A092315 -id:A092315 - cp's OEIS frontend

A056053 a(n) = smallest odd number 2m+1 such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

1, 3, 15, 113, 837, 6183, 45691, 337607, 2494595, 18432707, 136200301, 1006391657, 7436284415, 54947122715, 406007372211, 3000011249847, 22167251422541, 163795064320249, 1210290918990281, 8942907496445513, 66079645178783351, 488266205223462461, 3607826381608149807

Offset: 0

Robert G. Wilson v, Jul 25 2000 and Jan 11 2004

nonn

a(2) = 15 and a(3) = 113 are related to the Borwein integrals. Concretely, a(2) = 15 is the smallest odd m such that the integral Integral_{x=-oo..oo} Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi, and a(3) = 113 is the smallest odd m such that the integral Integral_{x=-oo..oo} cos(x) * Product_{1<=k<=m, k odd} (sin(k*x)/(k*x)) dx is slightly less than Pi/2. See the Wikipedia link and the 3Blue1Brown video link below. - Jianing Song, Dec 10 2022

Calvin C. Clawson, "Mathematical Mysteries, The Beauty and Magic of Numbers," Plenum Press, NY and London, 1996, page 64.

Robert G. Wilson v, Table of n, a(n) for n = 0..500
Grant Sanderson, Researchers thought this was a bug (Borwein integrals), 3Blue1Brown video (2022).
Wikipedia, Borwein integral

Cf. A002387, A056054, A091463, A091464, A091465.

Cf. also A092318, A092317, A092315, A281355, A074599, A025547.

s = 0; k = 1; Do[ While[s = N[s + 1/k, 24]; s <= n, k += 2]; Print[k]; k += 2, {n, 1, 11}]

a(n) ~ floor((1/2)*A002387(2n)).
The next term is approximately the previous term * e^2.
a(n) = A092315(n)*2 + 1 = floor(exp(n*2-Euler)/4+1/8)*2+1 for all n (conjectured). - M. F. Hasler, Jan 24 2017
a(n) ~ exp(2*n - A350763) = (1/2)*exp(2*n - gamma), gamma = A001620. - A.H.M. Smeets, Apr 15 2022

Corrected by N. J. A. Sloane, Feb 16 2004
More terms from Robert G. Wilson v, Apr 17 2004
a(17) corrected - see correction in A092315. - Gerhard Kirchner, Jul 25 2020
a(0) prepended by Robert G. Wilson v, Oct 23 2024

A092318 a(n) = smallest m such that value of odd harmonic series Sum_{j=0..m} 1/(2j+1) is >= n.

Original entry on oeis.org

0, 7, 56, 418, 3091, 22845, 168803, 1247297, 9216353, 68100150, 503195828, 3718142207, 27473561357, 203003686105, 1500005624923, 11083625711270, 81897532160124, 605145459495140, 4471453748222756, 33039822589391675

Offset: 1

N. J. A. Sloane, Feb 16 2004

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..1000 (127 terms corrected by Gerhard Kirchner)

Apart from first term, same as A092315. Equals (A092317-1)/2.
Cf. A074599, A025547.
Cf. A281355 (= a(n) + 1) for a variant.

a[n_] := Floor[(Exp[2 n - EulerGamma] + 1/2)/4]; a[1] = 0; Array[a, 20] (* Robert G. Wilson v, Jan 25 2017 *)

A092318=n->floor(exp(2*n-Euler)/4+1/8)-(n<2) \\ Cf. comments in A092315. - M. F. Hasler, Jan 24 2017

a(n) = floor(exp(2*n-gamma)/4+1/8), for all n > 1. - M. F. Hasler and Robert G. Wilson v, Jan 22 2017

a(n) = floor(exp(2*n-gamma)/4), for all n > 1, see correction in A092315, Gerhard Kirchner, Jul 25 2020

More terms (computed from A092317) from M. F. Hasler, Jan 22 2017

a(17) corrected by Gerhard Kirchner, Jul 26 2020

A281355 a(n) = A092318(n) + 1: Number of terms of the odd harmonic series 1 + 1/3 + 1/5 + 1/7 + ... required to reach a sum >= n.

