A112628 -id:A112628 - 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)

A016802 a(n) = (4*n)^2.

Original entry on oeis.org

0, 16, 64, 144, 256, 400, 576, 784, 1024, 1296, 1600, 1936, 2304, 2704, 3136, 3600, 4096, 4624, 5184, 5776, 6400, 7056, 7744, 8464, 9216, 10000, 10816, 11664, 12544, 13456, 14400, 15376, 16384, 17424, 18496, 19600, 20736, 21904, 23104, 24336, 25600, 26896, 28224

Offset: 0

N. J. A. Sloane

A bisection of A016742. Sequence arises from reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

Also, sequence found by reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Sep 10 2011

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000 (first 200 terms from Ivan Panchenko).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A001539, A016742, A016754, A016814, A016826, A016838, A112628.

Cf. A008598, A074377, A139098, A195146.

a(n) = (4*n)^2; \\ Michel Marcus, Mar 04 2014

def A016802(n): return (4*n)**2 # Karl-Heinz Hofmann, Sep 11 2024

a(n) = 16*n^2 = 16*A000290(n). - Omar E. Pol, Dec 11 2008
a(n) = 8*A001105(n) = 4*A016742(n) = 2*A139098(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 16*(2*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/96.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/192.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/4)/(Pi/4).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/4)/(Pi/4) = 2*sqrt(2)/Pi (A112628). (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 16*x*(1 + x)/(1-x)^3.
E.g.f.: 16*x*(1 + x)*exp(x).
a(n) = n*A008598(n) = A195146(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A224268 Decimal expansion of Product_{n>=1} (1 - 1/(4n+1)^2).

Original entry on oeis.org

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

Offset: 0

Bruno Berselli, Apr 02 2013

			0.9270373386506859592169251735976300231087994117608834527929640225280...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.5.4, p. 34.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on the constants A096427 and A224268.

Cf. A064853, A085565, A093341, A327996.

Cf. product(1-1/(4n+r)^2, n>=1): A096427 (r=-1), A112628 (r=0), A179587-1 (r=2).

RealDigits[N[Product[1 - 1/(4 n + 1)^2, {n, 1, Infinity}], 90]][[1]] (* or, by the formula: *) RealDigits[Gamma[1/4]^2/(8 Sqrt[Pi]), 10, 90][[1]]

prodnumrat(1 - 1/(4*n+1)^2, 1) \\ Charles R Greathouse IV, Feb 07 2025

Equals Gamma(1/4)^2/(8*sqrt(Pi)) = L/(4*sqrt(2)), where L is the Lemniscate constant (A064853).
From Peter Bala, Feb 26 2019: (Start)
C = (Pi/4)*( Sum_{n = -inf..inf} exp(-Pi*n^2) )^2.
C = (-1)^m*2^(2*m+1)/Catalan(m) * Product_{n >= 1} ( 1 - (4*m + 3)^2/(4*n + 1)^2 ), for m = 0,1,2,....
C = Integral_{x = 0..1} 1/sqrt(1 + x^4) dx.
C = (1/sqrt(2))*Integral_{x = 0..1} 1/sqrt(1 - x^4) dx.
C = (3/2)*Integral_{x = 0..1} sqrt(1 + x^4) dx - sqrt(2)/2.
C = (1/8)*Integral_{x = 0..1} 1/(x - x^2)^(3/4) dx.
C = Sum_{n >= 0} binomial(-1/2,n)/(4*n + 1) = Sum_{n >= 0} binomial(2*n,n)/4^n * 1/(4*n + 1).
C = (1/2)*Sum_{n >= 0} (-1)^n*binomial(-3/4,n)/(4*n + 1).
Continued fraction: 1 - 1/(5 + 20/(1 + 30/(3 + ... + (4*n)*(4*n + 1)/(1 + (4*n + 1)*(4*n + 2)/(3 + ... ))))).
C = A085565/sqrt(2). C = Pi/(4*A096427). (End)
Equals A093341/2 = A327996^2. - Hugo Pfoertner, Oct 31 2024

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

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A016802 a(n) = (4*n)^2.

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A224268 Decimal expansion of Product_{n>=1} (1 - 1/(4n+1)^2).

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A091392 Product_{ p | n } (1 + Legendre(-2,p) ).

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A371604 Decimal expansion of 5 * sqrt(3 - phi) / (2 * Pi).

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