cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A003881 Decimal expansion of Pi/4.

Original entry on oeis.org

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

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Author

Keywords

Comments

Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Least positive solution to sin(x) = cos(x). - Franklin T. Adams-Watters, Jun 17 2014
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023

Examples

			0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - _Peter Bala_, Nov 15 2016
		

References

  • Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 437.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 6.3 and 8.4, pp. 429 and 492.
  • Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
  • J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 119.

Crossrefs

Cf. A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A001622.

Programs

  • Haskell
    -- see link: Literate Programs
    import Data.Char (digitToInt)
    a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
       machin = 4 * arccot 5 unity - arccot 239 unity
       unity = 10 ^ (len + 10)
       arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
         arccot' x unity summa xpow n sign
        | term == 0 = summa
        | otherwise = arccot'
          x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
        where term = xpow `div` n
    -- Reinhard Zumkeller, Nov 20 2012
    
  • Magma
    R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018
  • Maple
    evalf(Pi/4) ;
  • Mathematica
    RealDigits[N[Pi/4,6! ]]  (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
    (* PROGRAM STARTS *)
    (* Define the nested radicals a_k by recurrence *)
    a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
    (* Example of Pi/4 approximation at K = 100 *)
    Print["The actual value of Pi/4 is"]
    N[Pi/4, 40]
    Print["At K = 100 the approximated value of Pi/4 is"]
    K := 100;  (* the truncating integer *)
    N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
    (* Error terms for Pi/4 approximations *)
    Print["Error terms for Pi/4"]
    k := 1; (* initial value of the index k *)
    K := 10; (* initial value of the truncating integer K *)
    sqn := {}; (* initiate the sequence *)
    AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
    While[K <= 30,
    AppendTo[sqn, {K,
       N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
        N}]; K++]
    Print[MatrixForm[sqn]]
    (* Sanjar Abrarov, Jan 09 2017 *)
  • PARI
    Pi/4 \\ Charles R Greathouse IV, Jul 07 2014
    
  • SageMath
    # Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
    def FastLeibniz(n):
        b = 2^(2*n-1); c = b; s = 0
        for k in range(n-1,-1,-1):
            t = 2*k+1
            s = s + c/t if is_even(k) else s - c/t
            b *= (t*(k+1))/(2*(n-k)*(n+k))
            c += b
        return s/c
    A003881 = RealField(3333)(FastLeibniz(1330))
    print(A003881)  # Peter Luschny, Nov 20 2012
    

Formula

Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals lim_{n->oo} n*A001586(n-1)/A001586(n) (conjecture). - Mats Granvik, Feb 23 2011
Equals Integral_{x=0..1} 1/(1+x^2) dx. - Gary W. Adamson, Jun 22 2003
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals (-digamma(1/4) + digamma(3/4))/4. - Jean-François Alcover, May 31 2013
Equals Sum_{n>=0} (-1)^n/(2*n+1). - Geoffrey Critzer, Nov 03 2013
Equals Integral_{x=0..1} Product_{k>=1} (1-x^(8*k))^3 dx [cf. A258414]. - Vaclav Kotesovec, May 30 2015
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
From Peter Bala, Nov 15 2016: (Start)
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
From Peter Bala, Nov 05 2019: (Start)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
From Amiram Eldar, Aug 19 2020: (Start)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
With offset 1, equals 5 * Pi / 2. - Sean A. Irvine, Aug 19 2021
Equals (1/2)!^2 = Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
Equals Sum_{n>=1} eta(n)/2^n, where eta(n) is the Dirichlet eta function. - Antonio Graciá Llorente, Oct 04 2024
Equals Product_{k>=2} ((k + 1)^(k*(2*k + 1))*(k - 1)^(k*(2*k - 1)))/k^(4*k^2). - Antonio Graciá Llorente, Apr 12 2025
Equals Integral_{x=sqrt(2)..oo} dx/(x*sqrt(x^2 - 1)). - Kritsada Moomuang, May 29 2025

Extensions

a(98) and a(99) corrected by Reinhard Zumkeller, Nov 20 2012

A101455 a(n) = 0 for even n, a(n) = (-1)^((n-1)/2) for odd n. Periodic sequence 1,0,-1,0,...