Original entry on oeis.org

1, 8, 57, 419, 3092, 22846, 168804, 1247298, 9216354, 68100151, 503195829, 3718142208, 27473561358, 203003686106, 1500005624924, 11083625711271, 81897532160125, 605145459495141, 4471453748222757, 33039822589391676, 244133102611731231, 1803913190804074904, 13329215764452299411

Offset: 1

N. J. A. Sloane, Jan 22 2017, following a suggestion from Jerry Polfer

nonn

A different way of listing the sums mentioned in A092318.

A092318 is the main entry for this problem.

Gerhard Kirchner, Table of n, a(n) for n = 1..1000 [replacing an older version by Vincenzo Librandi]
Jerry V Polfer, found a new sequence, not sure if interesting, and how to properly submit, Seqfan list, Jan. 22, 2017 (and follow-up messages).
Albert Säfström, n describes the expected number of loops created by tieing together two random loose ends of a(n) ropes until there is none left.

Cf. A002387, A092318.

a[n_]:=Floor[Exp[2*n-EulerGamma]/4+1]-Boole[n==1]; Array[a,23] (* Stefano Spezia, Jun 25 2024 *)

Lim_{n -> oo} a(n)/exp(2*n) = 1/4e^gamma ~ 0.140364870891721292456...;

a(n) = floor(exp(2*n-gamma)/4+1), for all given values a(n) > 1. - M. F. Hasler and Robert G. Wilson v, Jan 23 2017 [corrected by Gerhard Kirchner, Jul 25 2020]

More explicit, self-contained definition by M. F. Hasler, Jan 22 2017
More terms (computed using A056053) from M. F. Hasler, Jan 23 2017
a(17) corrected in data and 127 terms in the b-file, according to the corrections in A092315, Gerhard Kirchner, Jul 27 2020

A092317 a(n) = smallest odd number 2m+1 such that the partial sum Sum_{j=0..m} 1/(2j+1) of the odd harmonic series is >= n.

Original entry on oeis.org

1, 15, 113, 837, 6183, 45691, 337607, 2494595, 18432707, 136200301, 1006391657, 7436284415, 54947122715, 406007372211, 3000011249847, 22167251422541, 163795064320249, 1210290918990281, 8942907496445513, 66079645178783351

Offset: 1

N. J. A. Sloane, Feb 16 2004

nonn

Except for first term, same as A056053. Equals 2*A092318 + 1. Cf. A074599, A025547

a(n) ~ C*exp(2n) with C = 0.2807297417834425... - M. F. Hasler, Jan 22 2017

More terms (via A056053) from M. F. Hasler, Jan 22 2017

a(17) corrected - see correction in A092315. Gerhard Kirchner, Jul 25 2020

A289183 a(n) is the greatest m such that 2*H(n) > H(m), where H(n) is the n-th harmonic number.

Original entry on oeis.org

3, 10, 21, 35, 53, 74, 99, 128, 160, 196, 235, 277, 324, 374, 427, 484, 545, 609, 676, 748, 822, 901, 983, 1068, 1157, 1250, 1346, 1446, 1549, 1656, 1766, 1880, 1998, 2119, 2244, 2372, 2504, 2639, 2778, 2921, 3067, 3216, 3369, 3526, 3686, 3850, 4018, 4189

Offset: 1

Joseph Wheat, Jun 27 2017

nonn

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Harmonic Number

Cf. A000217, A002387, A002805, A092315.

