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.

Showing 1-10 of 10 results.

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)

A153071 Decimal expansion of L(3, chi4), where L(s, chi4) is the Dirichlet L-function for the non-principal character modulo 4.

Original entry on oeis.org

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

Views

Author

Stuart Clary, Dec 17 2008

Keywords

Examples

			L(3, chi4) = Pi^3/32 = 0.9689461462593693804836348458469186...
		

References

  • Bruce C. Berndt, Ramanujan's Notebooks, Part II, Springer-Verlag, 1989. See page 293, Entry 25 (iii).
  • Leonhard Euler, Introductio in Analysin Infinitorum, First Part, Articles 175, 284 and 287.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4.1, p. 20.

Crossrefs

Cf. A003881 (beta(1)=Pi/4), A006752 (beta(2)=Catalan), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).

Programs

  • Mathematica
    nmax = 1000; First[ RealDigits[Pi^3/32, 10, nmax] ]
  • PARI
    Pi^3/32 \\ Michel Marcus, Aug 15 2018

Formula

chi4(k) = Kronecker(-4, k); chi4(k) is 0, 1, 0, -1 when k reduced modulo 4 is 0, 1, 2, 3, respectively; chi4 is A101455.
Series: L(3, chi4) = Sum_{k>=1} chi4(k) k^{-3} = 1 - 1/3^3 + 1/5^3 - 1/7^3 + 1/9^3 - 1/11^3 + 1/13^3 - 1/15^3 + ...
Series: L(3, chi4) = Sum_{k>=0} tanh((2k+1) Pi/2)/(2k+1)^3. [Ramanujan; see Berndt, page 293]
Closed form: L(3, chi4) = Pi^3/32 = 1/A331095.
Equals Sum_{n>=0} (-1)^n/(2*n+1)^3. - Jean-François Alcover, Mar 29 2013
Equals Product_{k>=3} (1 - tan(Pi/2^k)^4) (Groenman, 1990). - Amiram Eldar, Apr 03 2022
Equals Integral_{x=0..1} arcsinh(x)*arccos(x)/x dx (Kobayashi, 2021). - Amiram Eldar, Jun 23 2023
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(3), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p^3)^(-1). (End)

Extensions

Offset corrected by R. J. Mathar, Feb 05 2009

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

A327135 Decimal expansion of Sum_{k>=1} Kronecker(-7,k)/k^3.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A175629 and s = 3.

Examples

			1 + 1/2^3 - 1/3^3 + 1/4^3 - 1/5^3 - 1/6^3 + 1/8^3 + 1/9^3 - 1/10^3 + 1/11^3 - 1/12^3 - 1/13^3 + ... = 32*Pi^3/(343*sqrt(7)) = 1.0933430694...
		

Crossrefs

Cf. A175629.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^3, where d is a fundamental discriminant: A251809 (d=-8), this sequence (d=-7), A153071 (d=-4), A129404 (d=-3), A002117 (d=1), A328723 (d=5), A329715 (d=8), A329716 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(-7,k)/k^s: A326919 (s=1), A103133 (s=2), this sequence (s=3).

Programs

  • Mathematica
    RealDigits[32*Pi^3/(343*Sqrt[7]), 10, 102] // First
  • PARI
    default(realprecision, 100); 32*Pi^3/(343*sqrt(7))

Formula

Equals 32*Pi^3/(343*sqrt(7)).
Equals (zeta(3,1/7) + zeta(3,2/7) - zeta(3,3/7) + zeta(3,4/7) - zeta(3,5/7) - zeta(3,6/7))/343.
Equals (polylog(3,u) + polylog(3,u^2) - polylog(3,u^3) + polylog(3,u^4) - polylog(3,u^5) - polylog(3,u^6))/sqrt(-7), where u = exp(2*Pi*i/7) is a 7th primitive root of unity, i = sqrt(-1).
Equals (polygamma(2,1/7) + polygamma(2,2/7) - polygamma(2,3/7) + polygamma(2,4/7) - polygamma(2,5/7) - polygamma(2,6/7))/(-686).
Equals 1/(Product_{p prime == 1, 2 or 4 (mod 7)} (1 - 1/p^3) * Product_{p prime == 3, 5 or 6 (mod 7)} (1 + 1/p^3)). - Amiram Eldar, Dec 17 2023

A309710 Decimal expansion of Sum_{k>=1} Kronecker(-8,k)/k^2.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A188510 and s = 2.