Original entry on oeis.org

0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0
Offset: 0

Views

Author

Gerald McGarvey, Jan 20 2005

Keywords

Comments

Called X(n) (i.e., Chi(n)) in Hardy and Wright (p. 241), who show that X(n*m) = X(n)*X(m) for all n and m (i.e., X(n) is completely multiplicative) since (n*m - 1)/2 - (n - 1)/2 - (m - 1)/2 = (n - 1)*(m - 1)/2 == 0 (mod 2) when n and m are odd.
Same as A056594 but with offset 1.
From R. J. Mathar, Jul 15 2010: (Start)
The sequence is the non-principal Dirichlet character mod 4. (The principal character is A000035.)
Associated Dirichlet L-functions are for example L(1,chi) = Sum_{n>=1} a(n)/n = A003881, or L(2,chi) = Sum_{n>=1} a(n)/n^2 = A006752, or L(3,chi) = Sum_{n>=1} a(n)/n^3 = A153071. (End)
a(n) is a strong elliptic divisibility sequence t_n as given in [Kimberling, p. 16] where x = 0, y = -1, z is arbitrary. - Michael Somos, Nov 27 2019

Examples

			G.f. = x - x^3 + x^5 - x^7 + x^9 - x^11 + x^13 - x^15 + x^17 - x^19 + x^21 + ...
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986, page 139, k=4, Chi_2(n).
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 5th ed., Oxford Univ. Press, 1979, p. 241.

Crossrefs

Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), A289741 (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), this sequence (d=-4), A102283 (d=-3), A080891 (d=5), A091337 (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), A322796 (d=24).

Programs

  • GAP
    a := [1, 0];; for n in [3..10^2] do a[n] := a[n-2]; od; a; # Muniru A Asiru, Feb 02 2018
    
  • Magma
    m:=75; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x/(1+x^2))); // G. C. Greubel, Aug 23 2018
    
  • Maple
    a := n -> `if`(n mod 2=0, 0, (-1)^((n-1)/2)):
    seq(a(n), n=1..10^3); # Muniru A Asiru, Feb 02 2018
  • Mathematica
    a[ n_] := {1, 0, -1, 0}[[ Mod[ n, 4, 1]]]; (* Michael Somos, Jan 13 2014 *)
    LinearRecurrence[{0, -1}, {1, 0}, 75] (* G. C. Greubel, Aug 23 2018 *)
  • PARI
    {a(n) = if( n%2, (-1)^(n\2))}; /* Michael Somos, Sep 02 2005 */
    
  • PARI
    {a(n) = kronecker( -4, n)}; /* Michael Somos, Mar 30 2012 */
    
  • Python
    def A101455(n): return (0,1,0,-1)[n&3] # Chai Wah Wu, Jun 21 2024

Formula

Multiplicative with a(2^e) = 0, a(p^e) = (-1)^((p^e-1)/2) otherwise. - Mitch Harris May 17 2005
Euler transform of length 4 sequence [0, -1, 0, 1]. - Michael Somos, Sep 02 2005
G.f.: (x - x^3)/(1 - x^4) = x/(1 + x^2). - Michael Somos, Sep 02 2005
G.f. A(x) satisfies: 0 = f(A(x), A(x^2)) where f(u, v) = v - u^2 * (1 + 2*v). - Michael Somos, Aug 04 2011
a(n + 4) = a(n), a(n + 2) = a(-n) = -a(n), a(2*n) = 0, a(2*n + 1) = (-1)^n for all n in Z. - Michael Somos, Aug 04 2011
a(n + 1) = A056594(n). - Michael Somos, Jan 13 2014
REVERT transform is A126120. STIRLING transform of A009454. BINOMIAL transform is A146559. BINOMIAL transform of A009116. BIN1 transform is A108520. MOBIUS transform of A002654. EULER transform is A111335. - Michael Somos, Mar 30 2012
Completely multiplicative with a(p) = 2 - (p mod 4). - Werner Schulte, Feb 01 2018
a(n) = (-(n mod 2))^binomial(n, 2). - Peter Luschny, Sep 08 2018
a(n) = sin(n*Pi/2) = Im(i^n) where i is the imaginary unit. - Jianing Song, Sep 09 2018
From Jianing Song, Nov 14 2018: (Start)
a(n) = ((-4)/n) (or more generally, ((-4^i)/n) for i > 0), where (k/n) is the Kronecker symbol.
E.g.f.: sin(x).
Dirichlet g.f. is the Dirichlet beta function.
a(n) = A091337(n)*A188510(n). (End)

Extensions

a(0) prepended by Jianing Song, Nov 14 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

Views

Author

Eric W. Weisstein, Apr 19 2004

Keywords

Comments

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

Examples

			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)
		

References

  • 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.

Crossrefs

Programs

  • Maple
    simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) );  # Peter Bala, Mar 09 2015
  • Mathematica
    RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* Michael De Vlieger, Sep 23 2016 and slightly modified by Robert G. Wilson v, Jul 23 2018 *)
  • PARI
    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

Formula

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)

A016755 Odd cubes: a(n) = (2*n + 1)^3.

Original entry on oeis.org

1, 27, 125, 343, 729, 1331, 2197, 3375, 4913, 6859, 9261, 12167, 15625, 19683, 24389, 29791, 35937, 42875, 50653, 59319, 68921, 79507, 91125, 103823, 117649, 132651, 148877, 166375, 185193, 205379, 226981, 250047, 274625, 300763, 328509, 357911, 389017, 421875
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A010014. - Jani Melik, May 20 2013
Terms end in the repeating sequence 1, 7, 5, 3, 9, ... - Melvin Peralta, Jul 08 2015

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3.

Crossrefs

Programs

Formula

Sum_{n >= 0} 1/a(n) = 7 * zeta(3) / 8.
G.f.: (1+23*x+23*x^2+x^3)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 02 2012
a(n) = A000578(A005408(n)). - Michel Marcus, Jul 09 2015
E.g.f.: exp(x)*(1 + 26*x + 36*x^2 + 8*x^3). See A154537, row n=3. - Wolfdieter Lang, Mar 12 2017
From Bruce J. Nicholson, Dec 08 2019: (Start)
a(n) = 24 * A000330(n) + A005408(n).
a(n) = 2 * A005917(n+1) - A005408(n). (End)
Sum_{n>=0} (-1)^n/a(n) = Pi^3/32 (A153071). - Amiram Eldar, Oct 10 2020
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/12)*(1 + sqrt(2)*cosh(sqrt(3)*Pi/4)) (Chamberland and Straub, 2013). - Amiram Eldar, Jan 26 2024

A233091 Decimal expansion of Sum_{i>=0} 1/(2*i+1)^3.

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Dec 04 2013

Keywords

Comments

This constant is irrational. - Charles R Greathouse IV, Feb 03 2025

Examples

			1.0517997902646449997247708913225187419193630057979365215682376109241...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3, p. 42.

Crossrefs

Cf. A002117: zeta(3); A197070: 3*zeta(3)/4; A233090: 5*zeta(3)/8.
Cf. A153071: sum( i >= 0, (-1)^i/(2*i+1)^3 ).
Cf. A251809: sum( i >= 0, (-1)^floor(i/2)/(2*i+1)^3 ).
Cf. A016755.

Programs

  • Mathematica
    RealDigits[7 Zeta[3]/8, 10, 90][[1]]
  • PARI
    7*zeta(3)/8 \\ Stefano Spezia, Oct 31 2024

Formula

Equals 7*zeta(3)/8.
Also equals -(1/16)*PolyGamma(2, 1/2). - Jean-François Alcover, Dec 18 2013
Equals Integral_{x=0..Pi/2} x * log(tan(x)) dx. - Amiram Eldar, Jun 29 2020
Equals Integral_{x=0..1} arcsin(x)*arccos(x)/x dx. - Amiram Eldar, Aug 03 2020

A050470 a(n) = Sum_{d|n, n/d == 1 (mod 4)} d^2 - Sum_{d|n, n/d == 3 (mod 4)} d^2.

Original entry on oeis.org

1, 4, 8, 16, 26, 32, 48, 64, 73, 104, 120, 128, 170, 192, 208, 256, 290, 292, 360, 416, 384, 480, 528, 512, 651, 680, 656, 768, 842, 832, 960, 1024, 960, 1160, 1248, 1168, 1370, 1440, 1360, 1664, 1682, 1536, 1848, 1920, 1898, 2112, 2208, 2048, 2353, 2604
Offset: 1

Views

Author

N. J. A. Sloane, Dec 23 1999

Keywords

Comments

Number 7 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).
Multiplicative because it is the Dirichlet convolution of A000290 = n^2 and A101455 = [1 0 -1 0 1 0 -1 ...], which are both multiplicative. - Christian G. Bower, May 17 2005

Examples

			G.f. = q + 4*q^2 + 8*q^3 + 16*q^4 + 26*q^5 + 32*q^6 + 48*q^7 + 64*q^8 + ...
		

Crossrefs

Programs

  • Haskell
    a050470 n = a050461 n - a050465 n  -- Reinhard Zumkeller, Mar 06 2012
    
  • Magma
    Basis( ModularForms( Gamma1(4), 3), 51) [2]; /* Michael Somos, May 17 2015 */
    
  • Mathematica
    a[ n_] := SeriesCoefficient[ q (QPochhammer[ q^2]^3 (QPochhammer[ q^4] / QPochhammer[ q])^2)^2, {q, 0, n}]; (* Michael Somos, May 17 2015 *)
    a[ n_] := SeriesCoefficient[ (EllipticTheta[ 3, 0, q] EllipticTheta[ 2, 0, q]^2 / 4)^2, {q, 0, n}]; (* Michael Somos, May 17 2015 *)
    a[ n_] := If[ n < 1, 0, Sum[ d^2 Mod[n/d, 2] (-1)^Quotient[n/d, 2], {d, Divisors@n}]]; (* Michael Somos, May 17 2015 *)
    s[n_] := If[OddQ[n], (-1)^((n-1)/2), 0]; (* A101455 *)
    f[p_, e_] := (p^(2*e+2) - s[p]^(e+1))/(p^2 - s[p]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2023 *)
  • PARI
    {a(n) = if( n<1, 0, sumdiv( n, d, d^2 * (n/d%2) * (-1)^(n/d\2)))};
    
  • PARI
    {a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( eta(x^2 + A)^6 * (eta(x^4 + A) / eta(x + A))^4, n))}; /* Michael Somos, May 17 2015 */
    
  • Python
    from math import prod
    from sympy import factorint
    def A050470(n): return prod((p**(e+1<<1)-(m:=(0,1,0,-1)[p&3]))//(p**2-m) for p, e in factorint(n).items()) # Chai Wah Wu, Jun 21 2024

Formula

G.f.: Sum_{n>=1} n^2*x^n/(1+x^(2*n)). - Vladeta Jovovic, Oct 16 2002
From Michael Somos, Aug 08 2005: (Start)
Euler transform of period 4 sequence [ 4, -2, 4, -6, ...].
Expansion of eta(q^2)^6 * eta(q^4)^4 / eta(q)^4 in powers of q.
G.f.: x Product_{k>0} (1 + x^k)^4 * (1 - x^(2*k))^2 * (1 - x^(4*k))^4.
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = u*w * (u - 8*v) * (v - 4*w) - v^2 * (v - 8*w)^2. (End)
G.f.: Sum_{k>0} Kronecker(-4, k) * x^k * (1 + x^k) / (1 - x^k)^3. - Michael Somos, Sep 02 2005
Expansion of q * phi(q)^2 * psi(q^2)^4 in powers of q where phi(), psi() are Ramanujan theta functions. - Michael Somos, Aug 15 2007
G.f. is a period 1 Fourier series which satisfies f(-1 / (4 t)) = (1/2) (t/i)^3 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A120030.
a(n) = A050461(n) - A050465(n). - Reinhard Zumkeller, Mar 06 2012
Multiplicative with a(p^e) = ((p^2)^(e+1) - Chi(p)^(e+1))/(p^2 - Chi(p)), Chi = A101455. - Jianing Song, Oct 30 2019
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Pi^3/32 (A153071). - Amiram Eldar, Nov 04 2023
a(n) = Sum_{d|n} (n/d)^2*sin(d*Pi/2). - Ridouane Oudra, Sep 26 2024

A175571 Decimal expansion of the Dirichlet beta function of 5.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Jul 15 2010

Keywords

Comments

The value of the Dirichlet L-series L(m=4,r=2,s=4), see arXiv:1008.2547.

Examples

			0.99615782807708806400631936...
		

References

  • L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 308.

Crossrefs

Cf. A003881 (beta(1)=Pi/4), A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A101455.

Programs

  • Maple
    DirichletBeta := proc(s) 4^(-s)*(Zeta(0,s,1/4)-Zeta(0,s,3/4)) ; end proc: x := DirichletBeta(5) ; x := evalf(x) ;
  • Mathematica
    RealDigits[ DirichletBeta[5], 10, 105] // First (* Jean-François Alcover, Feb 20 2013, updated Mar 14 2018 *)
  • PARI
    5*Pi^5/1536 \\ Charles R Greathouse IV, Jan 31 2018
    
  • PARI
    beta(x)=(zetahurwitz(x,1/4)-zetahurwitz(x,3/4))/4^x
    beta(5) \\ Charles R Greathouse IV, Jan 31 2018

Formula

Equals 5*Pi^5/1536 = Sum_{n>=1} A101455(n)/n^5, where Pi^5 = A092731. [corrected by R. J. Mathar, Feb 01 2018]
Equals Sum_{n>=0} (-1)^n/(2*n+1)^5. - Jean-François Alcover, Mar 29 2013
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p^5)^(-1). - Amiram Eldar, Nov 06 2023

A175572 Decimal expansion of the Dirichlet beta function of 4.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Jul 15 2010

Keywords

Comments

This is the value of the Dirichlet L-series for A101455 at s=4, see arXiv:1008.2547, L(m=4,r=2,s=4).

Examples

			0.988944551741105336108422633...
		

References

  • L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (308).

Crossrefs

Cf. A003881 (beta(1)=Pi/4), A006752 (beta(2)=Catalan), A153071 (beta(3)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A101455.

Programs

  • Maple
    DirichletBeta := proc(s) 4^(-s)*(Zeta(0,s,1/4)-Zeta(0,s,3/4)) ; end proc: x := DirichletBeta(4) ; x := evalf(x) ;
  • Mathematica
    RealDigits[ DirichletBeta[4], 10, 105] // First (* Jean-François Alcover, Feb 11 2013, updated Mar 14 2018 *)
  • PARI
    beta(x)=(zetahurwitz(x,1/4)-zetahurwitz(x,3/4))/4^x
    beta(4) \\ Charles R Greathouse IV, Jan 31 2018

Formula

Equals Sum_{n>=1} A101455(n)/n^4. [corrected by R. J. Mathar, Feb 01 2018]
Equals (PolyGamma(3, 1/4) - PolyGamma(3, 3/4))/1536. - Jean-François Alcover, Jun 11 2015
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p^4)^(-1). - Amiram Eldar, Nov 06 2023

A175570 Decimal expansion of the Dirichlet beta function of 6.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Jul 15 2010

Keywords

Examples

			0.998685222218438135441600...
		

References

  • L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 308.

Crossrefs

Cf. A003881 (beta(1)=Pi/4), A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A101455.

Programs

  • Maple
    DirichletBeta := proc(s) 4^(-s)*(Zeta(0,s,1/4)-Zeta(0,s,3/4)) ; end proc: x := DirichletBeta(6) ; x := evalf(x) ;
  • Mathematica
    RealDigits[ DirichletBeta[6], 10, 105] // First (* Jean-François Alcover, Feb 11 2013, updated Mar 14 2018 *)
  • PARI
    beta(x)=(zetahurwitz(x,1/4)-zetahurwitz(x,3/4))/4^x
    beta(6) \\ Charles R Greathouse IV, Jan 31 2018
    
  • PARI
    sumpos(n=1,(12288*n^5 - 30720*n^4 + 33280*n^3 - 19200*n^2 + 5808*n - 728)/(16777216*n^12 - 100663296*n^11 + 270532608*n^10 - 429916160*n^9 + 449249280*n^8 - 324796416*n^7 + 166445056*n^6 - 60899328*n^5 + 15793920*n^4 - 2833920*n^3 + 334368*n^2 - 23328*n + 729),1) \\ Charles R Greathouse IV, Feb 01 2018

Formula

Equals Sum_{n>=1} A101455(n)/n^6. [see arxiv:1008.2547, L(m=4,r=2,s=6)] [corrected by R. J. Mathar, Feb 01 2018]
Equals (PolyGamma(5, 1/4) - PolyGamma(5, 3/4))/491520. - Jean-François Alcover, Jun 11 2015
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p^6)^(-1). - Amiram Eldar, Nov 06 2023

A251809 Decimal expansion of 3*sqrt(2)*Pi^3/128.

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Dec 10 2014

Keywords

Comments

Equals the value of the Dirichlet L-series of the non-principal character modulo 8 (A188510) at s=3. - Jianing Song, Nov 16 2019

Examples

			1.027722585936858567879256618002255767210100318536997465331084755185...
		

References

  • L. B. W. Jolley, Summation of series, Dover Publications Inc. (New York), 1961, p. 64 (formula 340).

Crossrefs

Cf. A153071: Sum_{i >= 0} (-1)^i/(2i+1)^3.
Cf. A233091: Sum_{i >= 0} 1/(2i+1)^3.

Programs

  • Magma
    R:= RealField(); 3*Sqrt(2)*Pi(R)^3/128; // G. C. Greubel, Jul 27 2018
  • Mathematica
    RealDigits[3 Sqrt[2] Pi^3/128, 10, 90][[1]]
  • PARI
    3*sqrt(2)*Pi^3/128 \\ G. C. Greubel, Jul 27 2018
    

Formula

Equals Sum_{i >= 0} (-1)^floor(i/2)/(2i+1)^3 = +1 +1/3^3 -1/5^3 -1/7^3 +1/9^3 +1/11^3 - ...
Equals Sum_{i >= 1} A188510(i)/i^3 = Sum_{i >= 1} Kronecker(-8,i)/i^3. - Jianing Song, Nov 16 2019
Equals 1/(Product_{p prime == 1 or 3 (mod 8)} (1 - 1/p^3) * Product_{p prime == 5 or 7 (mod 8)} (1 + 1/p^3)). - Amiram Eldar, Dec 17 2023
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