A247719 -id:A247719 - cp's OEIS frontend

A033715 Number of integer solutions (x, y) to the equation x^2 + 2y^2 = n.

1, 2, 2, 4, 2, 0, 4, 0, 2, 6, 0, 4, 4, 0, 0, 0, 2, 4, 6, 4, 0, 0, 4, 0, 4, 2, 0, 8, 0, 0, 0, 0, 2, 8, 4, 0, 6, 0, 4, 0, 0, 4, 0, 4, 4, 0, 0, 0, 4, 2, 2, 8, 0, 0, 8, 0, 0, 8, 0, 4, 0, 0, 0, 0, 2, 0, 8, 4, 4, 0, 0, 0, 6, 4, 0, 4, 4, 0, 0, 0, 0, 10, 4, 4, 0, 0, 4, 0, 4, 4, 0, 0, 0, 0, 0, 0, 4, 4, 2, 12, 2, 0, 8, 0

Offset: 0

N. J. A. Sloane

Theta series of lattice C2 with Gram matrix [ 1, 0; 0, 2]. a(n) is nonzero if and only if n is in A002479. - Michael Somos, Dec 15 2011
Number 17 of the 74 eta-quotients listed in Table I of Martin (1996).
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Denoted by |a_4(n)| in Kassel and Reutenauer 2015. - Michael Somos, Jun 16 2015

			G.f. = 1 + 2*q + 2*q^2 + 4*q^3 + 2*q^4 + 4*q^6 + 2*q^8 + 6*q^9 + 4*q^11 + 4*q^12 + ...

Bruce C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag, 1991, see p. 114 Entry 8(iii).
J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer-Verlag, 1999, p. 102, eq. 9.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 3, p. 19.
Nathan J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 78, Eq. (32.24).
J. W. L. Glaisher, Table of the excess of the number of (8k+1)- and (8k+3)-divisors of a number over the number of (8k+5)- and (8k+7)-divisors, Messenger Math., 31 (1901), 82-91.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 346.

John Cannon, Table of n, a(n) for n = 0..10000
George E. Andrews, Richard Lewis, and Zhi-Guo Liu, An identity relating a theta series to a sum of Lambert series, Bull. London Math. Soc., 33 (2001), 25-31.
Michael Gilleland, Some Self-Similar Integer Sequences.
Michael D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211.
Christian Kassel and Christophe Reutenauer, The zeta function of the Hilbert scheme of n points on a two-dimensional torus, arXiv:1505.07229v3 [math.AG], 2015. [Note that a later version of this paper has a different title and different contents, and the number-theoretical part of the paper was moved to the publication which is next in this list.]
Christian Kassel and Christophe Reutenauer, Complete determination of the zeta function of the Hilbert scheme of n points on a two-dimensional torus, arXiv:1610.07793 [math.NT], 2016.
Yves Martin, Multiplicative eta-quotients, Trans. Amer. Math. Soc. 348 (1996), no. 12, 4825-4856, see page 4852, Table I.
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references).
Michael Somos, Index to Yves Martin's list of 74 multiplicative eta-quotients and their A-numbers, 2016.
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A002325, A002479, A028572, A033761, A082564, A112603, A113411, A133692, A139093, A226225, A226240, A242609, A245572, A247719.
Cf. A000122, A000700, A010054, A121373.
Number of integer solutions to f(x,y) = n where f(x,y) is the principal binary quadratic form with discriminant d: A004016 (d=-3), A004018 (d=-4), A002652 (d=-7), this sequence (d=-8), A028609 (d=-11), A028641 (d=-19), A138811 (d=-43).

