A103762 -id:A103762 - cp's OEIS frontend

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.

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

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)) ).

A192881 Number of terms for the shortest Egyptian fraction representation of 1 starting with 1/n.

Original entry on oeis.org

1, 3, 5, 8, 10, 11, 13, 15, 17, 19

Offset: 1

Teena Carroll, Jul 11 2011

An Egyptian fraction representation of a rational number a/b is a list of distinct unit fractions with sum a/b.

			Since 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1, we see that a(3) <= 5.  We know the maximum sum of 4 distinct unit fractions (1/3 or less) is 19/20, so this shows a(3)=5. An Egyptian fraction decomposition of 1 starting with 1/4 must have at least 8 terms; however, the expressions need not be unique, as all three of 1 = 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/230 + 1/57960, 1 = 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/231 + 1/27720 and 1 = 1/4 + 1/5 + 1/6 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 achieve this bound. - _Teena Carroll_, _Haoqi Chen_ and _Javier Múgica_

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.
M. N. Bleicher, A new algorithm for the expansion of Egyptian fractions, J. Numb. Theory 4 (1972) 342-382
Javier Múgica, decompositions achieving the terms in this sequence.
Index entries for sequences related to Egyptian fractions

Cf. A103762, A294651.

a(n) >= A103762(n) - n + 1.

Two more terms from Javier Múgica, Dec 18 2017

A242679 Bobo numbers: Numbers k with the property that floor(e*k) = least m with Sum_{j=k..m} 1/j > 1.

Original entry on oeis.org

4, 11, 18, 25, 32, 36, 43, 50, 57, 64, 71, 75, 82, 89, 96, 103, 114, 121, 128, 135, 142, 146, 153, 160, 167, 174, 185, 192, 199, 206, 213, 217, 224, 231, 238, 245, 256, 263, 270, 277, 284, 288, 295, 302, 309, 316, 327, 334, 341, 348, 355, 359, 366, 373, 380, 387, 398, 405, 412, 419, 426, 430, 437, 444, 451, 458, 469, 476, 483, 490, 497

Offset: 1

Steven J. Kifowit, May 20 2014

nonn

These are the numbers n for which A103762(n) = floor(e*n).
If frac(e*n) > (e-1)/2, then n is a Bobo number, but not every Bobo number has this property. The exceptions are in A277603.
In Bobo's article (see Bobo link), the Bobo numbers through 2105 are listed. There is a typo: the number 143 is given in place of the correct number 142.
These numbers are mentioned in the comments associated with A103762. Differences between consecutive Bobo numbers are indeed 4, 7, or 11. An elementary proof is given in the Clancy/Kifowit link.

Steven J. Kifowit, 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.
D. T. Clancy and S. J. Kifowit, A closer look at Bobo's sequence, College Math. J. 45 (2014), 199-206.
Steve Kifowit, Bobo Numbers, Bobbers, and Bears—Experiences, Undergraduate Research, Preprint, 2016.

Cf. A103762, A277603.

is(n)=my(e=exp(1),s); if(frac(e*n)>(e-1)/2, return(1)); s=sum(j=n,e*n\1-1, 1/j); s<=1 && s+e*n\1>1 \\ Charles R Greathouse IV, Sep 17 2016

A277603 Exceptional Bobo numbers: terms of A242679 that satisfy frac[e*A242679(n)]<(e-1)/2.

Original entry on oeis.org

36, 9045, 5195512, 5311399545, 8488859795196, 25466579385587, 19542965851120621, 58628897553361862, 61250772004870841520, 183752316014612524559, 250769086731739376780337, 752307260195218130341010, 1299515735021702625544976020, 3898547205065107876634928059

Offset: 1

Steven J. Kifowit, Oct 22 2016

nonn

The exceptional Bobo numbers (EBNs) are very rare relative to the Bobo numbers (A242679).