s = HarmonicNumber@ Range[10^4]; Table[Position[s, k_ /; k < 2 HarmonicNumber@ n][[-1, 1]], {n, 48}] (* Michael De Vlieger, Jun 27 2017 *)
(* The following program searches for such n that f(n) <> a(n) *)
f[n_] := Floor[n^2*E^(EulerGamma + 1/n) - (1/2 + (1/6)*E^(EulerGamma))];
harmonic[n_] := Log[n] + EulerGamma + 1/(2 n) - Sum[BernoulliB[2 k]/(2 k*n^(2 k)), {k, 1, 10}];
Select[Range[100000], 2*harmonic[#] < harmonic[f[#]] &]
(* Vaclav Kotesovec, Jul 17 2017 *)

a(n) = {my(m=1); hn = sum(k=1, n, 1/k); hm = 1; until(hm > 2*hn, m++; hm+=1/m); m--;} \\ Michel Marcus, Jul 19 2017

from sympy import harmonic
def a(n):
  hn2 = 2 * harmonic(n)
  m = n
  while harmonic(m) <= hn2: m += 1
  return m - 1
print([a(n) for n in range(1, 49)]) # Michael S. Branicky, Mar 10 2021

From Jon E. Schoenfield, Jul 13 2017: (Start)
It seems that, for the vast majority of values of n > 1, f(n) = floor(n^2 * exp(gamma + 1/n) - C), where gamma is the Euler-Mascheroni constant (A001620) and C = 1/2 + (1/6)*exp(gamma) = 0.7968454029983663308727506838511965915282742..., is equal to a(n); f(n) = a(n) for all n in [2..10000] except n=66: f(66)=7876, but a(66)=7875. [Thanks to Vaclav Kotesovec for identifying the value of C.]
Is there any n > 66 at which f(n) and a(n) differ?
(End)
From Vaclav Kotesovec, Jul 17 2017: (Start)
f(39087) = 2721180603, but a(39087) = 2721180602;
f(517345) = 476697560917, but a(517345) = 476697560916;
f(2013005) = 7217245877275, but a(2013005) = 7217245877274;
No other such numbers below 10000000.
(End)
After 2013005, the only other numbers n < 4*10^9 at which f(n) and a(n) differ are 10240491 and 80968833. - Jon E. Schoenfield, Aug 05 2017

More terms from Michael De Vlieger, Jun 27 2017

A092267 Values 2m_0+1 = 1, 2m_1, 2m_2+1, ... associated with divergent series T shown below.

Original entry on oeis.org

1, 454, 45891, 547208496, 3013267310449, 1961694770407970734, 589785633779065944213245, 20963601300674244910397534828794, 344117353602393170461608383214200982125

Offset: 0

N. J. A. Sloane, Feb 16 2004

T = 1
- (1/2 + 1/4 + 1/6 + ... + 1/(2m_1))
+ (1/3 + 1/5 + 1/7 + ... + 1/(2m_2+1))
- (1/(2m_1+2) + 1/(2m_1+4) + ... + 1/(2m_3)
+ (1/(2m_2+3) + 1/(2m_2+5) + ... + 1/(2m_4+1))
- (1/(2m_3+2) + 1/(2m_3+4) + ... + 1/(2m_5)
+ (1/(2m_4+3) + 1/(2m_4+5) + ... + 1/(2m_6+1))
- ...
where the partial sums of the terms from 1 through the end of rows 0, 1, ... are respectively 1, just < -2, just > 3, just < -4, just > 5, etc.
Every positive number appears exactly once as a denominator in T.
The series T is a divergent rearrangement of the conditionally convergent series Sum_{ j>=1} (-1)^j/j which has the entire real number system as its set of limit points.

			1 - (1/2 + 1/4 + 1/6 + ... + 1/454) = -2.002183354..., which is just less than -2; so a(1) = 2m_1 = 454.
1 - (1/2 + 1/4 + 1/6 + ... + 1/454) + (1/3 + 1/5 + ... + 1/45891) = 3.000021113057..., which is just greater than 3; so a(1) = 2m_2 + 1 = 45891.

B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Holden-Day, San Francisco, 1964; see p. 55.

Cf. A092324 (essentially the same), A002387, A056053, A092318, A092317, A092315.

Cf. A092273.

a(2) and a(3) from Hugo Pfoertner, Feb 17 2004

a(4) onwards from Hans Havermann, Feb 18 2004

A092324 Values m_0 = 0, m_1, m_2, ... associated with divergent series T shown below.

