A068465 -id:A068465 - cp's OEIS frontend

A068466 Decimal expansion of Gamma(1/4).

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

Offset: 1

Benoit Cloitre, Mar 10 2002

Nesterenko proves that this constant is transcendental (he cites Chudnovsky as the first to show this); in fact it is algebraically independent of Pi and e^Pi over Q. - Charles R Greathouse IV, Nov 11 2013

			3.6256099082219083119306851558676720029951676828800654674333...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.5.4, p. 33.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 43, equation 43:4:13 at page 414.

Harry J. Smith, Table of n, a(n) for n = 1..1000
William Duke and Özlem Imamoḡlu, Special values of multiple gamma functions, Journal de théorie des nombres de Bordeaux, Vol. 18. No. 1 (2006), pp. 113-123.
Greg J. Fee and Simon Plouffe, Gamma(1/4) to 25000 digits.
Yu. V. Nesterenko, Modular functions and transcendence questions, Sbornik: Mathematics, Vol. 187, No. 9 (1996), pp. 1319-1348. (English translation)
Simon Plouffe, Gamma(1/4) to 250000 digits.
Dan Romik, On Viazovska's modular form inequalities, arXiv:2303.13427 [math.NT], 2023.
Eric Weisstein's World of Mathematics, Gamma Function.
Wikipedia, Particular values of the Gamma function: General rational arguments.
Index to sequences related to the Gamma function.
Index entries for transcendental numbers.

Cf. A014549, A248557.

R:= RealField(100); SetDefaultRealField(R); Gamma(1/4); // G. C. Greubel, Mar 10 2018

Maple
```
evalf(GAMMA(1/4));
```

RealDigits[Gamma[1/4], 10, 110][[1]] (* Bruno Berselli, Dec 13 2012 *)

default(realprecision, 1080); x=gamma(1/4); for (n=1, 1000, d=floor(x); x=(x-d)*10; write("b068466.txt", n, " ", d)); \\ Harry J. Smith, Apr 19 2009

From Amiram Eldar, Jun 12 2021: (Start)
Equals sqrt(2*sqrt(2*Pi^3)*G), where G is Gauss's constant (A014549).
Equals (2*Pi)^(3/4) * Product_{k>=1} tanh(k*Pi/2) (Duke and Imamoḡlu, 2006). (End)
Gamma(1/4) * A068465 = A063448. - R. J. Mathar, May 22 2024
Equals Product_{n>=1} exp((2*(6*n + 1)*(1 - beta(n)) - (eta(n) - 1))/(4*n)), where eta(n) and beta(n) are the Dirichlet eta and beta functions, respectively. - Antonio Graciá Llorente, Sep 05 2024

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

Original entry on oeis.org

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)

A016838 a(n) = (4n + 3)^2.

Original entry on oeis.org

9, 49, 121, 225, 361, 529, 729, 961, 1225, 1521, 1849, 2209, 2601, 3025, 3481, 3969, 4489, 5041, 5625, 6241, 6889, 7569, 8281, 9025, 9801, 10609, 11449, 12321, 13225, 14161, 15129, 16129, 17161, 18225

Offset: 0

N. J. A. Sloane

If Y is a fixed 2-subset of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
A bisection of A016754. Sequence arises from reading the line from 9, in the direction 9, 49, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Using (n,n+1) to generate a Pythagorean triangle with sides of lengths xJ. M. Bergot, Jul 17 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..860
Milan Janjic, Two Enumerative Functions
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001539, A016742, A016754, A016802, A016814, A016826, A068465, A087197.

