A202537 -id:A202537 - cp's OEIS frontend

A202543 Decimal expansion of the number x satisfying e^(x/2) - e^(-x/2) = 1.

9, 6, 2, 4, 2, 3, 6, 5, 0, 1, 1, 9, 2, 0, 6, 8, 9, 4, 9, 9, 5, 5, 1, 7, 8, 2, 6, 8, 4, 8, 7, 3, 6, 8, 4, 6, 2, 7, 0, 3, 6, 8, 6, 6, 8, 7, 7, 1, 3, 2, 1, 0, 3, 9, 3, 2, 2, 0, 3, 6, 3, 3, 7, 6, 8, 0, 3, 2, 7, 7, 3, 5, 2, 1, 6, 4, 4, 3, 5, 4, 8, 8, 2, 4, 0, 1, 8, 8, 5, 8, 2, 4, 5, 4, 4, 6, 9, 4, 9

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.
W. Gawronski et al. in their paper - see ref. below - obtained the asymptotics for the Chebyshev-Stirling numbers. In the algebraic description of the respective "asymptotic coefficients" the number x = 2*log phi, where phi is the golden section, play the central role. - Roman Witula, Feb 02 2015
Also two times the Lévy measure for the continued fraction of the golden section, i.e., A202543/log(2) is the mean number of bits gained from the next convergent of the continued fraction representation. (See also Dan Lascu in links.) - A.H.M. Smeets, Jun 06 2018

			0.9624236501192068949955178268487368462703686...

W. Gawronski, L. L. Littlejohn, and T. Neuschel, Asymptotics of Stirling and Chebyshev-Stirling numbers of the second kind, arXiv:1308.6803 [math.CO], 2013.
W. Gawronski, L. L. Littlejohn, and T. Neuschel, Asymptotics of Stirling and Chebyshev-Stirling numbers of the second kind, Studies in Applied Mathematics by MIT 133 (2014), 1-17.
Dan Lascu, A Gauss-Kuzmintype problem for a family of continued fraction expansions, Journal of Number Theory 133 (2013), 2153-2181.
Index entries for transcendental numbers

Cf. A001622, A002390, A104457, A202537, A202543.

u = 1/2; v = 1/2;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, 0, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .9, 1}, WorkingPrecision -> 110]
RealDigits[r]    (* A202543 *)
RealDigits[ Log[ (3+Sqrt[5])/2], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)
RealDigits[ FindRoot[ Exp[x/2] == 1 +  Exp[-x/2] , {x, 0}, WorkingPrecision -> 128][[1, 2]]][[1]] (* Robert G. Wilson v, Jun 13 2018 *)

2*asinh(1/2) \\ Michel Marcus, Jun 24 2018, after A002390

Equals 2*A002390. - A.H.M. Smeets, Jun 06 2018
From Amiram Eldar, Aug 21 2020: (Start)
Equals log(A104457) = log(1 + A001622).
Equals 2*arcsinh(1/2). [corrected by Georg Fischer, Jul 12 2021]
Equals Sum_{k>=0} (-1)^k*binomial(2*k,k)/((2*k+1)*16^k). (End)
Equals Pi*i + Sum_{k>=0} arctanh(phi^(2^k))/2^k, with phi = A001622 and i = sqrt(-1). - Antonio Graciá Llorente, Feb 13 2025

Typo in name fixed by Jean-François Alcover, Feb 27 2013

A224868 a(1) = greatest k such that H(k) - H(4) < 1/3 + 1/4; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(4); and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

7, 11, 17, 26, 39, 58, 86, 127, 187, 275, 404, 593, 870, 1276, 1871, 2743, 4021, 5894, 8639, 12662, 18558, 27199, 39863, 58423, 85624, 125489, 183914, 269540, 395031, 578947, 848489, 1243522, 1822471, 2670962, 3914486, 5736959, 8407923, 12322411, 18059372

Offset: 1

Clark Kimberling, Jul 23 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A224868, (x,y) = (3,4); it appears that the sequence a(n) is linearly recurrent with signature (2,-1,1,-1). Possibly the constant at A202537 is the limit of the sequences H(c(n))-H(c(n-1)). Possibly the constant at A092526 is the limit of c(n)/c(n-1).

			The first three values (a(1),a(2),a(3)) = (7,11,17) match the beginning of the following inequality chain (and partition of {1/m: m>=3}):
1/3+1/4 > 1/5+1/6+1/7 > 1/8+1/9+1/10+1/11 > 1/12+ ... +1/17 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A224820.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 4; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A224868 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 5]  (* A202537? *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 5]  (* A092526? *)
(* Peter J. C. Moses, Jul 23 2013 *)

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) (conjectured).

