A136617 -id:A136617 - cp's OEIS frontend

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

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

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.

Original entry on oeis.org

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

A081881 Pack bins of size 1 sequentially with items of size 1/1, 1/2, 1/3, 1/4, ... . Sequence gives values of n for which 1/n starts a new bin.

Original entry on oeis.org

1, 2, 4, 10, 26, 69, 186, 504, 1369, 3720, 10111, 27483, 74705, 203068, 551995, 1500477, 4078718, 11087104, 30137872, 81923228, 222690421, 605335323, 1645472007, 4472856655, 12158484965, 33050188741, 89839727480, 244209698681, 663830786257, 1804479163453, 4905082919846

Offset: 1

Wouter Meeussen, Apr 13 2003

For n >= 3, it appears that a(n) = round((a(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) - ... + 7709321041217/(16320*k^32), where gamma is the Euler-Mascheroni constant, A001620). - Jon E. Schoenfield, Mar 30 2018

			1/1; 1/2+1/3, 1/4+1/5+1/6+1/7+1/8+1/9 are all just less than or equal to 1; so first four terms are 1, 2, 4, 10.
Lower and upper indices of bin contents are {1,1}, {2,3}, {4,9}, {10,25}, {26,68}, {69,185}, {186,503}, {504,1368}, {1369,3719}, {3720,10110}, {10111,27482}, ...

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

Cf. A002387, A136616, A136617, A295572, A300897.

res ={}; FoldList[If[ #1+#2 > 1, AppendTo[res, #2];#2, #1+#2]&, 0, Table[1/k, {k, 1, 1000}]]; 1/res
lst = {1, 2}; n = 2; Do[s = 0; While[s = N[s + 1/n, 64]; s < 1, n++ ]; AppendTo[lst, n]; Print@n, {i, 25}]; lst (* Robert G. Wilson v, Aug 19 2008 *)

default(realprecision, 10^4); e=exp(1);
A136616(k) = floor(e*k + (e-1)/2 + (e-1/e)/(24*k+12));
lista(nn) = {my(k=1); print1(k); for(n=2, nn, k=A136616(k-1)+1; print1(", ", k)); } \\ Jinyuan Wang, Feb 20 2020

a(n) is asymptotic to C*exp(n) where C=0.1688... - Benoit Cloitre, Apr 14 2003
C = 0.16885635666714420373167977550090103410150395689764... (cf. A300897). - Jon E. Schoenfield, Apr 12 2018
a(n) = 1 + (A136616^(n-1))(0), where (f^0)(x)=x, (f^(n+1))(x) = f((f^n)(x)) for any function f. - Rainer Rosenthal, Feb 16 2008, Apr 05 2020

a(13)-a(25) from Robert G. Wilson v, Aug 19 2008

More terms from Jinyuan Wang, Feb 20 2020

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 *)

A327600 a(n) is the largest k such that the sum of k consecutive reciprocals 1/p_n + ... + 1/p_(n+k-1) does not exceed 1 (where p_n = n-th prime).

Original entry on oeis.org

2, 8, 26, 65, 143, 252, 423, 650, 976, 1391, 1865, 2478, 3168, 3980, 4977, 6136, 7419, 8828, 10476, 12278, 14294, 16612, 19123, 21905, 24903, 28055, 31493, 35319, 39485, 44101, 49115, 54102, 59467, 65142, 71314, 77648, 84503, 91719, 99302, 107364

Offset: 1

Leon Bykov, Sep 18 2019

nonn

The sequence is nondecreasing, since 1/p_n + ... + 1/p_(n+k-1) > 1/p_(n+1) + ... + 1/p_(n+k).

			For n = 1, since 1/p_1 + 1/p_2 = 1/2 + 1/3 = 5/6 <= 1 while 1/p_1 + 1/p_2 + 1/p_3 = 1/2 + 1/3 + 1/5 = 31/30 > 1, a(1) = 2.

Analog of A136617. Cf. A137368.

a(n) = A137368(n) - 1.

cp's OEIS Frontend

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

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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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A081881 Pack bins of size 1 sequentially with items of size 1/1, 1/2, 1/3, 1/4, ... . Sequence gives values of n for which 1/n starts a new bin.

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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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A327600 a(n) is the largest k such that the sum of k consecutive reciprocals 1/p_n + ... + 1/p_(n+k-1) does not exceed 1 (where p_n = n-th prime).

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