[(4*n+3)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A016838:=n->(4*n + 3)^2; seq(A016838(n), n=0..50); # Wesley Ivan Hurt, Feb 24 2014

Table[(4*n + 3)^2, {n, 0, 40}] (* Vaclav Kotesovec, Nov 14 2017 *)

a(n)=(4*n+3)^2 \\ Charles R Greathouse IV, Jun 17 2017

Denominators of first differences Zeta[2,(4n-1)/4]-Zeta[2,(4(n+1)-1)/4]. - Artur Jasinski, Mar 03 2010
From George F. Johnson, Oct 03 2012: (Start)
G.f.: (9+22*x+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+1) = (A004767(n))^2.
(End)
Sum_{n>=0} 1/a(n) = Pi^2/16 - G/2, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=0} (1 - 1/a(n)) = Gamma(3/4)^2/sqrt(Pi) = A068465^2 * A087197. - Amiram Eldar, Feb 01 2021

A175573 Decimal expansion of Pi^(1/4)/Gamma(3/4).

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Jul 15 2010

Entry 34 a of chapter 11 of Ramanujan's second notebook. Entry 34 b is A085565.

			1.0864348112133080145753161...

G. C. Greubel, Table of n, a(n) for n = 1..5000
Bruce C. Berndt, Chapter 11 of Ramanujan's second notebook, Bull. Lond. Math. Soc., Vol. 15, No. 4 (1983), 273-320.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 27, equation 27:13:18 at page 258.
Wikipedia, Theta function.

Cf. A068465, A085565, A092040, A096427, A175574, A247217, A273081, A273082, A273083, A273084, A319332, A327995.

C := ComplexField(); [(Pi(C))^(1/4)/Gamma(3/4)]; // G. C. Greubel, Nov 05 2017

Maple

Pi^(1/4)/GAMMA(3/4) ; evalf(%) ;

Mathematica

RealDigits[ Pi^(1/4)/Gamma[3/4], 10, 105][[1]] (* Jean-François Alcover, Jul 04 2013 *)

PARI

Pi^(1/4)/gamma(3/4) \\ G. C. Greubel, Nov 05 2017

PARI

2*suminf(k=0,exp(-Pi)^(k^2))-1 \\ Hugo Pfoertner, Sep 17 2018

Formula

Equals A092040 / A068465.

Equals Sum_{n=-oo..oo} exp(-Pi*n^2), or also EllipticTheta(3, 0, exp(-Pi)). - Jean-François Alcover, Jul 04 2013

Equals sqrt(A175574). - Amiram Eldar, Jul 04 2023

Equals Gamma(1/4)/(sqrt(2)*Pi^(3/4)). - Vaclav Kotesovec, Jul 04 2023

Equals Product_{k>=1} tanh((1/2 + i/2)*Pi*k), i=sqrt(-1). - _Antonio Graciá Llorente, Mar 20 2024

Equals Product_{k>=0} (1/2)*(((k+1/2)/(k+1))^(1/2)+((k+1)/(k+1/2))^(1/2)). - Antonio Graciá Llorente, Jul 23 2024

Equals (1/A096427)^2 (see Spanier and Oldham at p. 258). - Stefano Spezia, Dec 31 2024

Equals 2*A319332 = 1/A327995. - Hugo Pfoertner, Dec 31 2024

A084254 Decimal expansion of Sum_{k>=1} 1/(k(exp(2Pi*k)-1)).

Original entry on oeis.org

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

Offset: 0

Benoit Cloitre, Jun 21 2003

			0.00187268244976854611563857947996139886916289565261...

Bruce C. Berndt, Ramanujan Notebook part II, Infinite series, Springer Verlag, 1989, pp. 280-281.

Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020-2022, p. 6.
Simon Plouffe, Identities inspired by Ramanujan Notebooks (part 2), April 2006.
Linas Vepštas, On Plouffe's Ramanujan identities, The Ramanujan Journal, Vol. 27 (2012), pp. 387-408; arXiv preprint, arXiv:math/0609775 [math.NT], 2006-2010.

Cf. A000203, A003881, A068465, A094640.