Examples

			1 + 1/3^2 - 1/5^2 - 1/7^2 + 1/9^2 + 1/11^2 - 1/13^2 - 1/15^2 + ...= 1.0647341710...
		

Crossrefs

Cf. A188510.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^2, where d is a fundamental discriminant: this sequence (d=-8), A103133 (d=-7), A006752 (d=-4), A086724 (d=-3), A013661 (d=1), A328717 (d=5), A328895 (d=8), A258414 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(-8,k)/k^s: A093954 (s=1), this sequence (s=2), A251809 (s=3).

Programs

  • Mathematica
    (PolyGamma[1, 1/8] + PolyGamma[1, 3/8] - PolyGamma[1, 5/8] - PolyGamma[1, 7/8])/64 // RealDigits[#, 10, 102] & // First

Formula

Equals (zeta(2,1/8) + zeta(2,3/8) - zeta(2,5/8) - zeta(2,7/8))/64, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(2,u) + polylog(2,u^3) - polylog(2,-u) - polylog(2,-u^3))/sqrt(-8), where u = sqrt(2)/2 + i*sqrt(2)/2 is an 8th primitive root of unity, i = sqrt(-1).
Equals (polygamma(1,1/8) + polygamma(1,3/8) - polygamma(1,5/8) - polygamma(1,7/8))/64.
Equals 1/(Product_{p prime == 1 or 3 (mod 8)} (1 - 1/p^2) * Product_{p prime == 5 or 7 (mod 8)} (1 + 1/p^2)). - Amiram Eldar, Dec 17 2023

A329715 Decimal expansion of Sum_{k>=1} Kronecker(8,k)/k^3.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A091337 and s = 3.

Examples

			1 - 1/3^3 - 1/5^3 + 1/7^3 + 1/9^3 - 1/11^3 - 1/13^3 + 1/15^3 + ... = 0.9583804545...
		

Crossrefs

Cf. A091337.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^3, where d is a fundamental discriminant: A251809 (d=-8), A327135 (d=-7), A153071 (d=-4), A129404 (d=-3), A002117 (d=1), A328723 (d=5), this sequence (d=8), A329716 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(8,k)/k^s: A196525 (s=1), A328895 (s=2), this sequence (s=3).

Programs

  • Mathematica
    (PolyGamma[2, 1/8] - PolyGamma[2, 3/8] - PolyGamma[2, 5/8] + PolyGamma[2, 7/8])/(-1024) // RealDigits[#, 10, 102] & // First

Formula

Equals (zeta(3,1/8) - zeta(3,3/8) - zeta(3,5/8) + zeta(3,7/8))/512, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(3,u) - polylog(3,u^3) - polylog(3,-u) + polylog(3,-u^3))/sqrt(8), where u = sqrt(2)/2 + i*sqrt(2)/2 is an 8th primitive root of unity, i = sqrt(-1).
Equals (polygamma(2,1/8) - polygamma(2,3/8) - polygamma(2,5/8) + polygamma(2,7/8))/(-1024).
Equals 1/(Product_{p prime == 1 or 7 (mod 8)} (1 - 1/p^3) * Product_{p prime == 3 or 5 (mod 8)} (1 + 1/p^3)). - Amiram Eldar, Dec 17 2023

A328723 Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^3.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A080891 and s = 3.

Examples

			1 - 1/2^3 - 1/3^3 + 1/4^3 + 1/6^3 - 1/7^3 - 1/8^3 + 1/9^3 + ... = 0.8548247666...
		