A := Basis( ModularForms( Gamma1(8), 1), 105); A[1] + 2*A[2] + 2*A[3]; /* Michael Somos, Aug 29 2014 */

d:=proc(r,m,n) local i,t1; t1:=0; for i from 1 to n do if n mod i = 0 and i-r mod m = 0 then t1:=t1+1; fi; od: t1; end; [seq(2*(d(1,8,n)+d(3,8,n)-d(5,8,n)-d(7,8,n)),n=1..120)];

a[ n_] := SeriesCoefficient[ EllipticTheta[ 3, 0, q] EllipticTheta[ 3, 0, q^2], {q, 0, n}]; (* Michael Somos, Sep 09 2012 *)
a[ n_] := If[ n < 1, Boole[ n == 0], 2 DivisorSum[ n, KroneckerSymbol[ -2, #] &]]; (* Michael Somos, Aug 29 2014 *)
a[ n_] := SeriesCoefficient[ (QPochhammer[ q^2] QPochhammer[ q^4])^3 / (QPochhammer[ q] QPochhammer[ q^8])^2, {q, 0, n}]; (* Michael Somos, Aug 29 2014 *)

{a(n) = if( n<1, n==0, 2 * (issquare(n) - issquare(2*n) + 2 * sum( i=1, sqrtint(n\2), issquare(n - 2*i^2))))};

{a(n) = if( n<1, n==0, 2 * sumdiv( n, d, kronecker( -2, d)))}; /* Michael Somos, Aug 23 2005 */

{a(n) = if( n<1, n==0, 2 * qfrep([ 1, 0; 0, 2], n)[n])}; /* Michael Somos, Aug 23 2005 */

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^-2 * eta(x^2 + A)^3 * eta(x^4 + A)^3 * eta(x^8 + A)^-2, n))};

Q = DiagonalQuadraticForm(ZZ, [1,2]); Q.representation_number_list(104); # Peter Luschny, Jun 20 2014

Fine gives an explicit formula for a(n) in terms of the divisors of n.
Euler transform of period 8 sequence [ 2, -1, 2, -4, 2, -1, 2, -2, ...].
Expansion of (eta(q^2) * eta(q^4))^3 / (eta(q) * eta(q^8))^2 in powers of q.
Coefficients in expansion of Sum_{i,j=-inf..inf} q^(i^2 + 2*j^2).
G.f. = s(2)^3*s(4)^3/(s(1)^2*s(8)^2), where s(k) := subs(q=q^k, eta(q)), where eta(q) is Dedekind's function, cf. A010815. [Fine]
G.f.: 1 + 2 * Sum_{k>0} Kronecker(-2, n) * x^k / (1 - x^k) = 1 + 2 * Sum_{k>0} (x^k + x^(3*k)) / (1 + x^(4*k)).
G.f.: theta_3(q) * theta_3(q^2) = Product_{k>0} (1 + x^(2*k)) * ((1 + x^k) * (1 - x^(2*k)) / (1 + x^(4*k)))^2.
From Michael Somos, Oct 23 2006: (Start)
Moebius transform is period 8 sequence [ 2, 0, 2, 0, -2, 0, -2, 0, ...].
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = (u1 - 3*u3) * (u1 - u2 - u3 + u6) - (u2 - 3*u6) * (u1 - 2*u2 - u3 + 2*u6). (End)
a(n) = 2 * A002325(n) unless n = 0.
G.f. is a period 1 Fourier series which satisfies f(-1 / (8 t)) = 8^(1/2) (t/i) f(t) where q = exp(2 Pi i t). - Michael Somos, Sep 09 2012
From Michael Somos, Aug 29 2014: (Start)
Expansion of phi(q) * phi(q^2) in powers of q where phi() is a Ramanujan theta function.
a(2*n) = a(n). a(2*n + 1) = 2 * A113411(n). (End)
From Michael Somos, May 17 2015: (Start)
a(n) = A028572(4*n) = A133692(2*n) = A139093(8*n) = A226225(8*n) = A226240(4*n) = A242609(4*n) = A245572(4*n) / 3 = (-1)^floor((n + 1)/2) * A082564(n).
a(8*n + 5) = a(8*n + 7) = 0. a(8*n + 1) = 2 * A112603(n). a(8*n + 3) = 4 * A033761(n). (End)
a(0) = 1, a(n) = 2 * b(n) for n > 0, where b() is multiplicative with b(2^e) = 1, b(p^e) = e + 1 if p == 1, 3 (mod 8), b(p^e) = (1 + (-1)^e)/2 if p == 5, 7 (mod 8). - Jianing Song, Sep 04 2018 [Corrected by Jeremy Lovejoy, Nov 12 2024]
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=0..m} a(k) = Pi/sqrt(2) = 2.221441... (A247719). - Amiram Eldar, Dec 16 2023

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)

A063448 Decimal expansion of Pi * sqrt(2).