Cf. A255695 (S(1,1)), A255697 (S(1,4)), A255698 (S(3,1)), A255699 (S(3,2)), A255700 (S(3,4)), A255701 (S(5,1)), A255702 (S(5,2)), A255703 (S(5,4)).

digits = 104; S[1, 2] = NSum[1/(n*(Exp[2*Pi*n] - 1)), {n, 1, Infinity}, WorkingPrecision -> digits+10, NSumTerms -> digits]; RealDigits[S[1, 2], 10, digits] // First (* Jean-François Alcover, Mar 02 2015 *)
Join[{0,0},RealDigits[Log[4/Pi]/4 - Pi/12 + Log[Gamma[3/4]], 10, 100][[1]]] (* Amiram Eldar, May 21 2022 *)

PARI
```
1/4*log(4/Pi)-Pi/12+log(gamma(3/4))
```

Equals log(4/Pi)/4 - Pi/12 + log(Gamma(3/4)).
From Jean-François Alcover, Mar 02 2015: (Start)
This is the case k=1, m=2 of the Plouffe sum S(k,m) = Sum_{n >= 1} 1/(n^k*(exp(m*Pi*n)-1)).
Pi = 72*S(1,1) - 96*S(1,2) + 24*S(1,4). (End)
Equals Sum_{k>=1} sigma(k)/(k*exp(2*Pi*k)). - Amiram Eldar, Jun 05 2023

A091670 Decimal expansion of Gamma(1/4)^4/(4*Pi^3).

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Jan 27 2004

Watson's first triple integral.
This is also the value of F. Morley's series from 1902 Sum_{k=0..n} (risefac(k,1/2)/k!)^3 = hypergeometric([1/2,1/2,1/2],[1,1],1) with the rising factorial risefac(n,x). See A277232, also for the Hardy reference and a MathWorld link. - Wolfdieter Lang, Nov 11 2016
This constant is transcendental due to a result of Nesterenko, who proves that Gamma(1/4) is algebraically independent of Pi. - Charles R Greathouse IV, Aug 19 2025

			1.39320392968567685918424626032536824265748121751561787897...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.9, p. 324.

G. C. Greubel, Table of n, a(n) for n = 1..10000
A. M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 256, 6.1.17 , p. 557, 15.1.26.
M. L. Glasser, I. J. Zucker, Extended Watson integrals for the cubic lattices, Proc. Nat. Acad. Sci., Vol. 74, No. 5 (1977), p. 1800-1801.
Paul J. Nahin, Inside interesting integrals, Undergrad. Lecture Notes in Physics, Springer (2020), (6.5.1)
Yu. V. Nesterenko, Modular functions and transcendence questions, Sbornik: Mathematics, Vol. 187, No. 9 (1996), pp. 1319-1348. (English translation)
Tito Piezas III, Watson's triple integrals.
Eric Weisstein's World of Mathematics, Watson's Triple Integrals.
I. J. Zucker, 70+years of the Watson integrals, J. Stat. Phys., Vol. 145, No. 3 (2011), pp. 591-612.
Index entries for transcendental numbers.

Cf. A091671, A091672, A277232, A293238 (inverse), A068466.

SetDefaultRealField(RealField(100)); R:= RealField(); Gamma(1/4)^4/(4*Pi(R)^3); // G. C. Greubel, Oct 26 2018

Pi/GAMMA(3/4)^4 ; evalf(%) ; # R. J. Mathar, Jun 17 2016

RealDigits[ N[ Gamma[1/4]^4/(4*Pi^3), 102]][[1]] (* Jean-François Alcover, Nov 12 2012, after Eric W. Weisstein *)

1/agm(sqrt(1/2),1)^2 \\ Charles R Greathouse IV, Mar 03 2016

From Joerg Arndt, Nov 27 2010: (Start)
Equals 1/agm(1,sqrt(1/2))^2.
Equals Gamma(1/4)^4 / (4*Pi^3) = Pi / (Gamma(3/4))^4 = hypergeom([1/2,1/2],[1],1/2)^2, see the two Abramowitz - Stegun references. (End)
Equals the square of A175574. Equals A000796/A068465^4. - R. J. Mathar, Jun 17 2016
Equals hypergeom([1/2,1/2,1/2],[1,1],1) - Wolfdieter Lang, Nov 12 2016
Equals Sum_{k>=0} binomial(2*k,k)^3/2^(6*k). - Amiram Eldar, Aug 26 2020

A025749 4th-order Patalan numbers (generalization of Catalan numbers).

