A081881 -id:A081881 - cp's OEIS frontend

A295572 First differences of A081881.

1, 2, 6, 16, 43, 117, 318, 865, 2351, 6391, 17372, 47222, 128363, 348927, 948482, 2578241, 7008386, 19050768, 51785356, 140767193, 382644902, 1040136684, 2827384648, 7685628310, 20891703776, 56789538739, 154369971201, 419621087576, 1140648377196, 3100603756393

Offset: 1

N. J. A. Sloane, Nov 30 2017, following a suggestion from Loren Booda

nonn

See A081881 and A295571 for discussion.

If the harmonic series is divided into the longest possible consecutive groups so that the sum of each group is <= 1, then a(n) is the number of terms in the n-th group. - Pablo Hueso Merino, Feb 16 2020

			From _Pablo Hueso Merino_, Feb 16 2020: (Start)
a(1) = 1 because 1 <= 1, 1 is one term (if you added 1/2 the sum would be greater than 1).
a(2) = 2 because 1/2 + 1/3 = 0.8333... <= 1, 1/2 and 1/3 are two terms (if you added 1/4 the sum would be greater than one).
a(3) = 6 because 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 = 0.9956... <= 1, it is a sum of six terms. (End)

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

Cf. A081881, A295571, A300897, A331028, A331030.

a[1]=1;
a[n_]:= a[n]= Module[{sum = 0}, r = 1 + Sum[a[k], {k, n-1}];
   x = r;
   While[sum <= 1, sum += 1/x++];
   p = x-2;
   p -r +1];
Table[a[n], {n, 10}] (* Pablo Hueso Merino, Feb 16 2020 *)

a(1) = 1, a(n) = (max(m) : Sum_{s=r..m} 1/s <= 1)-r+1, r = Sum_{k=1..n-1} a(k). - Pablo Hueso Merino, Feb 16 2020

a(n) ~ c * exp(n), where c = (exp(1)-1) * A300897 = 0.290142809280953235916025... - Vaclav Kotesovec, Apr 05 2020

More terms from Jinyuan Wang, Feb 20 2020

A295571 a(n) = A081881(n) - 1.

Original entry on oeis.org

0, 1, 3, 9, 25, 68, 185, 503, 1368, 3719, 10110, 27482, 74704, 203067, 551994, 1500476, 4078717, 11087103, 30137871, 81923227, 222690420, 605335322, 1645472006, 4472856654, 12158484964, 33050188740, 89839727479, 244209698680, 663830786256, 1804479163452, 4905082919845

Offset: 1

N. J. A. Sloane, Nov 30 2017

nonn

The blocks of fractions described in A081881 extend from 1/A081881(k) through 1/a(k+1) and contain A295572(k) terms. For example the third block is 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/6 and has length 6.

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

Cf. A081881, A295572.

More terms from Jinyuan Wang, Feb 20 2020

A300897 Decimal expansion of lim_{n->infinity} A081881(n)/exp(n).

Original entry on oeis.org

1, 6, 8, 8, 5, 6, 3, 5, 6, 6, 6, 7, 1, 4, 4, 2, 0, 3, 7, 3, 1, 6, 7, 9, 7, 7, 5, 5, 0, 0, 9, 0, 1, 0, 3, 4, 1, 0, 1, 5, 0, 3, 9, 5, 6, 8, 9, 7, 6, 4, 9, 2, 2, 2, 3, 7, 7, 2, 2, 5, 5, 2, 2, 7, 1, 4, 1, 7, 5, 3, 3, 0, 3, 0, 3, 3, 0, 1, 2, 8, 7, 7, 6, 7, 1, 0, 7

Offset: 0

Jon E. Schoenfield, Apr 12 2018

			0.16885635666714420373167977550090103410150395689764...

Cf. A081881, A295572.

A103762 a(n) = least k with Sum_{j = n..k} 1/j >= 1.

