cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Views

Author

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

Keywords

Comments

A different way of listing the sums mentioned in A092318.
A092318 is the main entry for this problem.

Links

Crossrefs

Cf. A002387, A092318.

Programs

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

Formula

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]

Extensions

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

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.

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

References

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

Links

Crossrefs

Cf. A002387, A056054, A091463, A091464, A091465.
Cf. also A092318, A092317, A092315, A281355, A074599, A025547.

Programs

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

Formula

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

Extensions

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

A092315 a(n) is the smallest m such that the partial sum of the odd harmonic series Sum_{j=0..m} 1/(2j+1) is > n.

Original entry on oeis.org

1, 7, 56, 418, 3091, 22845, 168803, 1247297, 9216353, 68100150, 503195828, 3718142207, 27473561357, 203003686105, 1500005624923, 11083625711270, 81897532160124, 605145459495140, 4471453748222756, 33039822589391675, 244133102611731230, 1803913190804074903
Offset: 1

Views

Author

N. J. A. Sloane, Feb 16 2004

Keywords

Comments

From Gerhard Kirchner, May 21 2020: (Start)
The terms a(n), evaluated by the formula, should pass the test OH(a(n))=n and OH(a(n)-1)=n-1, where OH(m) is the odd harmonic series, see above.
Another formula, see link Asymptotic formulas, formula 1, is OH(m) = (log(4*m)+gamma)/2+1/(2*m)-11/(48*m^2)+1/(8*m^3)-127*t/(1920*m^4), 0
The Maxima code includes both tests and creates a b-file in the current directory. For n<=1000, the case "Precision too low" does not occur. (End)
a(2) = 7 and a(3) = 56 are related to the Borwein integrals. Concretely, a(2) = 7 is the smallest m such that the integral Integral_{x=-oo..oo} Product_{k=0..m} (sin((2*k+1)*x)/((2*k+1)*x)) dx is slightly less than Pi, and a(3) = 56 is the smallest m such that the integral Integral_{x=-oo..oo} cos(x) * Product_{k=0..m} (sin((2*k+1)*x)/((2*k+1)*x)) dx is slightly less than Pi/2. See the Wikipedia link and the 3Blue1Brown video link below. - Jianing Song, Dec 10 2022

Links

Crossrefs

Except for first term, same as A092318. Equals (A056053-1)/2.
Cf. A074599, A025547, A281355.

Programs

  • Mathematica
    A092315[n_] := Floor[Exp[2*n - EulerGamma]/4]; Table[A092315[n], {n, 1, 22}] (* Robert P. P. McKone, Jul 13 2021 *)
  • Maxima
    block(
fpprec:1000, gam: %gamma, nmax:1000,
fl: openw("bfile1000.txt"),
OH(k,t):=(log(4*k)+gam)/2+1/(2*k)-11/(48*k^2)+1/(8*k^3)-127*t/(1920*k^4),
printf(fl, "1 1"),   newline(fl),
for n from 2 thru nmax do
(u: bfloat(exp(2*n-gam)/4), k: floor(u),
x0: bfloat(OH(k,0)), x01: bfloat(OH(k,1)), x1: bfloat(OH(k-1,0)),
n0: floor(x0), n01: floor(x01), n1: floor(x1),  m: n,
if n0=n and n01=n and n1=n-1 then
         (h: concat(n, " ", k), printf(fl, h),  newline(fl)) else n: nmax),
if mGerhard Kirchner, Jul 23 2020 */
/* The first nmax terms are saved as a b-file */

Formula

a(n) = floor(exp(2*n-gamma)/4+1/8) for all n >= 1 (conjectured; see also comments in A002387). - M. F. Hasler, Jan 22 2017
a(n) = floor(exp(2*n-gamma)/4). - Gerhard Kirchner, Jul 23 2020

Extensions

More terms from M. F. Hasler, Jan 24 2017
a(17) in the data section and 127 terms in the b-file corrected by Gerhard Kirchner, Jul 23 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

Views

Author

N. J. A. Sloane, Feb 16 2004

Keywords

Crossrefs

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

Formula

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

Extensions

More terms (via A056053) from M. F. Hasler, Jan 22 2017
a(17) corrected - see correction in A092315. Gerhard Kirchner, Jul 25 2020

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

Views

Author

N. J. A. Sloane, Feb 16 2004

Keywords

Comments

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.

Examples

			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.

References

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

Crossrefs

Cf. A092324 (essentially the same), A002387, A056053, A092318, A092317, A092315.
Cf. A092273.

Extensions

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

Views

Author

N. J. A. Sloane, Feb 16 2004

Keywords

Comments

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.

Examples

			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.

References

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

Crossrefs

Cf. A092267 (essentially the same), A002387, A056053, A092318, A092317, A092315.
Cf. A092273.

Extensions

a(2) and a(3) from Hugo Pfoertner, Feb 17 2004
a(4) onwards from Hans Havermann, Feb 18 2004

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

Views

Author

Gerhard Kirchner, Nov 12 2020

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

Cf. A056053, A092315, A092318, A281355.

Programs

  • Mathematica
    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 *)
  • Maxima
    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));

Formula

Conjecture: a(n) = ceiling(n*exp(2)+(exp(2)+exp(-2))/(24*n)), verified for n<=3000.
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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