Original entry on oeis.org

1, 1, 6, 56, 616, 7392, 93632, 1230592, 16612992, 228890112, 3204461568, 45445091328, 651379642368, 9419951751168, 137262154088448, 2013178259963904, 29694379334467584, 440175505428578304, 6553724191936610304, 97960930026841964544, 1469413950402629468160

Offset: 0

Olivier Gérard

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Thomas M. Richardson, The Super Patalan Numbers, J. Int. Seq. 18 (2015), Article 15.3.3; arXiv preprint, arXiv:1410.5880 [math.CO], 2014.

Cf. A034176, A048779, A068465.

a[n_] := (4^(n-1)*Sum[ Binomial[n+k-1, n-1]*Sum[ Binomial[j, n-3*k+2*j-1] * 4^(j-k) * Binomial[k, j] * 3^(-n+3*k-j+1) * 2^(n-3*k+j-1) * (-1)^(n-3*k+2*j-1), {j, 0, k}], {k, 1, n-1}])/n; a[0] = a[1] = 1; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Mar 05 2013, after Vladimir Kruchinin *)
a[n_] := 16^(n-1) * Pochhammer[3/4, n-1]/n!; a[0] = 1; Array[a, 21, 0] (* Amiram Eldar, Aug 20 2025 *)

a(n):=(4^(n-1)*sum(binomial(n+k-1,n-1)*sum(binomial(j,n-3*k+2*j-1)*4^(j-k)*binomial(k,j)*3^(-n+3*k-j+1)*2^(n-3*k+j-1)*(-1)^(n-3*k+2*j-1),j,0,k),k,1,n-1))/n; /* Vladimir Kruchinin, Apr 01 2011 */

a(n) = 2^(n-1) * A048779(n), n > 1.
From Wolfdieter Lang: (Start)
G.f.: (5-(1-16*x)^(1/4))/4.
a(n) = 4^(n-1)*3*A034176(n-1)/n!, n >= 2, where 3*A034176(n-1) = (4*n-5)(!^4) = Product_{j=2..n} (4*j - 5). (End)
a(n) = (4^(n-1) * Sum_{k=1..n-1} binomial(n+k-1,n-1) * Sum_{j=0..k} binomial(j,n-3*k+2*j-1)*4^(j-k)*binomial(k,j)*3^(-n+3*k-j+1)*2^(n-3*k+j-1)*(-1)^(n-3*k+2*j-1))/n. - Vladimir Kruchinin, Apr 01 2011
n*a(n) + 4*(-4*n+5)*a(n-1) = 0. - R. J. Mathar, Apr 05 2018
a(n) ~ 16^(n-1) / (Gamma(3/4) * n^(5/4)). - Amiram Eldar, Aug 20 2025

A175576 Decimal expansion Pi^(3/2)/Gamma(3/4)^2.

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Jul 15 2010

Entry 34 e of chapter 11 of Ramanujan's second notebook.
In addition, Pi^(3/2) / Gamma(3/4)^2 is the area of the unit "squircle" as defined in MathWorld. (Note that 8*Gamma(5/4)^2 / sqrt(Pi) is the same constant.) - Jean-François Alcover, Feb 24 2011
Real period of the elliptic curve y^2 = x*(x - 1)*(x - 1/2). See Rouse. - Peter Bala, Dec 06 2024

			3.708149354602743836867700694390520092435197647...

G. C. Greubel, Table of n, a(n) for n = 1..5000
Bruce C. Berndt, Chapter 11 of Ramanujan's second notebook, Bull. Lond. Math. Soc. vol 15 no 4 (1983) 273-320.
Jeremy Rouse, Hypergeometric functions and elliptic curves, Ramanujan Journal, Vol. 12 (2006), pp. 197-205.
Eric Weisstein's World of Mathematics, Squircle

Cf. A062539, A068465, A175476, A175575, A290570.

