A244644 -id:A244644 - cp's OEIS frontend

A093954 Decimal expansion of Pi/(2*sqrt(2)).

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

Offset: 1

Eric W. Weisstein, Apr 19 2004

The value is the length Pi*sqrt(2)/4 of the diagonal in the square with side length Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) = A003881. The area of the circumcircle of this square is Pi*(Pi*sqrt(2)/8)^2 = Pi^3/32 = A153071. - Eric Desbiaux, Jan 18 2009
This is the value of the Dirichlet L-function of modulus m=8 at argument s=1 for the non-principal character (1,0,1,0,-1,0,-1,0). See arXiv:1008.2547. - R. J. Mathar, Mar 22 2011
Archimedes's-like scheme: set p(0) = sqrt(2), q(0) = 1; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, 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
The area of a circle circumscribing a unit-area regular octagon. - Amiram Eldar, Nov 05 2020

			1.11072073453959156175397...
From _Peter Bala_, Mar 03 2015: (Start)
Asymptotic expansion at n = 5000.
The truncated series Sum_{k = 0..5000 - 1} (-1)^floor(k/2)/(2*k + 1) = 1.110(6)207345(42)591561(18)3970(5238)1.... The bracketed digits show where this decimal expansion differs from that of Pi/(2*sqrt(2)). The numbers 1, -3, 57, -2763 must be added to the bracketed numbers to give the correct decimal expansion to 30 digits: Pi/(2*sqrt(2)) = 1.110(7)207345(39)591561(75)3970 (2475)1.... (End)
From _Peter Bala_, Nov 24 2016: (Start)
Case m = 1, n = 1:
Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
We appear to have the following asymptotic expansion for the tails of this series: for N divisible by 4, Sum_{k >= N/2} (-1)^floor(k/2)/((2*k - 1)*(2*k + 1)*(2*k + 3)) ~ 1/N^3 - 14/N^5 + 691/N^7 - 62684/N^9 - ..., where the coefficient sequence [1, 0, -14, 0, 691, 0, -62684, ...] appears to come from the e.g.f. (1/2!)*cosh(x)/cosh(2*x)*sinh(x)^2 = x^2/2! - 14*x^4/4! + 691*x^6/6! - 62684*x^8/8! + .... Cf. A019670.
For example, take N = 10^5. The truncated series Sum_{k = 0..N/2 -1} (-1)^(1+floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.27768018363489(8)89043849(11)61878(80026)6163(351171)58.... The bracketed digits show where this decimal expansion differs from that of (1/4)*Pi/(2*sqrt(2)). The numbers -1, 14, -691, 62684 must be added to the bracketed numbers to give the correct decimal expansion: (1/4)*Pi/(2*sqrt(2)) = 0.27768018363489(7) 89043849(25)61878(79335)6163(413855)58... (End)

J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press, 2006, p. 149.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4.1, p. 20.
L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 76, page 16.
Joel L. Schiff, The Laplace Transform: Theory and Applications, Springer-Verlag New York, Inc. (1999). See p. 149.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 53.

Harry J. Smith, Table of n, a(n) for n = 1..20000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
R. J. Mathar, Table of Dirichlet L-series and prime zeta modulo functions for small moduli, arXiv:1008.2547 [math.NT], 2010-2015, table 7 and section 2.2, value of L(m=8,r=4,s=1).
Michael Penn, Newton's sum (2023), YouTube video.
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 149.
Eric Weisstein's World of Mathematics, Bifoliate.
Index entries for transcendental numbers.

Cf. A161684 (continued fraction).
Cf. A002388, A019670.
Cf. A000281, A063448, A247719, A193887, A244976, A181048, A181049.
Cf. A003881, A251809, A188510.
Cf. A352324, A248897, A019669.

simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) );  # Peter Bala, Mar 09 2015

RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* Michael De Vlieger, Sep 23 2016 and slightly modified by Robert G. Wilson v, Jul 23 2018 *)

default(realprecision, 20080); x=Pi*sqrt(2)/4; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093954.txt", n, " ", d)); \\ Harry J. Smith, Jun 17 2009