G.f.: (7 - 3 x + 2 x^2 - 4 x^3)/(1 - 2 x + x^2 - x^3 + x^4) (conjectured).

A202541 Decimal expansion of the number x satisfying e^(2x) - e^(-2x) = 1.

Original entry on oeis.org

2, 4, 0, 6, 0, 5, 9, 1, 2, 5, 2, 9, 8, 0, 1, 7, 2, 3, 7, 4, 8, 8, 7, 9, 4, 5, 6, 7, 1, 2, 1, 8, 4, 2, 1, 1, 5, 6, 7, 5, 9, 2, 1, 6, 7, 1, 9, 2, 8, 3, 0, 2, 5, 9, 8, 3, 0, 5, 0, 9, 0, 8, 4, 4, 2, 0, 0, 8, 1, 9, 3, 3, 8, 0, 4, 1, 1, 0, 8, 8, 7, 2, 0, 6, 0, 0, 4, 7, 1, 4, 5, 6, 1, 3, 6, 1, 7, 3, 7

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.

Archimedes's-like scheme: set p(0) = 1/(2*sqrt(5)), q(0) = 1/4; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (arithmetic mean of reciprocals, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - A.H.M. Smeets, Jul 12 2018

			0.24060591252980172374887945671218421156759216719...

R. S. Melham and A. G. Shannon, Inverse trigonometric and hyperbolic summation formulas involving generalized Fibonacci numbers, Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
D. Zagier, Algebraic numbers close to both 0 and 1, Mathematics of Computation, Vol. 61, No. 203 (1993), pp. 485-491.
Index entries for transcendental numbers

Cf. A001622, A002390, A202537.

u = 2; v = 2;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -2, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .2, .3}, WorkingPrecision -> 110]
RealDigits[r]   (* A202541 *)
RealDigits[ Log[ (1+Sqrt[5])/2 ] / 2, 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)
RealDigits[ FindRoot[ Exp[2x] - Exp[-2x] == 1, {x, 1}, WorkingPrecision -> 128][[1, 2]], 10, 111][[1]] (* Robert G. Wilson v, Jul 23 2018 *)

asinh(1/2)/2 \\ Michel Marcus, Jul 12 2018

Equals (1/2)*arcsinh(1/2) or (1/2)*log(phi), phi being the golden ratio. - A.H.M. Smeets, Jul 12 2018
Equals Sum_{k>=1} (-1)^(k+1) * arctanh(1/Fibonacci(3*k)^2) (Melham and Shannon, 1995). - Amiram Eldar, Oct 04 2021
Equals A002390/2. - Alois P. Heinz, Jul 14 2022
Equals arctanh(sqrt(5)-2). - Amiram Eldar, Feb 09 2024

A202538 Decimal expansion of the number x satisfying e^x-e^(-3x)=1.

Original entry on oeis.org

3, 2, 2, 2, 8, 4, 6, 1, 5, 9, 7, 1, 0, 3, 0, 0, 6, 0, 0, 3, 6, 2, 3, 5, 4, 8, 6, 2, 8, 9, 1, 3, 9, 2, 3, 5, 4, 5, 5, 4, 4, 3, 1, 1, 4, 8, 0, 7, 4, 6, 3, 8, 6, 8, 3, 0, 3, 7, 2, 4, 5, 0, 6, 7, 0, 1, 4, 1, 5, 2, 2, 6, 1, 2, 9, 4, 3, 3, 8, 1, 6, 4, 6, 7, 8, 0, 0, 8, 9, 8, 7, 3, 2, 7, 2, 1, 6, 4, 6

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.

			0.32228461597103006003623548628913923545544311...

Index entries for transcendental numbers

Cf. A202537.

u = 1; v = 3;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -2, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .3, .4}, WorkingPrecision -> 110]
RealDigits[r]  (* A202538 *)
RealDigits[ Log[ Root[#^4 - #^3 - 1&, 2]], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)

log(polrootsreal(x^4-x^3-1)[2]) \\ Charles R Greathouse IV, May 15 2019

A202540 Decimal expansion of the number x satisfying e^(3x)-e^(-x)=1.