Original entry on oeis.org

1, 4, 7, 10, 12, 15, 18, 20, 23, 26, 29, 31, 34, 37, 39, 42, 45, 48, 50, 53, 56, 58, 61, 64, 67, 69, 72, 75, 77, 80, 83, 86, 88, 91, 94, 97, 99, 102, 105, 107, 110, 113, 116, 118, 121, 124, 126, 129, 132, 135, 137, 140, 143, 145, 148, 151, 154, 156, 159, 162

Offset: 1

David W. Wilson, Apr 14 2008

nonn

a(n) = A136617(n) + n for n > 1. Also a(n) = A136616(n-1) + 1 for n > 1.
If you compare this to floor(e*n) = A022843, 2,5,8,10,13,16,..., it appears that floor(e*n)-a(n) = 1,1,1,0,1,1,1,1,1,1,0,..., initially consisting of 0's and 1's. The places where the 0's occur are 4, 11, 18, 25, 32, 36, 43, 50, 57, 64, 71, ... whose differences seem to be 4, 7 or 11.
There are some rather sharp estimates on this type of differences between harmonic numbers in Theorem 3.2 of the Sintamarian reference, which may help to uncover such a pattern. - R. J. Mathar, Apr 15 2008
a(n) = round(e*(n-1/2)) with the exception of the terms of A277603; at those values of n, a(n) = round(e*(n-1/2)) + 1. - Jon E. Schoenfield, Apr 03 2018

T. D. Noe, Table of n, a(n) for n = 1..1000
E. R. Bobo, A sequence related to the harmonic series, College Math. J. 26 (1995), 308-310.
D. T. Clancy and S. J. Kifowit, A closer look at Bobo's sequence, College Math. J. 45 (2014), 199-206.
A. Sintamarian, A generalization of Euler's constant, Numer. Algor. 46 (2007), pp. 141-151.

Cf. A136616, A136617, A242679 (Bobo numbers).

Cf. A081881, A277603, A289183. - Jon E. Schoenfield, Mar 31 2018

i = 0; s = 0; Table[While[s < 1, i++; s = s + 1/i]; s = s - 1/n; i, {n, 100}] (* T. D. Noe, Jun 26 2012 *)

default(realprecision, 10^5); e=exp(1);
a(n) = if(n<2, 1, floor(e*n+(1-e)/2+(e-1/e)/(24*n-12))); \\ Jinyuan Wang, Mar 06 2020

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)) ).

A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

Original entry on oeis.org

2, 5, 13, 36, 98, 266, 723, 1965, 5342, 14521, 39472, 107296, 291661, 792817, 2155100, 5858169, 15924154, 43286339, 117664468, 319845186, 869429357, 2363354022, 6424262292, 17462955450, 47469234471, 129034757473, 350752836478, 953445061679, 2591732385596

Offset: 1

Pablo Hueso Merino and Alejandro Argüelles Trujillo, Jan 07 2020

nonn

a(n) = A046171(n+1) through a(5), and grows similarly for n > 5.

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). 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) - ..., where gamma is the Euler-Mascheroni constant, A001620). Cf. A081881. - Jon E. Schoenfield, Jan 10 2020

			a(1)=2 because 1 + 1/2 = 1.5 > 1,
a(2)=5 because 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 1.0928... > 1,
etc.

Cf. A002387, A046171, A081881, A096005, A180224, A295572, A331028.

lista(lim=oo)={my(s=0, p=0); for(i=1, lim, s+=1/i; if(s>1, print1(i-p, ", "); s=0; p=i))} \\ Andrew Howroyd, Jan 08 2020

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

a(1)=2, a(n) = (min(p) : Sum_{s=r..p} 1/s > 1)-r+1, r=Sum_{k=1..n-1} a(k).

a(25)-a(29) from Jon E. Schoenfield, Jan 10 2020

cp's OEIS Frontend

A295572 First differences of A081881.

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A295571 a(n) = A081881(n) - 1.

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A300897 Decimal expansion of lim_{n->infinity} A081881(n)/exp(n).

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A103762 a(n) = least k with Sum_{j = n..k} 1/j >= 1.

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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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A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

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