Equals 1/A112628.
Equals Integral_{x=0..oo} 1/(x^4+1) dx. - Jean-François Alcover, Apr 29 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
The integer sequences A(n) := 2^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k ) both satisfy the second order recurrence equation u(n) = (12*n^2 + 1)*u(n-1) - 4*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/(2*sqrt(2)) = 1 + 1/(12 - 4*3^3/(49 - 4*2*5^3/(109 - 4*3*7^3/(193 - ... - 4*(n - 1)*(2*n - 1)^3/((12*n^2 + 1) - ... ))))). Cf. A002388 and A019670. (End)
From Peter Bala, Mar 03 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} (-1)^floor(k/2)/(2*k + 1) = limit (n -> infinity) Sum_{k = -n .. n - 1} (-1)^k/(4*k + 1). See Wells.
We conjecture the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + ..., where n is a multiple of 4 and the sequence of unsigned coefficients [1, 3, 57, 2763, ...] is A000281. An example with n = 5000 is given below. (End)
From Peter Bala, Sep 21 2016: (Start)
c = 2 * Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)) = A181048 + A181049. The asymptotic expansion conjectured above follows from the asymptotic expansions given in A181048 and A181049.
c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
From Peter Bala, Nov 24 2016: (Start)
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 2^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 2*m*j). Cf. A003881.
In the particular case m = 1 the result has the equivalent form: for n a nonnegative integer, Pi/(2*sqrt(2)) = 2^n*(2*n)!*Sum_{k >= 0} (-1)^(n+k)*(8*k + 4)* 1/Product_{j = -n..n+1} (4*k + 2*j + 1). The case m = 1, n = 1 is considered in the Example section below.
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 4^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 4*m*j). (End)
Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - Peter Bala, Nov 01 2019
Equals Sum_{k>=1} A188510(k)/k = Sum_{k>=1} Kronecker(-8,k)/k = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - 1/15 + ... - Jianing Song, Nov 16 2019
From Amiram Eldar, Jul 16 2020: (Start)
Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
Equals Integral_{x=0..oo} dx/(x^2 + 2).
Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - Bernard Schott, May 19 2022
Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - Peter Bala, Jul 22 2022
Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - Terry D. Grant, Mar 17 2023
Equals 1/Product_{p prime} (1 - Kronecker(-8,p)/p), where Kronecker(-8,p) = 0 if p = 2, 1 if p == 1 or 3 (mod 8) or -1 if p == 5 or 7 (mod 8). - Amiram Eldar, Dec 17 2023
Equals A068465*A068467. - R. J. Mathar, Jun 27 2024
From Stefano Spezia, Jun 05 2025: (Start)
Equals Sum_{k>=1} (-1)^(k+1)(1/(4*k - 3) + 1/(4*k - 1)).
Equals Product_{k=0..oo} (1 + (-1)^k/(2*k + 3)).
Equals Integral_{x=0..oo} 1/(2*x^2 + 1).
Equals Integral_{x=0..1} 1/((1 + x^2)*sqrt(1 - x^2)). (End)

A014549 Decimal expansion of 1 / M(1,sqrt(2)) (Gauss's constant).

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, N. J. A. Sloane

On May 30, 1799, Gauss discovered that this number is also equal to (2/Pi)*Integral_{t=0..1} 1/sqrt(1-t^4).

M(a,b) is the limit of the arithmetic-geometric mean iteration applied repeatedly starting with a and b: a_0 = a, b_0 = b, a_{n+1} = (a_n + b_n)/2, b_{n+1} = sqrt(a_n*b_n).

			0.8346268416740731862814297327990468...

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, page 5.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 1.5.4 and 6.1, pp. 34, 420.
J. R. Goldman, The Queen of Mathematics, 1998, p. 92.

Harry J. Smith, Table of n, a(n) for n = 0..20000
Markus Faulhuber, An application of hypergeometric functions to heat kernels on rectangular and hexagonal tori and a "Weltkonstante"-or-how Ramanujan split temperatures, The Ramanujan Journal volume 54, pages 1-27 (2021). See pp. 4 and 24.
Markus Faulhuber, Anupam Gumber, and Irina Shafkulovska, The AGM of Gauss, Ramanujan's corresponding theory, and spectral bounds of self-adjoint operators, arXiv:2209.04202 [math.CA], 2022, p. 2.
Alessandro Languasco and Pieter Moree, Euler constants from primes in arithmetic progression, arXiv:2406.16547 [math.NT], 2024. See p. 10.
Eric Weisstein's World of Mathematics, Gauss's Constant.
Eric Weisstein's World of Mathematics, Arithmetic-Geometric Mean.
Index entries for transcendental numbers.