Original entry on oeis.org

1, 9, 9, 4, 6, 0, 5, 7, 8, 2, 4, 3, 0, 0, 5, 3, 5, 1, 4, 8, 8, 5, 7, 7, 7, 1, 8, 3, 8, 4, 9, 4, 9, 1, 7, 8, 3, 9, 2, 7, 7, 6, 9, 2, 6, 2, 0, 8, 1, 2, 4, 9, 2, 4, 0, 1, 5, 3, 6, 4, 5, 4, 7, 1, 6, 8, 0, 8, 6, 6, 4, 3, 9, 3, 8, 4, 3, 2, 8, 5, 4, 8, 7, 9, 2, 7, 9, 9, 8, 0, 3, 6, 1, 6, 3, 6, 4, 6, 4

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.

			0.19946057824300535148857771838494917839277692...

Index entries for transcendental numbers

Cf. A202537.

u = 3; v = 1;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -2, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .1, .2}, WorkingPrecision -> 110]
RealDigits[r]     (* A202540 *)
RealDigits[ Log[ Root[#^4 - # - 1&, 2]], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)

log(polrootsreal(x^4-x-1)[2]) \\ Charles R Greathouse IV, May 14 2019

A202539 Decimal expansion of the number x satisfying e^(2x)-e^(-x)=1.

Original entry on oeis.org

2, 8, 1, 1, 9, 9, 5, 7, 4, 3, 2, 2, 9, 6, 1, 8, 4, 6, 5, 1, 2, 0, 5, 0, 7, 6, 4, 0, 6, 7, 8, 7, 8, 2, 9, 9, 7, 9, 2, 0, 2, 3, 2, 2, 5, 7, 4, 4, 0, 6, 6, 4, 6, 2, 6, 7, 5, 7, 3, 0, 3, 3, 4, 3, 1, 8, 0, 3, 8, 4, 5, 3, 0, 6, 2, 1, 2, 0, 8, 9, 1, 3, 2, 2, 9, 8, 7, 7, 0, 7, 4, 7, 5, 4, 9, 4, 0, 5, 4

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.

			0.281199574322961846512050764067878299792023...

Index entries for transcendental numbers

Cf. A202537.

u = 2; v = 1;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -2, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .2, .3}, WorkingPrecision -> 110]
RealDigits[r]  (* A202539 *)
RealDigits[ Log[ Root[#^3 - # - 1&, 1]], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)

log(polrootsreal(x^3-x-1)[1]) \\ Charles R Greathouse IV, May 15 2019

Equals log((v^2+12)/(6*v)) with v = (108+12*sqrt(69))^(1/3). - Alois P. Heinz, Jul 14 2022

A202542 Decimal expansion of the number x satisfying e^(3x)-e^(-3x)=1.

Original entry on oeis.org

1, 6, 0, 4, 0, 3, 9, 4, 1, 6, 8, 6, 5, 3, 4, 4, 8, 2, 4, 9, 9, 2, 5, 2, 9, 7, 1, 1, 4, 1, 4, 5, 6, 1, 4, 1, 0, 4, 5, 0, 6, 1, 4, 4, 4, 7, 9, 5, 2, 2, 0, 1, 7, 3, 2, 2, 0, 3, 3, 9, 3, 8, 9, 6, 1, 3, 3, 8, 7, 9, 5, 5, 8, 6, 9, 4, 0, 7, 2, 5, 8, 1, 3, 7, 3, 3, 6, 4, 7, 6, 3, 7, 4, 2, 4, 1, 1, 5, 8

Offset: 0

Clark Kimberling, Dec 21 2011

See A202537 for a guide to related sequences. The Mathematica program includes a graph.

			x=0.1604039416865344824992529711414561410450614447...

Index entries for transcendental numbers

Cf. A202537.

u = 3; v = 3;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -1, 1}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .2, .3}, WorkingPrecision -> 110]
RealDigits[r]    (* A202542 *)
RealDigits[ Log[ (1+Sqrt[5])/2 ] / 3, 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)

log((sqrt(5)+1)/2)/3 \\ Charles R Greathouse IV, May 14 2019

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A202543 Decimal expansion of the number x satisfying e^(x/2) - e^(-x/2) = 1.

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A224868 a(1) = greatest k such that H(k) - H(4) < 1/3 + 1/4; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(4); and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

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A202541 Decimal expansion of the number x satisfying e^(2x) - e^(-2x) = 1.

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A202538 Decimal expansion of the number x satisfying e^x-e^(-3x)=1.

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Mathematica

PARI

A202540 Decimal expansion of the number x satisfying e^(3x)-e^(-x)=1.

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Author

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Comments

Examples

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Programs

Mathematica

PARI

A202539 Decimal expansion of the number x satisfying e^(2x)-e^(-x)=1.

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Author

Keywords

Comments

Examples

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Programs

Mathematica

PARI

Formula

A202542 Decimal expansion of the number x satisfying e^(3x)-e^(-3x)=1.

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