Exceptional Bobo numbers come in two varieties. Type-1 EBNs are given by the recurrence E(0)=1,E(1)=1,E(k)=(2*k-1)*(2*E(k-1)-1)+E(k-2) for k=3,5,7,... These are derived from the denominators of the odd-indexed convergents of the continued fraction expansion of (e-1)/2 = [0;1,6,10,14,18,...]. The Type-2 EBNs are derived from the Type-1 EBNs. They have the form n*m-(m-1)/2 where n is a Type-1 EBN and m>=3 is an odd integer. However, not every number of this form is an EBN.

S. J. Kifowit, A. Mitchell, and S. Zandi, Exceptional Bobo Numbers, in preparation 2016

Steven J. Kifowit, Table of n, a(n) for n = 1..99
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.
Steve Kifowit, Bobo Numbers, Bobbers, and Bears—Experiences, Undergraduate Research, Preprint, 2016.

Cf. A103762, A242679 (Bobo numbers).

A180224 a(n+1) is the least k such that 1/(a(n)+1) + 1/(a(n)+2) + ... + 1/k > 1, with a(1) = 1.

Original entry on oeis.org

1, 4, 12, 34, 94, 257, 700, 1904, 5177, 14074, 38258, 103997, 282695, 768446, 2088854, 5678095, 15434664, 41955768, 114047603, 310013528, 842704141, 2290707355, 6226788179, 16926165158, 46010087176, 125068383898, 339971115266, 924137304830, 2512065642722, 6828502388509

Offset: 1

Pierre CAMI, Aug 16 2010

nonn

			1/2 = 0.5, 1/2 + 1/3 = 0.833..., 1/2 + 1/3 + 1/4 = 1.0833... > 1, so a(2) = 4.

Cf. A103762.

default(realprecision, 10^5); e=exp(1);
lista(nn) = {my(k=1); print1(k); for(n=2, nn, print1(", ", k=floor(e*k+(e+1)/2+(e-1/e)/(24*(n+1/2))))); } \\ Jinyuan Wang, Mar 05 2020

a(n) = A103762(a(n-1) + 1) for n > 1. - Jinyuan Wang, Mar 05 2020

Name clarified by and more terms from Jinyuan Wang, Mar 05 2020

A213908 Minimal number of terms in the series 1/n + 1/(n+1) + 1/(n+2) + ... to obtain a sum >= 1.

Original entry on oeis.org

1, 3, 5, 7, 8, 10, 12, 13, 15, 17, 19, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 49, 51, 53, 55, 56, 58, 60, 62, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 80, 82, 84, 86, 87, 89, 91, 92, 94, 96, 98, 99, 101, 103, 104, 106, 108, 110

Offset: 1

Franz Vrabec, Jun 24 2012

nonn

			a(3)=5 because 1/3 + 1/4 + 1/5 + 1/6 < 1 (4 terms), but 1/3 + 1/4 + 1/5 + 1/6 + 1/7 >= 1 (5 terms).

T. D. Noe, Table of n, a(n) for n = 1..1000

i = 0; s = 0; Table[While[s < 1, i++; s = s + 1/i]; s = s - 1/n; i - n + 1, {n, 100}] (* T. D. Noe, Jun 26 2012 *)
With[{nn=100},Table[Position[Accumulate[1/Range[n,2 nn]],?(#>=1&),1,1],{n,nn}]]//Flatten (* _Harvey P. Dale, May 09 2021 *)

a(n) = A103762(n) - n + 1. - T. D. Noe, Jun 26 2012

cp's OEIS Frontend

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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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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A192881 Number of terms for the shortest Egyptian fraction representation of 1 starting with 1/n.

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A242679 Bobo numbers: Numbers k with the property that floor(e*k) = least m with Sum_{j=k..m} 1/j > 1.

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A277603 Exceptional Bobo numbers: terms of A242679 that satisfy frac[e*A242679(n)]<(e-1)/2.

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A180224 a(n+1) is the least k such that 1/(a(n)+1) + 1/(a(n)+2) + ... + 1/k > 1, with a(1) = 1.

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A213908 Minimal number of terms in the series 1/n + 1/(n+1) + 1/(n+2) + ... to obtain a sum >= 1.

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