Cf. A053002, A053003, A053004, A062539, A093341, A244644, A248557.

SetDefaultRealField(RealField(100)); R:= RealField(); Sqrt(Pi(R)/2)/Gamma(3/4)^2; // G. C. Greubel, Aug 17 2018

evalf(1/GaussAGM(1, sqrt(2)), 144);  # Alois P. Heinz, Jul 05 2023

RealDigits[Gamma[1/4]^2/(2*Pi^(3/2)*Sqrt[2]), 10, 105][[1]] (* or: *)
RealDigits[1/ArithmeticGeometricMean[1, Sqrt[2]], 10, 105][[1]] (* Jean-François Alcover, Dec 13 2011, updated Nov 11 2016, after Eric W. Weisstein *)
First[RealDigits[N[EllipticTheta[4, Exp[-Pi]]^2, 90]]] (* Stefano Spezia, Sep 29 2022 *)

default(realprecision, 20080); x=10*agm(1, sqrt(2))^-1; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b014549.txt", n, " ", d)); \\ Harry J. Smith, Apr 20 2009

1/agm(sqrt(2),1) \\ Charles R Greathouse IV, Feb 04 2015

sqrt(Pi/2)/gamma(3/4)^2 \\ Charles R Greathouse IV, Feb 04 2015

from mpmath import mp, agm, sqrt
mp.dps=105
print([int(z) for z in list(str(1/agm(sqrt(2)))[2:-1])]) # Indranil Ghosh, Jul 11 2017

Equals (lim_{k->oo} p(k))/(1+i) and (lim_{k->oo} q(k))/(1+i), where i is the imaginary unit, p(0) = 1, q(0) = i, p(k+1) = 2*p(k)*q(k)/(p(k)+q(k)) and q(k+1) = sqrt(p(k)*q(k)) for k >= 0. - A.H.M. Smeets, Jul 26 2018
Equals the infinite quotient product (3/4)*(6/5)*(7/8)*(10/9)*(11/12)*(14/13)*(15/16)*... . - James Maclachlan, Jul 28 2019
Equals (9/15)*hypergeom([1/2, 3/4], [9/4], 1). - Peter Bala, Mar 03 2022
Equals A062539 / Pi. - Amiram Eldar, May 04 2022
From Stefano Spezia, Sep 29 2022: (Start)
Equals theta4(exp(-Pi))^2.
Equals sqrt(2)*A093341/Pi. (End)
Equals Sum_{k>=0} (-1)^k * binomial(2*k,k)^2/16^k. - Amiram Eldar, Jul 04 2023
From Gerry Martens, Jul 31 2023: (Start)
Equals 2*Gamma(5/4)/(sqrt(Pi)*Gamma(3/4)).
Equals hypergeom([1/4, -2/4], [1], 1). (End)
Equals A248557^2. - Hugo Pfoertner, Jun 28 2024

Extended to 105 terms by Jean-François Alcover, Dec 13 2011

a(104) corrected by Andrew Howroyd, Feb 23 2018

A188615 Decimal expansion of Brocard angle of side-silver right triangle.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Apr 05 2011

The Brocard angle is invariant of the size of the side-silver right triangle ABC.  The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(silver ratio)=(1+sqrt(2)).  This is the unique right triangle matching the continued fraction [2,2,2,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there are exactly 2 removable subtriangles at each stage.  (This is analogous to the removal of 2 squares at each stage of the partitioning of the silver rectangle as a nest of squares.)
Archimedes's-like scheme: set p(0) = 1/(2*sqrt(2)), q(0) = 1/3; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, 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
This angle is also the half-angle at the summit of the Kelvin wake pattern traced by a boat. - Robert FERREOL, Sep 27 2019

			Brocard angle: 0.3398369094541219370963925133917640663882 approx.
Brocard angle: 19.471220634490691369245999 degrees, approx.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.
Wikipedia, Kelvin wake pattern
Index entries for transcendental numbers.

Cf. A188614, A188543, A152149.

[Arccos(Sqrt(8/9))]; // G. C. Greubel, Nov 18 2017

r=1+2^(1/2);
b=1; a=r*b; c=(a^2+b^2)^(1/2);
area=(1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);
brocard=ArcCot[(a^2+b^2+c^2)/(4area)];
N[brocard, 130]
RealDigits[N[brocard,130]][[1]]
N[180 brocard/Pi,130] (* degrees *)
RealDigits[ArcCos[Sqrt[8/9]], 10, 50][[1]] (* G. C. Greubel, Nov 18 2017 *)

acos(sqrt(8/9)) \\ Charles R Greathouse IV, May 02 2013

(Brocard angle) = arccot((a^2+b^2+c^2)/(4*area(ABC))) = arccot(sqrt(8)).
Also equals arcsin(1/3) or arccsc(3). - Jean-François Alcover, May 29 2013
Equals Integral_{x=sqrt(2)/2..sqrt(2)} dx/(x^2 + 1). - Kritsada Moomuang, May 29 2025

A244645 Decimal expansion of the sum of the reciprocals of the octagonal numbers (A000567).

Original entry on oeis.org

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

Offset: 1

Robert G. Wilson v, Jul 03 2014

			1.2774090575596367311949534921024332115566344803902472326934919840751515151955452...

Lawrence Downey, Boon W. Ong, and James A. Sellers, Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers, Coll. Math. J., 39, no. 5 (2008), 391-394.
Wikipedia, Polygonal number

Cf. A000038, A013661, A244639, A244644, A244646, A244647, A244648, A244649, A000567.

RealDigits[ Sum[1/(3n^2 - 2n), {n, 1 , Infinity}], 10, 111][[1]]

sumpos(n=1, 1/(3*n^2 - 2*n)) \\ Michel Marcus, Sep 12 2016

sumnumrat(1/(3*n-2)/n,1) \\ Charles R Greathouse IV, Feb 08 2023

Equals Sum_{n>=1} 1/(3*n^2 - 2*n).

Equals Pi/(4*sqrt(3)) + 3*log(3)/4. - Vaclav Kotesovec, Jul 05 2014

A244646 Decimal expansion of the sum of the reciprocals of the 9-gonal (or enneagonal or nonagonal) numbers (A001106).

Original entry on oeis.org

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

Offset: 1

Robert G. Wilson v, Jul 03 2014

			1.2433209261537129892066077396310142821358441010300996244152817525...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Ravi P. Agarwal, Pythagoreans Figurative Numbers: The Beginning of Number Theory and Summation of Series, Journal of Applied Mathematics and Physics, Vol.9, No.8 (2021), pp. 2038-2113. See p. 2076.
Wikipedia, Polygonal number.

Cf. A000038, A001106, A013661, A001620, A244639, A244644, A244645, A244647, A244648, A244649.

RealDigits[ Sum[2/(7n^2 - 5n), {n, 1 , Infinity}], 10, 111][[1]]

Equals Sum_{n>=1} 2/(7n^2 - 5n).
Equals (2*log(14) + 4*(cos(Pi/7)*log(cos(3*Pi/14)) + log(sin(Pi/7))*sin(Pi/14) - log(cos(Pi/14)) * sin(3*Pi/14)) + Pi*tan(3*Pi/14))/5. - Vaclav Kotesovec, Jul 04 2014
Equals 14/25 - (2/5)*(gamma + psi(-5/7)), where gamma is Euler's constant (A001620) and psi(x) is the digamma function (Agarwal, 2021), psi(-5/7) = psi(2/7)+7/5 = -2.285517..., see A354628. - Amiram Eldar, Nov 12 2021

A244649 Decimal expansion of the sum of the reciprocals of the Dodecagonal numbers (A051624).

Original entry on oeis.org

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

Offset: 1

Robert G. Wilson v, Jul 03 2014

From Wolfdieter Lang, Nov 09 2017: (Start)
In the Downey et al. link this is the instance k = 5 of the formula given there for S_{2*k+2}. A simpler formula is given in the Koecher reference as (5/4)*v_5(1) on p. 192. See the Kotesovec formula given below.
The partial sums are given in A294520/A294521. (End)

			1.1779560579226638587351739680918874184458572345666798028425228573...

Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, pp. 189 - 193.

Lawrence Downey, Boon W. Ong, and James A. Sellers, Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers, Coll. Math. J., 39, no. 5 (2008), 391-394.
Wikipedia, Polygonal number

Cf. A000038, A013661, A051624, A244639, A244644, A244645, A244646, A244647, A244648, A294520/A294521.

RealDigits[ Sum[1/(5n^2 - 4n), {n, 1 , Infinity}], 10, 111][[1]]

Equals Sum_{n>=1} 1/(5n^2 - 4n).
Equals Pi/8*sqrt(1+2/sqrt(5)) + (5*log(5) + sqrt(5)*log((3+sqrt(5))/2))/16. - Vaclav Kotesovec, Jul 04 2014
This is the value given in  the Koecher reference (see a comment above), and rewritten with the golden section phi = (1 + sqrt(5))/2 this becomes
  ((5/2)*log(5) + (2*phi - 1)*(log(phi) + (Pi/5)*sqrt(3 + 4*phi)))/8. - Wolfdieter Lang, Nov 09 2017

A244648 Decimal expansion of the sum of the reciprocals of the hendecagonal numbers (A051682).

Original entry on oeis.org

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

Offset: 1

Robert G. Wilson v, Jul 03 2014

			1.195434116529627974352499234698499354884682627084658062386021603017...

Wikipedia, Polygonal number

Cf. A000038, A013661, A244639, A244644, A244645, A244646, A244647, A244649.

RealDigits[ Sum[2/(9n^2 - 7n), {n, 1 , Infinity}], 10, 111][[1]]

Sum_{n=1..infinity} 2/(9n^2 - 7n).

Equals (5*log(3) + Pi*cot(2*Pi/9) - 4*cos(2*Pi/9)*log(cos(Pi/18)) + 4*cos(Pi/9)*log(sin(2*Pi/9)) - 4*log(sin(Pi/9))*sin(Pi/18))/7. - Vaclav Kotesovec, Jul 04 2014

A129200 Decimal expansion of arcsinh(1/4).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Jul 27 2008

Archimedes's-like scheme: set p(0) = 1/sqrt(17), 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

			.24746646154726345294478154978835928925376690309856769646911...

Muniru A Asiru, Table of n, a(n) for n = 0..2000
Index entries for transcendental numbers

SetDefaultRealField(RealField(100)); Argsinh(1/4); // G. C. Greubel, Nov 11 2018

RealDigits[ArcSinh[1/4], 10, 111][[1]] (* Robert G. Wilson v, Jul 23 2018 *)

asinh(1/4) \\ Michel Marcus, Jul 12 2018

Equals log((1 + sqrt(17))/4). - Jianing Song, Jul 12 2018

A195621 Decimal expansion of arccsc(4).

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Sep 23 2011

Archimedes's-like scheme: set p(0) = 1/sqrt(15), q(0) = 1/4; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, 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

			arccsc(4) = arcsin(1/4) = 0.25268025514207865...

Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 35, page 338.

Muniru A Asiru, Table of n, a(n) for n = 0..2000
Index entries for transcendental numbers.

Cf. A195628, A244644.

SetDefaultRealField(RealField(100)); Arcsin(1/4); // G. C. Greubel, Nov 11 2018

(See A195628.)
RealDigits[ ArcCsc@4, 10, 111][[1]] (* Robert G. Wilson v, Jul 23 2018 *)

asin(1/4) \\ Michel Marcus, Jul 12 2018

Equals arccos(sqrt(15)/4) = arctan(1/sqrt(15)). - Amiram Eldar, Jul 11 2023

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

cp's OEIS Frontend

A093954 Decimal expansion of Pi/(2*sqrt(2)).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A014549 Decimal expansion of 1 / M(1,sqrt(2)) (Gauss's constant).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

Python

Formula

Extensions

A188615 Decimal expansion of Brocard angle of side-silver right triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A244645 Decimal expansion of the sum of the reciprocals of the octagonal numbers (A000567).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A244646 Decimal expansion of the sum of the reciprocals of the 9-gonal (or enneagonal or nonagonal) numbers (A001106).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A244649 Decimal expansion of the sum of the reciprocals of the Dodecagonal numbers (A051624).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica