A002387 -id:A002387 - cp's OEIS frontend

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

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

A082912 Least k such that H(k) > 10^n, where H(k) is the harmonic number Sum_{i=1..k} 1/i.

Original entry on oeis.org

2, 12367, 15092688622113788323693563264538101449859497

Offset: 0

Robert G. Wilson v, Apr 14 2003

"In 1968 John W. Wrench Jr calculated the exact minimum number of terms needed for the series to sum past 100; that number is 15 092 688 622 113 788 323 693 563 264 538 101 449 859 497. Certainly, he did not add up the terms.

"Imagine a computer doing so and suppose that it takes it 10^-9 seconds to add each new term to the sum and that we set it adding and let it continue doing so indefinitely. The job will have been completed in not less than 3.5 * 10^17 (American) billion years." Havil.

Julian Havil, Gamma, Exploring Euler's Constant, Princeton University Press, Princeton and Oxford, 2003, page 23.

R. Baillie, Fun With Very Large Numbers, arXiv preprint arXiv:1105.3943, 2011
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, Harmonic Number
Lin Zhang, A Likelihood Ratio Test of Independence of Components for High-dimensional Normal Vectors, MS Thesis, Univ. Minnesota, 2013.

Cf. A002387, A001620.

f[n_] := Floor[Exp[n - EulerGamma] - 1/2] + 1; Table[ f[10^n], {n, 0, 2}]

H(k) ~= log(k) + Euler's Gamma Constant (A001620) + 1/(2k).

a(n) = A002387(10^n). - Joerg Arndt, Jul 13 2015

A082913 Least k such that H(k) > 2^n, where H(k) is the harmonic number Sum_{i=1..k} 1/i.

Original entry on oeis.org

2, 4, 31, 1674, 4989191, 44334502845080, 3500783582875029181027036603, 21827907538883637012326748457700300661358717434156476363

Offset: 0

Robert G. Wilson v, Apr 14 2003

nonn

Julian Havil, Gamma, Exploring Euler's Constant, Princeton University Press, Princeton and Oxford, 2003, page 23.
Murray Schechter, Summation of divergent series by computer, Amer. Math. Monthly, 91:10 (1984), 629-632. See Table 1.

Cf. A002387, A001620.

f[n_] := Floor[Exp[n - EulerGamma] - 1/2] + 1; Table[ f[ 2^n], {n, 0, 7}]

H(k) ~= log(k) + Euler's Gamma Constant (A001620) + 1/(2k).

a(n) = A002387(2^n). - Joerg Arndt, Jul 13 2015

A087460 Least n such that H(n) is closer to an integer than any H(j) with j < n; where H(n) is the harmonic number sum_{i=0..n} 1/i.

Original entry on oeis.org

2, 3, 4, 10, 11, 30, 83, 226, 4549, 91379, 91380, 248396, 248397, 675213, 4989190, 4989191, 13562026, 13562027, 36865412, 100210580, 2012783315, 5471312310, 40427833595, 40427833596, 109894245428, 812014744421, 812014744422, 2207284924202, 2207284924203

Offset: 2

Robert G. Wilson v, Sep 03 2003

nonn

H(36865412) = 18.000000003719931082993704481490195538573320586002

The inequality, 0.5*log(n^2+n)+gamma < H(n) < 0.5*log(n^2+n)+gamma +1/(6*n^2+6*n) (see Villarino link), where gamma is the Euler-Mascheroni constant, can be used to determine terms of this sequence without directly computing the harmonic numbers. - Steven J. Kifowit, May 26 2015

Steven J. Kifowit, Table of n, a(n) for n = 2..50
M. B. Villarino, Ramanujan's Approximation to the nth Partial Sum of the Harmonic Series, arXiv:math/0402354v5 [math.CA], 2005.

Cf. A002387, A004796.

d = 1; s = 1; n = 2; Do[ While[s = N[s + 1/n, 50]; Abs[Round[s] - s] > d, n++ ]; Print[n]; d = Abs[Round[s] - s]; n++, {i, 2, 18}]

Corrected and extended by Steven J. Kifowit, May 26 2015

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

A214966 Array T(m,n) = greatest k such that 1/n + ... + 1/(n+k-1) <= m, by rising antidiagonals.

Original entry on oeis.org

1, 3, 2, 10, 9, 4, 30, 29, 16, 6, 82, 81, 48, 22, 7, 226, 225, 134, 67, 28, 9, 615, 614, 370, 188, 86, 35, 11, 1673, 1672, 1012, 517, 241, 105, 41, 12, 4549, 4548, 2756, 1413, 664, 295, 124, 47, 14, 12366, 12365, 7498, 3847, 1814, 811, 348, 143, 54

Offset: 1

Clark Kimberling, Sep 01 2012

Row 1: A136617.

Column 1: A115515 = -1 + A002387.

			Northwest corner (the array is read by northeast antidiagonals):
    1     2     4     6     7     9
    3     9    16    22    28    35
   10    29    48    67    86   105
   30    81   134   188   241   295
   82   225   370   517   664   811
  226   614  1012  1413  1814  2216

Clark Kimberling, Rising antidiagonals n = 1..60, flattened

Cf. A136617, A115515, A002387.

t = Table[1 + Floor[x /. FindRoot[HarmonicNumber[N[x + z, 150]] - HarmonicNumber[N[z - 1, 150]] == m, {x, Floor[-E^bm/2 + (-1 + E^m) z]}, WorkingPrecision -> 100]], {m, 1, #}, {z, 1, #}] &[12]
TableForm[t]
u = Flatten[Table[t[[i - j]][[j]], {i, 2, 12}, {j, 1, i - 1}]]
(* Peter J. C. Moses, Aug 29 2012 *)

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.

A332991 a(n) is the smallest prime power p^k (p prime, k >= 1) such that sum of reciprocals of prime powers up to p^k (prime power harmonic sum) exceeds n.

Original entry on oeis.org

2, 4, 19, 1307, 266655247

Offset: 0

Ilya Gutkovskiy, Mar 05 2020

The corresponding indices of a(n) in A246655 are 1, 3, 12, 240, ...

			a(1) = 4 because 1/2 + 1/3 = 0.8333... < 1 but 1/2 + 1/3 + 1/4 = 1.0833... > 1.

Eric Weisstein's World of Mathematics, Prime Power

Similar sequences: A002387 (for positive integers), A016088 (for primes), A076751 (for composite numbers), A333004 (for squarefree numbers).

Cf. A246655 (prime powers).

a(4) from Daniel Suteu, Mar 05 2020

cp's OEIS Frontend

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

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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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A082912 Least k such that H(k) > 10^n, where H(k) is the harmonic number Sum_{i=1..k} 1/i.

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A082913 Least k such that H(k) > 2^n, where H(k) is the harmonic number Sum_{i=1..k} 1/i.

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A087460 Least n such that H(n) is closer to an integer than any H(j) with j < n; where H(n) is the harmonic number sum_{i=0..n} 1/i.

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

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A214966 Array T(m,n) = greatest k such that 1/n + ... + 1/(n+k-1) <= m, by rising antidiagonals.

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

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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.