s>=n) {n++;document.write(Math.floor(i)+", ");}
}

s = 0; i = 0; Table[i++; While[s = s + 1/(i^(1/3)); s < n, i++]; i, {n, 100}] (* T. D. Noe, Nov 09 2013 *)
Module[{nn=300,c},c=Accumulate[1/Surd[Range[nn],3]];Table[Position[ c,?(#>=n&),1,1],{n,0,60}]]//Flatten (* _Harvey P. Dale, Aug 14 2021 *)

a(0) added by Jon Perry, Nov 10 2013

A339061 Least integer j such that H(k+j)>=n+1, where k is the least integer to satisfy H(k)>=n, and H(k) is the sum of the first k terms of the harmonic series.

Original entry on oeis.org

1, 3, 7, 20, 52, 144, 389, 1058, 2876, 7817, 21250, 57763, 157017, 426817, 1160207, 3153770, 8572836, 23303385, 63345169, 172190019, 468061001, 1272321714, 3458528995, 9401256521, 25555264765, 69466411833, 188829284972, 513291214021

Offset: 0

Matthew J. Bloomfield, Dec 21 2020

nonn

			Define H(0)=0, H(k) = Sum_{i=1..k} 1/i for k=1,2,3,...
a(0)=1: To reach n+1 from n=0 requires 1 additional term of the harmonic partial sum: H(0+1) = H(0) + 1/1 = H(1) = 1.
a(1)=3: To reach n+1 from n=1 requires 3 additional terms of the harmonic partial sum: H(1+3) = H(1) + 1/(1+1) + 1/(1+2) + 1/(1+3) = H(4) = 2.08333....
a(2)=7: To reach n+1 from n=2 requires 7 additional terms of the harmonic partial sum: H(4+7) = H(4) + 1/(4+1) + 1/(4+2) + ... + 1/(4+6) + 1/(4+7) = H(11) = 3.01987....
a(3)=20: To reach n+1 from n=3 requires 20 additional terms of the harmonic partial sum: H(11+20) = H(11) + 1/(11+1) + 1/(11+2) + ... + 1/(11+19) + 1/(11+20) = H(31) = 4.02724....

First differences of A004080.
Cf. A001113 (e), A001620 (gamma).
Cf. A001008/A002805 (harmonic numbers).
Some sequences in the same spirit as this: A331028, A002387, A004080.

#set size of search space
Max=10000000
#initialize sequence to empty
seq=vector(length=0)
#initialize partial sum to 0
partialsum=0
k=1
n=1
for(i in 1:Max){
   partialsum=partialsum+1/i
   if(partialsum>=n){
      seq=c(seq, k)
      k=0
      n=n+1
   }
   k=k+1
}
#print sequence numbers below Max
seq

a(n) ~ (e-1)*e^(n-gamma), where e is Euler's number and gamma is the Euler-Mascheroni constant.

Conjecture: a(n) = floor(1/2 + e^(n-gamma+1)) - floor(1/2 + e^(n-gamma)) for n > 1 where e is Euler's number and gamma is the Euler-Mascheroni constant. - Adam Hugill, Nov 06 2022

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cp's OEIS Frontend

A096618 Least k such that H(k) >= 10^n, where H(k) is the harmonic number Sum_{i=0..k} 1/i.

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A074469 Least m such that Sigma-Composite-Harmonic series Sum_{k=1..m} 1/A000203(A002808(k)) >= n.

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Mathematica

PARI

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A074470 Least m such that Phi-Composite-Harmonic series Sum_{k=1..m} 1/A000010(A002808(k)) >= n.

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Mathematica

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A094229 Numbers n such that d(n) >= n-th harmonic number H(n)=sum_{i=1..n}1/i.

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Mathematica

A226187 Least positive integer k such that 1 + 1/2 + ... + 1/k > n/3.

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Mathematica

PARI

Extensions

A258255 Least k such that n <= Sum_{i=1..k} 1/A258252(i), where A258252 are the numbers having lowest possible denominators for the sums of reciprocals.

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Comments

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A078141 a(n) = floor(exp(n - gamma)), where gamma is Euler's constant.

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PARI

A168214 Least k such that Sum_{i=n..k} 1/i >= n.

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Mathematica

PARI

VBA

Extensions

A231405 Least integer j such that Sum_{i=1..j} 1/i^(1/3) >= n.

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