Maple
```
Pi^(3/2)/GAMMA(3/4)^2 ; evalf(%) ;
```

RealDigits[Pi*EllipticTheta[3, 0, Exp[-Pi]]^2, 10, 50][[1]]
RealDigits[Pi^(3/2)/(Gamma[3/4])^2, 10, 50][[1]] (* G. C. Greubel, Feb 12 2017 *)

Pi^1.5/gamma(3/4)^2 \\ Charles R Greathouse IV, Jun 06 2016

Equals A175476 / A068465^2 = 1/A175575.
Equals Integral_{-oo, oo} 1/(1+2*x^2)^(3/4) or Integral_{-oo, oo} 1/sqrt(1+x^4). - Jean-François Alcover, Jun 04 2013
Equals sqrt(2)*L, where L is the lemniscate constant A062539. - Jean-François Alcover, Aug 11 2014
From Peter Bala, Mar 01 2022 : (Start)
Equals 3*Sum_{n >= 0} (1/(4*n+1) + 1/(4*n-3))*binomial(1/2,n). Cf. A290570.
Equals hypergeom([-1/2, 3/4, -3/4], [-1/4, 5/4], -1).
Equals 2*hypergeom([1/4, 3/4], [5/4], 1) = (16/5)*hypergeom([-1/4, -3/4], [5/4], 1). (End)
Equals 2 * A093341. - R. J. Mathar, Dec 08 2023
From Peter Bala, Dec 06 2024: (Start)
Equals Pi*hypergeom([1/2, 1/2], [1], 1/2).
Equals 2*Integral_{x = 0..Pi/2} 1/sqrt(1 - (1/2)*sin^2(x)) dx. See Rouse. (End)

A175574 Decimal expansion of sqrt(Pi) / (Gamma(3/4))^2.

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Jul 15 2010

Entry 34 c of chapter 11 of Ramanujan's second notebook.

This constant is also the ratio T(Pi/2)/T(0), where T(Pi/2) is the exact pendulum period for an amplitude of Pi/2 and T(0) the approximate period 2*Pi*sqrt(L/g) for small angles. - Jean-François Alcover, Aug 05 2014

			1.18034059901609622604533794..

G. C. Greubel, Table of n, a(n) for n = 1..5000
Bruce C. Berndt, Chapter 11 of Ramanujan's second notebook, Bull. Lond. Math. Soc., Vol. 15, No. 4 (1983), 273-320.
Claudio Carvalhaes and Patrick Suppes, Approximations for the period of the simple pendulum based on the arithmetic-geometric mean, American Journal of Physics 76 (2008), 1150-1154.

Cf. A002161, A068465, A085565, A096427, A175573.

sqrt(pi)/gamma(3/4)^2 % Altug Alkan, Dec 05 2015

Maple
```
sqrt(Pi)/GAMMA(3/4)^2 ; evalf(%) ;
```

First@ RealDigits[N[Sqrt@ Pi/Gamma[3/4]^2, 120]] (* Michael De Vlieger, Dec 06 2015 *)

sqrt(Pi)/gamma(3/4)^2 \\ Altug Alkan, Dec 05 2015

Equals A002161 /A068465^2.
Equals 2F1([1/2,1/2],[1],1/2) = 1/agm(1, sqrt(1/2)) = gamma(1/4)^2/(2*Pi^(3/2)).
Equals 2*sqrt(2)*K(-1)/Pi, where K is the complete elliptic integral of the first kind, K(-1) being A085565. - Jean-François Alcover, Jun 03 2014
Equals Product_{k>=1} (1-(-1)^k/(2*k)) =  3/2 * 3/4 * 7/6 * 7/8 * 11/10 * 11/12 * ... .  - Richard R. Forberg, Dec 05 2015
Reciprocal of A096427. Equals ( Sum_{n = -inf..inf} exp(-Pi*n^2) )^2, a rapidly converging series. For example, summing from n = -5 to n = 5 gives the constant correct to 49 decimal places. - Peter Bala, Mar 06 2019
Equals Sum_{k>=0} binomial(2*k,k)^2/2^(5*k). - Amiram Eldar, Aug 26 2020
Equals (3/2)*hypergeom([-1/4, 3/4], [3/2], 1). - Peter Bala, Mar 04 2022
Equals A175573^2. - Amiram Eldar, Jul 04 2023