Original entry on oeis.org

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

Offset: 1

Jason Earls, Jul 24 2001

Hypotenuse of the right triangle with legs Pi and Pi. - Zak Seidov, May 04 2005
Circumference of the circumcircle of the unit square. - Jonathan Sondow, Nov 23 2017
Half-perimeter of the closed curve with implicit Cartesian equation x^2 + y^2 = abs(x) + abs(y). - Stefano Spezia, Oct 20 2020

			4.4428829381583662470158809900606936986146216893756902230853...

Harry J. Smith, Table of n, a(n) for n = 1..20000
Peter Bala, Series for sqrt(2)*Pi
Eric Weisstein's World of Mathematics, Gauss Multiplication Formula.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers

Cf. A063447 (continued fraction), A093954, A153799, A193887, A244976, A247719.

RealDigits[N[Pi*Sqrt[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Mar 21 2011*)

PARI
```
\p 400; Pi * sqrt(2)
```

default(realprecision, 20080); x=Pi*sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b063448.txt", n, " ", d)) \\ Harry J. Smith, Aug 21 2009

# Use some guard digits when computing.
# BBP formula (1/8) P(1, 64, 12, (32, 0, 8, 0, 8, 0, -4, 0, -1, 0, -1, 0))
from decimal import Decimal as dec, getcontext
def BBPpisqrt2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        twk = dec(12 * k)
        s += f * ( dec(32) / (twk + 1) + dec(8)  / (twk + 3)
                 + dec(8)  / (twk + 5) - dec(4)  / (twk + 7)
                 - dec(1)  / (twk + 9) - dec(1)  / (twk + 11))
        f /= g
    return s / dec(8)
print(BBPpisqrt2(200))  # Peter Luschny, Nov 03 2023

Equals Gamma(1/4)*Gamma(3/4). - Jean-François Alcover, Nov 24 2014
From Amiram Eldar, Aug 06 2020: (Start)
Equals Integral_{x=0..oo} log(1 + 1/x^4) dx.
Equals Integral_{x=0..oo} log(1 + 2/x^2) dx.
Equals Integral_{x=-oo..oo} exp(x/4)/(exp(x) + 1) dx.
Equals Integral_{x=0..2*Pi} 1/(cos(x)^2 + 1) dx = Integral_{x=0..2*Pi} 1/(sin(x)^2 + 1) dx. (End)
From Andrea Pinos, Jul 03 2023: (Start)
Equals (Product_{k=1..4} Gamma(k/8)*Gamma(1 - k/8))^(1/4).
General result: 2*Pi/(4*y)^(1/(2*y)) = (Product_{k=1..y} Gamma(k/(2*y))*Gamma(1 - k/(2*y)) )^(1/y). (End)
From Peter Bala, Oct 22 2023: (Start)
sqrt(2)*Pi = 4 + 8*Sum_{n >= 0} (-1)^n/(16*n^2 + 32*n + 15). See A141759.
In the following the Eisenstein summation convention is assumed: that is,
Sum_{n = -oo..oo} means Limit_{N -> oo} Sum_{n = -N..N}:
sqrt(2)*Pi = 4*Sum_{n = -oo..oo} (-1)^n/(4*n + 1).
More generally, it appears that for k >= 0, k not of the form 4*m + 1,
sqrt(2)*Pi = -sign(cos(Pi*(k - 3)/4)) * 4*(2^floor(k/2))*k! * Sum_{n = -oo..oo} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 2*k + 1)) (verified up to k = 50).
sqrt(2)*Pi = (2^4)*Sum_{n >= 0} (-1)^n * (2*n + 1)/((4*n + 1)*(4*n + 3)) = 512/105 - (2^6)*4!*Sum_{n >= 0} (-1)^n * (2*n + 3)/((4*n + 1)*(4*n + 3)*...*(4*n + 11)).
sqrt(2)*Pi = 4 + (2^3)*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*(4*n + 5)) = 1408/315 - (2^5)*5!*Sum_{n >= 0} (-1)^n * (4*n + 1)/((4*n + 1)*(4*n + 3)*...*(4*n + 13)).
sqrt(2)*Pi = 16/3 - (2^4)*3!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*(4*n + 5)*(4*n + 7)) = 14848/3465 + (2^6)*7!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 3)*...*(4*n + 15)). (End)
From Peter Bala, Nov 19 2023: (Start)
sqrt(2)*Pi = 512*Sum_{k >= 1} (-1)^(k+1) * k^2/((16*k^2 - 1)*(16*k^2 - 9)).
This is the case n = 1 of the more general result: for n >= 1,
sqrt(2)*Pi = (-1)^(n+1) * 2^(n+7) * (2*n)!/(2*n - 1) * Sum_{k >= 1} (-1)^(k+1) * k^2/( Product_{i = 0..n} (16*k^2 - (2*i+1)^2) ). Cf. A334422. (End)
Equals Integral_{x=-oo..oo} (x^2 + 1)/(x^4 + 1) dx. - Kritsada Moomuang, Jun 04 2025

Edited by N. J. A. Sloane, May 05 2007

Corrected by Neven Juric, Apr 24 2008

A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 05 2010

Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - Peter Bala, Sep 23 2016

			0.2437477471996805241799750836323027110...
From _Peter Bala_, Sep 23 2016: (Start)
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181049. The numbers 1, -1, -3, 11, 57, -361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.

Cf. A113476, A001586, A093954, A181048, A212435, A247719.

C := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // G. C. Greubel, Nov 28 2017

Mathematica

First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* Michael De Vlieger, Oct 07 2015 *)

PARI

default(realprecision, 106); eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

PARI

(Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ G. C. Greubel, Nov 28 2017

Formula

Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).

Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - Peter Bala, Sep 23 2016

Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - Peter Bala, Dec 01 2021

From Peter Bala, Mar 03 2024: (Start)

Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.

Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).

Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)

Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - Vaclav Kotesovec, Jun 13 2024

A244976 Decimal expansion of Pi/(8*sqrt(2)).

Original entry on oeis.org

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

Offset: 0

Jean-François Alcover, Jul 09 2014

			0.277680183634897890438492561878793356163413855585980638942837225434777...

George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press (2006), Chapter 13 A Master Formula, p. 250.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Beta Function.
Index entries for transcendental numbers

Cf. A063448, A247719, A093954, A193887.

RealDigits[Pi/(8*Sqrt[2]), 10, 105] // First

Pi/(8*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

Equals Integral_{x=0..1} (x^2*(1 + x^2))/(1 + x^4)^2 dx.
Equals beta(3/2, 1/2)/(4*sqrt(2)), where 'beta' is Euler's beta function.
Equals Sum_{k >= 0} (-1)^k * (2*k + 1)/((4*k + 1)*(4*k + 3)). - Peter Bala, Sep 21 2016
Equals Integral_{x>=0} 1/(x^2 + 2)^2 dx. - Amiram Eldar, Nov 16 2021

A193887 Decimal expansion of Pi * sqrt(2)/8.

Original entry on oeis.org

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

Offset: 0

Alonso del Arte, Aug 07 2011

This number arises as an addend in one way of giving the closed form of sum(k>=0, (-1)^k/(4*k + 1) ), for example, in Spiegel et al. (2009).

			0.55536036726979578088...

Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

G. C. Greubel, Table of n, a(n) for n = 0..10000
Piotr Garbaczewski and Vladimir A. Stephanovich, Semigroup modeling of confined Levy flights, arXiv:1106.1530 [cond-mat.stat-mech], 2011, p. 8, equation 40.
Piotr Garbaczewski and Vladimir A. Stephanovich, Dynamics of Confined Lévy Flights in Terms of (Lévy) Semigroups, Acta Phys. Pol. B 43, 977-999 (2012). See p. 992.
Index entries for transcendental numbers.

Cf. A181048, A063448, A247719, A093954, A244976.

R:= RealField(); Pi(R)*Sqrt(2)/8; // G. C. Greubel, Feb 02 2018

RealDigits[(Pi Sqrt[2])/8, 10, 100][[1]]

Pi*sqrt(2)/8 \\ G. C. Greubel, Feb 02 2018

Equals Pi/(4*sqrt(2)).
Equals Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)). - Peter Bala, Sep 21 2016
From Amiram Eldar, Aug 15 2020: (Start)
Equals Integral_{x=0..oo} 1/(x^2 + 8) dx.
Equals Integral_{x=0..oo} 1/(8*x^2 + 1) dx.
Equals Integral_{x=0..oo} 1/(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) dx. (End)

A300727 Decimal expansion of the total harmonic distortion (THD) of the sawtooth signal filtered by a 2nd-order low-pass filter.

Original entry on oeis.org

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

Offset: 0

Iaroslav V. Blagouchine, Mar 11 2018

See formula (34) in Blagouchine & Moreau link.

			0.1811416137938290040802181055813016784438928351595635...

I. V. Blagouchine and E. Moreau, Analytic Method for the Computation of the Total Harmonic Distortion by the Cauchy Method of Residues. IEEE Trans. Commun., vol. 59, no. 9, pp. 2478-2491, 2011. PDF file.

Cf. A247719 (Pi/sqrt(2)), A300690, A300713, A300714.

format long; sqrt(sqrt(pi*(cot(pi/sqrt(2))*coth(pi/sqrt(2))^2-cot(pi/sqrt(2))^2*coth(pi/sqrt(2))-cot(pi/sqrt(2))-coth(pi/sqrt(2)))/((cot(pi/sqrt(2))^2+coth(pi/sqrt(2))^2)*sqrt(2))+(1/3)*pi^2-1))

evalf(sqrt(Pi*(cot(Pi/sqrt(2))*coth(Pi/sqrt(2))^2-cot(Pi/sqrt(2))^2*coth(Pi/sqrt(2))-cot(Pi/sqrt(2))-coth(Pi/sqrt(2)))/((cot(Pi/sqrt(2))^2+coth(Pi/sqrt(2))^2)*sqrt(2))+(1/3)*Pi^2-1), 120)

RealDigits[Sqrt[Pi*(Cot[Pi/Sqrt[2]]*Coth[Pi/Sqrt[2]]^2-Cot[Pi/Sqrt[2]]^2*Coth[Pi/Sqrt[2]]-Cot[Pi/Sqrt[2]]-Coth[Pi/Sqrt[2]])/((Cot[Pi/Sqrt[2]]^2+Coth[Pi/Sqrt[2]]^2)*Sqrt[2])+(1/3)*Pi^2-1], 10, 120][[1]]

s2=sqrt(2);
A=Pi/s2;
B=1+2/(exp(2*A)-1)
C=1/tan(A);
sqrt(Pi*(B^2*C-B*C^2-C-B)/((C^2+B^2)*s2) + Pi^2/3 - 1) \\ Charles R Greathouse IV, Mar 11 2018

Equals sqrt(Pi*(cot(Pi/sqrt(2))*coth(Pi/sqrt(2))^2-cot(Pi/sqrt(2))^2*coth(Pi/sqrt(2))-cot(Pi/sqrt(2))-coth(Pi/sqrt(2)))/((cot(Pi/sqrt(2))^2+coth(Pi/sqrt(2))^2)*sqrt(2))+(1/3)*Pi^2-1).

A193355 Decimal expansion of Pi/(2 + 2*sqrt(2)).

Original entry on oeis.org

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

Offset: 0

Frank M Jackson, Jul 24 2011

This is the first of the three angles (in radians) of a unique triangle that is right angled and where the angles are in a harmonic progression: Pi/(2+2*sqrt(2)) (this sequence), Pi/(2+sqrt(2)) (A193373), Pi/2 (A019669). The angles (in degrees) are approximately 37.279, 52.721, 90. The common difference between the denominators of the harmonic progression is sqrt(2).

			0.6506451422...

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for transcendental numbers

SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/(2 + 2*Sqrt(2)); // G. C. Greubel, Sep 29 2018

evalf(Pi/(2+2*sqrt(2)),120); # Muniru A Asiru, Sep 30 2018

N[Pi/(2 + 2*Sqrt[2]), 100]
RealDigits[Pi/(2 + 2*Sqrt[2]), 10, 100][[1]] (* G. C. Greubel, Sep 29 2018 *)

default(realprecision,100); Pi/(2+2*sqrt(2))

Equals Pi/(2+2*sqrt(2)).
Equals Integral_{x=0..Pi/2} cos(x)^2/(1 + sin(x)^2) dx = Integral_{x=0..Pi/2} sin(x)^2/(1 + cos(x)^2) dx. - Amiram Eldar, Aug 16 2020
Equals 4*Sum_{k >= 0} (-1)^k/((4*k + 1)*(4*k + 2)*(4*k + 3)). - Peter Bala, Jul 15 2024
Equals Integral_{x=0..1} sqrt(1 - x^2)/(1 + x^2) dx. - Kritsada Moomuang, Jun 05 2025
Equals A247719 - A019669 = A000796*A268683. - R. J. Mathar, Jul 22 2025

A193373 Decimal expansion of Pi/(2 + sqrt(2)).

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Jul 24 2011

			0.92015118451061011495470288824915603488985855468726...

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for transcendental numbers

Cf. A193355.

SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/(2 + Sqrt(2)); // G. C. Greubel, Sep 29 2018

evalf(Pi/(2+sqrt(2)),120); # Muniru A Asiru, Sep 30 2018

RealDigits[Pi/(2 + Sqrt[2]), 10, 100][[1]] (* G. C. Greubel, Sep 29 2018 *)

default(realprecision, 99); vecextract(eval(Vec(Str(Pi/(2+sqrt(2))))),"3..")

From Amiram Eldar, May 08 2021: (Start)
Equals Integral_{x>=0} log(1 + 1/(1 + 2*x^2)) dx.
Equals Integral_{x=0..Pi} 1/(1 + csc(x)^2) dx. (End)
Equals A000796 - A247719 = A000796*A268682. - R. J. Mathar, Jul 22 2025

A380142 Decimal expansion of the imaginary part of (-1)^sqrt(i), where i is the imaginary unit.

Original entry on oeis.org

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

Offset: 0

Hugo Pfoertner, Jan 23 2025

			0.0862950481802362811285347518373265409649389366268...

Paolo Xausa, Table of n, a(n) for n = 0..10000
Math Beast, Oxford university entrance exam question, YouTube video, 2025.

A380141 is the real part.

Cf. A247719.

evalf[140](Im((-1)^sqrt(I)));  # Alois P. Heinz, Jan 23 2025

First[RealDigits[Im[(-1)^Sqrt[I]], 10, 100, -1]] (* Paolo Xausa, Feb 26 2025 *)

PARI
```
imag((-1)^sqrt(I))
```

Equals sin(Pi/sqrt(2))/exp(Pi/sqrt(2)).

cp's OEIS Frontend

A033715 Number of integer solutions (x, y) to the equation x^2 + 2y^2 = n.

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

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A063448 Decimal expansion of Pi * sqrt(2).

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A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

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A244976 Decimal expansion of Pi/(8*sqrt(2)).

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A193887 Decimal expansion of Pi * sqrt(2)/8.

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