Original entry on oeis.org

0, 227, 22945, 273604248, 1506633655224, 980847385203985367, 294892816889532972106622, 10481800650337122455198767414397, 172058676801196585230804191607100491062

Offset: 0

N. J. A. Sloane, Feb 16 2004

T = 1
- (1/2 + 1/4 + 1/6 + ... + 1/(2m_1))
+ (1/3 + 1/5 + 1/7 + ... + 1/(2m_2+1))
- (1/(2m_1+2) + 1/(2m_1+4) + ... + 1/(2m_3)
+ (1/(2m_2+3) + 1/(2m_2+5) + ... + 1/(2m_4+1))
- (1/(2m_3+2) + 1/(2m_3+4) + ... + 1/(2m_5)
+ (1/(2m_4+3) + 1/(2m_4+5) + ... + 1/(2m_6+1))
- ...
where the partial sums of the terms from 1 through the end of rows 0, 1, ... are respectively 1, just < -2, just > 3, just < -4, just > 5, etc.
Every positive number appears exactly once as a denominator in T.
The series T is a divergent rearrangement of the conditionally convergent series Sum_{j>=1} (-1)^j/j which has the entire real number system as its set of limit points.
Comment from Hans Havermann: I calculated these with Mathematica. I used NSum[1/(2i), {i, 1, x}] for the even denominators, where I had to adjust the options to obtain maximal accuracy and N[(EulerGamma + Log[4] - 2)/2 + PolyGamma[0, 3/2 + y]/2, precision] for the odd denominators. The precision needed for the last term shown was around 45 digits.

			1 - (1/2 + 1/4 + 1/6 + ... + 1/454) = -2.002183354..., which is just less than -2; so a(1) = m_1 = 227.
1 - (1/2 + 1/4 + 1/6 + ... + 1/454) + (1/3 + 1/5 + ... + 1/45891) = 3.000021113057..., which is just greater than 3; so a(2) = m_2 = 22945.

B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Holden-Day, San Francisco, 1964; see p. 55.

Cf. A092267 (essentially the same), A002387, A056053, A092318, A092317, A092315.

Cf. A092273.

a(2) and a(3) from Hugo Pfoertner, Feb 17 2004

a(4) onwards from Hans Havermann, Feb 18 2004

A337748 T(m,n) is the least k such that the partial sum of the series Sum_{j=0..k} 1/(m*j+1) is > n, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 7, 10, 1, 17, 56, 30, 1, 43, 353, 418, 82, 1, 111, 2374, 7100, 3091, 226, 1, 289, 16497, 129644, 142624, 22845, 615, 1, 762, 116759, 2448436, 7078315, 2864677, 168803, 1673, 1, 2021, 835695, 47104189, 363380155, 386462913, 57538579, 1247297, 4549, 1, 5389, 6025478, 916451548, 19003186159, 53930396770, 21100160132, 1155693253, 9216353, 12366

Offset: 1

Gerhard Kirchner, Sep 18 2020

H(m,k) = Sum_{j=0..k} 1/(m*j+1) may be thought of as a generalized harmonic sequence: H(1,k) = H(k+1), where H(k) = Sum_{j=1..k} 1/j is the harmonic and H(2,k) the odd harmonic sequence. As a consequence, T(1,n) = A002387(n)-1 and T(2,n) = A092315(n). For m > 2, the sequence T(m,n) is not currently in the OEIS. Therefore, I give a detailed derivation of the formulas below, see link "Generalized harmonic series". The parameter c(m) in the formula for T(m,n): c(1) = gamma (Euler's constant), c(2) = gamma+1-log(2). For m > 2, c(m) has to be determined individually. The algorithm is described in the formula section though it is not only one formula.

In the Maxima code, two tests are implemented, see Formula. The first test will probably be passed without exception, the second one limits the terms. For greater T(m,n), the precision must be increased.

