A136616 -id:A136616 - 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

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

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)) ).

A225669 Slowest-growing sequence of odd primes whose reciprocals sum to 1.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 967, 101419, 2000490719, 106298338760698351, 586903266015193517540253132922939, 3494365451928289992209032562272585187947069047023572601254975717

Offset: 1

Jonathan Sondow, May 11 2013

nonn

See comments, references, and links in A075442 = slowest-growing sequence of primes whose reciprocals sum to 1.

a(n) = 3, 5, 7, 11, 13, 17, 19, 23, 967, ..., so A225671(2) = 23.

			Since 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/23 < 1, the first eight odd primes are members. The ninth is not, because adding 1/29 pushes the sum over 1.

Popular Computing (Calabasas, CA), Problem 175: A Sum of a Different Kind, Vol. 5 (No. 50, May 1977), p. PC50-8.

Amiram Eldar, Table of n, a(n) for n = 1..18

Cf. A000058, A075442, A046689, A136616, A181503, A225670, A225671.

a[n_] := a[n] = Block[{sm = Sum[1/(a[i]), {i, n - 1}]}, NextPrime[ Max[ a[n - 1], 1/(1 - sm)]]]; a[0] = 2; Array[a, 14]

A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

Original entry on oeis.org

1, 3, 8, 22, 60, 163, 443, 1204, 3273, 8897, 24184, 65739, 178698, 485751, 1320408, 3589241, 9756569, 26521104, 72091835, 195965925, 532690613, 1448003214, 3936080824, 10699376979, 29083922018, 79058296722, 214902731368, 584166189564, 1587928337892, 4316436745787

Offset: 1

Alejandro Argüelles Trujillo and Pablo Hueso Merino, Jan 07 2020

nonn

a(n) is equal to A024581(n) through a(10), and grows very similarly for n > 10.

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). Cf. A331030. - Jon E. Schoenfield, Jan 14 2020

			a(1)=1 because 1 >= 1,
a(2)=3 because 1/2 + 1/3 + 1/4 = 1.0833... >= 1, etc.

Cf. A004080, A024581, A136616, A331030.

Some sequences in the same spirit as this: A002387, A004080, A055980, A115515.

default(realprecision, 10^5); e=exp(1);
lista(nn) = {my(r=1); print1(r); for(n=2, nn, print1(", ", -r+(r=floor(e*r+(e+1)/2+(e-1/e)/(24*(r+1/2)))))); } \\ Jinyuan Wang, Mar 31 2020

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

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

a(20)-a(21) from Giovanni Resta, Jan 14 2020

More terms from Jinyuan Wang, Mar 31 2020

A363993 a(n) = greatest integer k such that 1/n + 1/(n + 1) +...+ 1/k < 2.

Original entry on oeis.org

3, 10, 18, 25, 32, 40, 47, 54, 62, 69, 77, 84, 91, 99, 106, 114, 121, 128, 136, 143, 150, 158, 165, 173, 180, 187, 195, 202, 210, 217, 224, 232, 239, 247, 254, 261, 269, 276, 283, 291, 298, 306, 313, 320, 328, 335, 343, 350, 357, 365, 372, 380, 387, 394, 402

Offset: 1

Clark Kimberling, Sep 06 2023

nonn

In general, if r > 0 and n > 1, then a(n) is the number k such that h(k) <= r + h(n-1) < h(k+1), where h(m) = m-th harmonic number. Since h(n) is approximately g + log(n+1/2), where g = Euler-Mascheroni constant (A001620), it is easy to prove that a(n) or a(n)-1 is the number floor[n*e^r-(1+e^r)/2]. Thus, the difference sequence of (a(n)) has at most two distinct numbers. For r = 2, the two numbers are 7 and 8.

			a(2) = 10 because 1/2 + 1/3 +...+ 1/10 < 2 < 1/2 + 1/3 +...+ 1/11.

Cf. A001620, A136616, A364609.

r = 2; h[n_] := HarmonicNumber[n];
a[n_] : = Select[Range[500], h[#] <= r + h[n - 1] < h[# + 1] & ]
Flatten[Table[a[n], {n, 1, 70}]]

a(n) = my(k=0); while (sum(i=n, n+k, 1/i) < 2, k++); n+k-1; \\ Michel Marcus, Sep 08 2023

from itertools import accumulate, count
from fractions import Fraction
def A363993(n): return next(x[0]+n-1 for x in enumerate(accumulate(Fraction(1,k) for k in count(n))) if x[1] >= 2) # Chai Wah Wu, Sep 07 2023

A364609 a(n) = greatest integer k such that 1/n + 1/(n + 1) + ... + 1/k < sqrt(2).

Original entry on oeis.org

1, 5, 9, 13, 18, 22, 26, 30, 34, 38, 42, 46, 50, 55, 59, 63, 67, 71, 75, 79, 83, 87, 92, 96, 100, 104, 108, 112, 116, 120, 124, 129, 133, 137, 141, 145, 149, 153, 157, 161, 166, 170, 174, 178, 182, 186, 190, 194, 198, 203, 207, 211, 215, 219, 223, 227, 231

Offset: 1

Clark Kimberling, Sep 06 2023

nonn

In general, if r > 0 and n > 1, then a(n) is the number k such that h(k) <= r + h(n-1) < h(k+1), where h(m) = m-th harmonic number. Since h(n) is approximately g + log(n+1/2), where g = Euler-Mascheroni constant (A001620), it is easy to prove that a(n) or a(n)-1 is the number floor(n*e^r - (1+e^r)/2). Thus, the difference sequence of (a(n)) has at most two distinct numbers; for r = sqrt(2), the two numbers are 4 and 5.

			a(3) = 9 because 1/3 + 1/4 + ... + 1/9 < sqrt(2) < 1/3 + 1/4 + ... + 1/10.

Cf. A001620, A136616, A357923, A363993.

r = Sqrt[2]; h[n_] := HarmonicNumber[n];
a[n_] : = Select[Range[500], h[#] <= r + h[n - 1] < h[# + 1] & ]
Flatten[Table[a[n], {n, 1, 70}]]

a(n) = my(k=0); while (sum(i=n, n+k, 1/i)^2 < 2, k++); n+k-1; \\ Michel Marcus, Sep 08 2023

from itertools import accumulate, count
from fractions import Fraction
def A364609(n): return next(x[0]+n-1 for x in enumerate(accumulate(Fraction(1,k) for k in count(n))) if x[1]**2 >= 2) # Chai Wah Wu, Sep 07 2023

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A136617 a(n) = largest k such that the sum of k consecutive reciprocals 1/n + ... + 1/(n+k-1) does not exceed 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A225669 Slowest-growing sequence of odd primes whose reciprocals sum to 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Python

Formula

Extensions

A363993 a(n) = greatest integer k such that 1/n + 1/(n + 1) +...+ 1/k < 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

A364609 a(n) = greatest integer k such that 1/n + 1/(n + 1) + ... + 1/k < sqrt(2).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI