A068467 -id:A068467 - 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)

A016814 a(n) = (4*n + 1)^2.

Original entry on oeis.org

1, 25, 81, 169, 289, 441, 625, 841, 1089, 1369, 1681, 2025, 2401, 2809, 3249, 3721, 4225, 4761, 5329, 5929, 6561, 7225, 7921, 8649, 9409, 10201, 11025, 11881, 12769, 13689, 14641, 15625, 16641, 17689, 18769, 19881, 21025, 22201, 23409, 24649, 25921, 27225, 28561, 29929

Offset: 0

N. J. A. Sloane

A bisection of A016754. Sequence arises from reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

G. C. Greubel, Table of n, a(n) for n = 0..10000 (terms 0..200 from Ivan Panchenko).
Leo Tavares, Illustration: Square trapeziums
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), this sequence (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).
Cf. A001539, A006752, A016742, A016802, A016813.
Cf. A016826, A016838, A017126, A068467, A087197.
Cf. A272399, A014105.

[(4*n+1)^2: n in [0..40]]; // G. C. Greubel, Dec 28 2022

A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013

(4*Range[0,40] +1)^2 (* or *) LinearRecurrence[{3,-3,1}, {1,25,81}, 40] (* Harvey P. Dale, Nov 20 2012 *)
Accumulate[32Range[0, 47] - 8] + 9 (* Alonso del Arte, Aug 19 2017 *)

a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015

[(4*n+1)^2 for n in range(41)] # G. C. Greubel, Dec 28 2022

a(n) = a(n-1) + 32*n - 8, n > 0. - Vincenzo Librandi, Dec 15 2010
From George F. Johnson, Sep 28 2012: (Start)
G.f.: (1 + 22*x + 9*x^2)/(1 - x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n) - 16)^2 ; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n) = (A016813(n))^2. (End)
Sum_{n>=0} 1/a(n) = G/2 + Pi^2/16, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=1} (1 - 1/a(n)) = 2*Gamma(5/4)^2/sqrt(Pi) = 2 * A068467^2 * A087197. - Amiram Eldar, Feb 01 2021
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A017078(n).
a(2*n+1) = A017126(n).
E.g.f.: (1 + 24*x + 16*x^2)*exp(x). (End)
a(n) = A272399(n+1) - A014105(n). - Leo Tavares, Dec 24 2023

A202623 Decimal expansion of (1/3)! = Gamma(4/3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Dec 29 2011

			0.89297951156924921121856431365822588137622979265243370031680...

Index to sequences related to the Gamma function

Cf. A068467, A073005.

4^(8/9)*%PI^(2/3)*THETA[3](0,%E^-(16*%PI/SQRT(3)))^(2/3)/(3^(1/4)*(2^(7/16)*(SQRT(2)-1)^(1/4)/((SQRT(3)-1)^(1/8)*(SQRT(3)-SQRT(2))^(1/4))+1/(2^(1/4)*SQRT(SQRT(3)+1))+1)^(2/3))
/* This is exact, but degrades to 50+ digits if you replace
THETA[3](0,%E^-(16*%PI/SQRT(3)))
by 1+2*%E^-(16*%PI/SQRT(3)) */
/* R. W. Gosper, Posting to Math Fun Mailing List, Dec 27 2011 */

Maple
```
evalf(GAMMA(4/3)) ;
```

RealDigits[(1/3)!,10,150][[1]] (* or *) RealDigits[Gamma[4/3],10,150] [[1]] (* Harvey P. Dale, Sep 03 2016 *)

A formula from R. W. Gosper, Posting to Math Fun Mailing List, Dec 27 2011:
Equals (1/3) * (2*2^(7/9)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1 + 1/(2^(1/4)*Sqrt[1 + Sqrt[3]]) + (2^(7/16)*((-1 + Sqrt[2])/(-Sqrt[2] + Sqrt[3]))^(1/4))/(-1+Sqrt[3])^(1/8)))^(2/3))/3^(1/4).
Equals Integral_{0..oo} exp(-x^3) dx. [Jean-François Alcover, Mar 29 2013]
Equals A073005/3. - R. J. Mathar, Jan 15 2021
Equals 3*Integral_{-1/e..0} (-LambertW(-1,x))^(1/3)-(-LambertW(x))^(1/3) dx. - Gleb Koloskov, Jun 07 2021

Corrected and extended by Harvey P. Dale, Sep 03 2016

A203125 Decimal expansion of (1/8)! = Gamma(9/8).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Dec 29 2011

			.94174269984970148808740373015189170307630244851863449262289...

Index to sequences related to the Gamma function

Cf. A068467, A202623, A203126, A203142.

RealDigits[(1/8)!,10,120][[1]] (* Harvey P. Dale, May 30 2014 *)

gamma(9/8) \\ Charles R Greathouse IV, Mar 25 2014

Equals A203142/8. - R. J. Mathar, Jan 15 2021
A203144 *this *A231863 *A011006 = A068467. - R. J. Mathar, Jan 15 2021
Equals Integral_{x=0..oo} exp(-x^8) dx. - Ilya Gutkovskiy, Sep 18 2021

A203126 Decimal expansion of (1/6)! = Gamma(7/6).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Dec 29 2011

			.92771933363003920070834948253462101856646651914547557693612...

Index to sequences related to the Gamma function

Cf. A068467, A175379, A202623, A203125.

RealDigits[(1/6)!,10,120][[1]] (* Harvey P. Dale, Aug 10 2013 *)

gamma(7/6) \\ Charles R Greathouse IV, Mar 25 2014

Equals A175379/6. - R. J. Mathar, Jan 15 2021
A073006 * this * A231863 * A329219 = A202623. - R. J. Mathar, Jan 15 2021
Equals Integral_{x=0..oo} exp(-x^6) dx. - Ilya Gutkovskiy, Sep 18 2021

A377731 Decimal expansion of 4 * sqrt(2Pi) Gamma(5/4) / (3 * Gamma(3/4)).

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Nov 05 2024

This constant appears in asymptotic formulas related to A211996 and A377732.

			2.47209956973516255791180046292701339495679843136235...

Jean-Marie De Koninck, A. Arthur Bonkli Razafindrasoanaivolala, and Hans Schmidt Ramiliarimanana, Integers with a sum of co-divisors yielding a square, Research in Number Theory, Vol. 10, No. 2 (2024), Article 30; author's copy.

Cf. A019727, A068465, A068467, A211996, A377732.

RealDigits[4 * Sqrt[2*Pi] * Gamma[5/4] / (3 * Gamma[3/4]), 10, 120][[1]]

4 * sqrt(2*Pi) * gamma(5/4) / (3 * gamma(3/4))

A256667 Decimal expansion of Integral_{x=0..Pi/2} sqrt(2-sin(x)^2) dx, an elliptic integral once studied by John Landen.

Original entry on oeis.org

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

Offset: 1

Jean-François Alcover, Apr 07 2015

Arclength on sine from origin to first maximum point. - Clark Kimberling, Jul 01 2020

			1.91009889451385600895238104108572164595498380732363736...

Mark Pinsky, Björn Birnir, Probability, Geometry and Integrable Systems (Cambridge University Press 2007), p. 289.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Eric Weisstein's MathWorld, Lemniscate Constant
Wikipedia, John Landen

Cf. A062539 (Lemniscate constant), A068465 (Gamma(3/4)), A068467 (Gamma(5/4)).

SetDefaultRealField(RealField(100)); R:= RealField(); (1/Sqrt(2*Pi(R)))*(Gamma(3/4)^2 + 4*Gamma(5/4)^2); // G. C. Greubel, Oct 07 2018

RealDigits[(1/Sqrt[2*Pi])*(Gamma[3/4]^2 + 4*Gamma[5/4]^2), 10, 105] // First

default(realprecision, 100); (1/sqrt(2*Pi))*(gamma(3/4)^2 + 4*gamma(5/4)^2) \\ G. C. Greubel, Oct 07 2018

Equals (1/sqrt(2*Pi))*(Gamma(3/4)^2 + 4*Gamma(5/4)^2).
Equals sqrt(2)*E(Pi/2 | 1/2), where E(phi|m) is the elliptic integral of the second kind.
Equals (L^2 + Pi)/(2*L), where L is the lemniscate constant 2.622...

cp's OEIS Frontend

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

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A016814 a(n) = (4*n + 1)^2.

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A202623 Decimal expansion of (1/3)! = Gamma(4/3).

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A203125 Decimal expansion of (1/8)! = Gamma(9/8).

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A203126 Decimal expansion of (1/6)! = Gamma(7/6).

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A377731 Decimal expansion of 4 * sqrt(2*Pi) * Gamma(5/4) / (3 * Gamma(3/4)).

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A256667 Decimal expansion of Integral_{x=0..Pi/2} sqrt(2-sin(x)^2) dx, an elliptic integral once studied by John Landen.

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A377731 Decimal expansion of 4 * sqrt(2Pi) Gamma(5/4) / (3 * Gamma(3/4)).

A256929 Decimal expansion of Sum_{k>=1} (zeta(2k)/k)(1/2)^(4*k).