			T(m,n) begins:
m=1: 1   3    10      30        82 ...
m=2: 1   7    56     418      3091 ...
m=3: 1  17   353    7100    142624 ...
m=4: 1  43  2374  129644   7078315 ...
m=5: 1 111 16497 2448436 363380155 ...
     .............................
Evaluation of c(3):
1. S(3,99) = 0.1472791427382, see formulas (P1) and (P2).
2. F(3,100) = 10^(-2)/9 + 10^(-4)/27 + 10^(-6)/243 - 10^(-8)/243 ..., see (P5) = 0.001114818889, see formulas (P3) and (P4).
The next summand is 10^(-10)/729 ~ 10^(-13), the precision is approximately  p = 13 digits.
3. beta(3) = 0.1483939616270.
4. c(3) = 3.1320337800204.
Evaluation of T(3,5):
5. Summing up directly with k = 142624: H(3,k-1) = 4.9999999, H(3,k) = 5.0000022. => T(3,5) = 142624.
6. Using the formula with k1 = exp(3*5-c(3))-5/6 = 142623.0446139:
   T(3,5) = floor(k1+1) = 142624.
7,1. floor(k1) = floor(k2) is satisfied with k2 = k1-1/(24*k1) = 142623.0446136.
7,2. With d = 0.0446 and p = 13, k1 < d*10^(-13) is satisfied, too.

Gerhard Kirchner, Table of n, a(n) for n = 1..496, 31 antidiagonals
Gerhard Kirchner, Generalized harmonic sequence

Cf. A002387, A092315.

block(p:100, km: 100, eps:10^(-p), lim:10^(p), cc:[],
/*"harm-bfile.txt" with terms < lim is saved*/
gam: bfloat(%gamma),
fpprec:p, load(newton1),
fl: openw("harm-bfile.txt"),
c(m):= block(x:0, delta: 1, r:0,
if m=1 then return(gam) else
  (for k from 1 thru km-1 do x: x+bfloat(1/(m*k*(m*k+1))),
   while delta=0 or delta >eps do
      (r: r+1, su:0,
      for j from 1 thru r-1 do su: su+a[r-j]*(-1)^j*binomial(r,j+1),
      ar: bfloat(-((-1)^r/m^(r+1)+su)/r),  ad: ar/km^r, a: append(a,[ar]),
      x: x+bfloat(ad), delta: abs(ad)), return(gam+m*(1-x)))),
mn:0, ok: true,  nr:0,
while ok do (mn: mn+1, a:[], cc: append(cc, [c(mn)]), m:mn+1,
   while m>1 and ok do  (m: m-1, n: mn+1-m,
   k1: exp(m*n-cc[m])-(m+2)/(2*m), k2: k1-1/(24*k1),
   ka: floor(k1+1), kb: floor(k2+1),
   if ka>1 then
     (d: kb-k2, if d>1-d then d: 1-d, if ka#kb or kb> d*lim then  ok: false),
  if ok then (nr: nr+1,  printf(fl, concat(nr, " ", ka)), newline(fl) ) ) ),
close(fl));

T(m,n) = floor(k1+1) with k1 = exp(m*n-c(m))-(m+2)/(2*m).
For avoiding bad terms, k1 has to pass two tests:
Test 1: floor(k1) = floor(k2) with k2 = k1-1/(24*k1).
Test 2: k1 < d*10^p, where p is the precision of c(m) (number of digits) and d the distance between k1 and the next integer floor(k1) or floor(k1)+1.
The parameter c(m):
c(m) = gamma+m*(1-beta(m)) with (P1) beta(m) = lim_{k->oo} S(m,k) and S(m,k) = Sum_{j=1..k}(1/(m*j)-1/(m*j+1)).
For faster convergence, the infinite sum must be split up: (P2) beta(m) = S(m,k-1) + F(m,k).
F(m,k) can be written as:
   (P3) F(m,k) = Sum_{r=1..s-1} b(r)/k^r + R(s) with R(s) ~ b(s)/k^s.
Example: For the precision p = 100 digits, k = 100 is large enough to find s with b(s)/k^s < 10^(-p) where s >= 73.
The coefficients b(r) are defined by the recurrence b(1) = 1/m^2 and, for r>1: (P4) b(r) = ((-1/m)^(r+1)-Sum_{j=1..r-1} (-1)^j*b(r-j)*binomial(r,j+1))/r.
              1            m-1           m^2-3*m+2            m^2-2*m+1
(P5) b(1) = -----, b(2) = -----, b(3) = -----------,  b(4) = -----------,
             m^2          2*m^3            6*m^4                4*m^5

A338857 With S(n,k) = Sum_{n<=j<=k} 1/(2*j+1), a(n)=k+1 such that S(n,k-1) < 1 <= S(n,k) for n>=0 and a(0)=1.