A-number typo for sqrt(Pi) corrected by R. J. Mathar, Aug 01 2010

A048779 Coefficients of power series for (1 - (1-8*x)^(1/4))/2.

Original entry on oeis.org

1, 3, 14, 77, 462, 2926, 19228, 129789, 894102, 6258714, 44379972, 318056466, 2299792908, 16755634044, 122874649656, 906200541213, 6716545187814, 50000947509282, 373691291911476, 2802684689336070, 21086865757861860, 159109987082048580, 1203701641403324040

Offset: 1

Jan Kristian Haugland

			G.f.: x + 3*x^2 + 14*x^3 + 77*x^4 + 462*x^5 + 2926*x^6 + 19228*x^7 + ...

Michael De Vlieger, Table of n, a(n) for n = 1..1112
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Related to Catalan numbers (A000108).

Cf. A004984, A068465.

[Round(8^(n-1)*Gamma(n-1/4)/(Gamma(3/4)*Gamma(n+1))): n in [1..40]]; // G. C. Greubel, Aug 09 2022

a[ n_]:= If[n<1, 0, (-1/2)Pochhammer[-1/4, n] 8^n/n!] (* Michael Somos, Jan 17 2014 *)
a[ n_]:= SeriesCoefficient[(1 -(1-8x)^(1/4))/2, {x,0,n}] (* Michael Somos, Jan 17 2014 *)

{a(n) = if( n<0, 0, polcoeff( (1 - (1 - 8*x + x * O(x^n))^(1/4)) / 2, n))} /* Michael Somos, Jan 17 2014 */

[8^(n-1)*binomial(n-5/4,-1/4)/n for n in (1..40)] # G. C. Greubel, Aug 09 2022

a(n) = 2^(n-1)*3*7*11*...*(4n-5)/n! = 2*a(n-1)*(32*a(n-2) + a(n-1))/(18*a(n-2) -a(n-1)).
a(n) = -A004984(n)/2.
D-finite with recurrence n*a(n) + 2*(5-4*n)*a(n-1) = 0. - R. J. Mathar, Oct 29 2012
G.f. A(x) =: y satisfies x = y * (1 - y) * (1 - 2*y + 2*y^2). - Michael Somos, Jan 17 2014
0 = a(n) * (64*a(n+1) - 18*a(n+2)) + a(n+1) * (2*a(n+1) + a(n+2)) unless n=0. - Michael Somos, Jan 17 2014
From Karol A. Penson, Dec 19 2015: (Start)
a(n) = 8^(n-1)*binomial(n-5/4, -1/4)/n.
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([3/4], [2], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n) = int(x^n*((2^(1/4)/(2*Pi*x^(1/4))*(1-x/8)^(1/4))), x=0..8), n=0,1,... . This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)
a(n) ~ 2^(3*n-3) / (Gamma(3/4) * n^(5/4)). - Amiram Eldar, Sep 01 2025

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A068466 Decimal expansion of Gamma(1/4).

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

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A016838 a(n) = (4n + 3)^2.

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A175573 Decimal expansion of Pi^(1/4)/Gamma(3/4).

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A084254 Decimal expansion of Sum_{k>=1} 1/(k*(exp(2*Pi*k)-1)).

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A091670 Decimal expansion of Gamma(1/4)^4/(4*Pi^3).

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A084254 Decimal expansion of Sum_{k>=1} 1/(k(exp(2Pi*k)-1)).