Original entry on oeis.org

1, 8, 15, 23, 30, 38, 45, 52, 60, 67, 74, 82, 89, 97, 104, 111, 119, 126, 134, 141, 148, 156, 163, 170, 178, 185, 193, 200, 207, 215, 222, 230, 237, 244, 252, 259, 267, 274, 281, 289, 296, 303, 311, 318, 326, 333, 340, 348, 355, 363, 370, 377, 385, 392, 400, 407, 414

Offset: 0

Gerhard Kirchner, Nov 12 2020

nonn

The following version of the well-known "camel and banana (dates)" problem, is an application of the sequence above:
A camel is to bring full water bags from oasis A to oasis B. It can carry the driver, one full and one empty bag. A full bag is just enough to supply the camel with water for one way from A to B.  What is the minimum reserve a(n) of full bags at oasis A if n full bags are to be delivered at B and depots may be installed along the way?
For details, see link "Transport problem".
n=0: The camel carries one bag which is full in A and empty in B.

			n=0: S(0,k-1)=1 for k=1.
  Thus a(0) = 1+0 = 1.
n=1: S(1,6)=1/3+1/5+...+1/11+1/13=0.995<1, S(1,7)=S(1,6)+1/15=1.022>1.
  Thus a(1) = 7+1 = 8.
n=2: S(2,13)=1/5+1/7+...+1/25+1/27=0.968<1, S(2,14)=S(2,13)+1/29=1.003>1.
  Thus a(2) = 14+1 = 15.

Gerhard Kirchner, Transport problem

Cf. A056053, A092315, A092318, A281355.

Block[{S}, S[n_, k_] := Sum[1/(2 j + 1), {j, n, k}]; {1}~Join~Array[Block[{k = 1}, While[Nand[S[#, k - 1] < 1 <= S[#, k]], k++]; k + 1] &, 56]] (* Michael De Vlieger, Nov 12 2020 *)

block(su: 0, k: 0, n: 1, nmax: 80,
    /*program returns the first nmax terms*/
    v: makelist(0, i, 0, nmax), v[1]: 1,
    while n<=nmax do
    (k: k+1, su: su+1/(2*k+1),
    if su>1 then
     (v[n+1]: k+1, su: su-1/(2*n+1), n: n+1)),
  return(v));

Conjecture: a(n) = ceiling(n*exp(2)+(exp(2)+exp(-2))/(24*n)), verified for n<=3000.

cp's OEIS Frontend

A056053 a(n) = smallest odd number 2m+1 such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

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A092318 a(n) = smallest m such that value of odd harmonic series Sum_{j=0..m} 1/(2j+1) is >= n.

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A281355 a(n) = A092318(n) + 1: Number of terms of the odd harmonic series 1 + 1/3 + 1/5 + 1/7 + ... required to reach a sum >= n.

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A092317 a(n) = smallest odd number 2m+1 such that the partial sum Sum_{j=0..m} 1/(2j+1) of the odd harmonic series is >= n.

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A289183 a(n) is the greatest m such that 2*H(n) > H(m), where H(n) is the n-th harmonic number.

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A092267 Values 2m_0+1 = 1, 2m_1, 2m_2+1, ... associated with divergent series T shown below.

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A092324 Values m_0 = 0, m_1, m_2, ... associated with divergent series T shown below.

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A337748 T(m,n) is the least k such that the partial sum of the series Sum_{j=0..k} 1/(m*j+1) is > n, read by ascending antidiagonals.

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