Crossrefs

Cf. A080891.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^3, where d is a fundamental discriminant: A251809 (d=-8), A327135 (d=-7), A153071 (d=-4), A129404 (d=-3), A002117 (d=1), this sequence (d=5), A329715 (d=8), A329716 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^s: A086466 (s=1), A328717 (s=2), this sequence (s=3).

Programs

  • Mathematica
    (PolyGamma[2, 1/5] - PolyGamma[2, 2/5] - PolyGamma[2, 3/5] + PolyGamma[2, 4/5])/(-250) // RealDigits[#, 10, 102] & // First

Formula

Equals (zeta(3,1/5) - zeta(3,2/5) - zeta(3,3/5) + zeta(3,4/5))/25, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(3,u) - polylog(3,u^2) - polylog(3,u^3) + polylog(3,u^4))/sqrt(5), where u = exp(2*Pi*i/5) is a 5th primitive root of unity, i = sqrt(-1).
Equals (polygamma(2,1/5) - polygamma(2,2/5) - polygamma(2,3/5) - polygamma(2,4/5))/(-250).
Equals 1/(Product_{p prime == 1 or 4 (mod 5)} (1 - 1/p^3) * Product_{p prime == 2 or 3 (mod 5)} (1 + 1/p^3)). - Amiram Eldar, Dec 17 2023

A329716 Decimal expansion of Sum_{k>=1} Kronecker(12,k)/k^3.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A110161 and s = 3.

Examples

			1 - 1/5^3 - 1/7^3 + 1/11^3 + 1/13^3 - 1/17^3 - 1/19^3 + 1/23^3 + ... = 0.9900400194...
		

Crossrefs

Cf. A110161.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^3, where d is a fundamental discriminant: A251809 (d=-8), A327135 (d=-7), A153071 (d=-4), A129404 (d=-3), A002117 (d=1), A328723 (d=5), A329715 (d=8), this sequence (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(12,k)/k^s: A196530 (s=1), A258414 (s=2), this sequence (s=3).

Programs

  • Mathematica
    (PolyGamma[2, 1/12] - PolyGamma[2, 5/12] - PolyGamma[2, 7/12] + PolyGamma[2, 11/12])/(-3456) // RealDigits[#, 10, 102] & // First

Formula

Equals (zeta(3,1/12) - zeta(3,5/12) - zeta(3,7/12) + zeta(3,11/12))/1728, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(3,u) - polylog(3,u^5) - polylog(3,-u) + polylog(3,-u^5))/sqrt(12), where u = (sqrt(3)+i)/2 is a 12th primitive root of unity, i = sqrt(-1).
Equals (polygamma(2,1/12) - polygamma(2,5/12) - polygamma(2,7/12) + polygamma(2,11/12))/(-3456).
Equals 1/(Product_{p prime == 1 or 11 (mod 12)} (1 - 1/p^3) * Product_{p prime == 5 or 7 (mod 12)} (1 + 1/p^3)). - Amiram Eldar, Dec 17 2023

A346907 Decimal expansion of 2 - 3 * Pi^3 / (2^6 * sqrt(2)).

Original entry on oeis.org

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

Views

Author

Sean A. Irvine, Aug 06 2021

Keywords

Examples

			0.9722774140631414321207433819977442327898996814630...
		

References

  • L. B. W. Jolley, Summation of Series, Dover, 1961, Eq. (326).

Crossrefs

Programs

  • Mathematica
    RealDigits[2-3*Pi^3/(2^6*Sqrt[2]), 10, 120][[1]] (* Amiram Eldar, Jun 13 2023 *)

Formula

Equals 2 - 3 * Pi^3 / (2^6 * sqrt(2)) = 2-A251809.
Equals 1 + Sum_{k>=1} ( (-1)^k/(4*k-1)^3 - (-1)^k/(4*k+1)^3 ).

A251967 Decimal expansion of 29*Pi^3/864.

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Dec 12 2014

Keywords

Examples

			1.040719934871174519778718908502245903778395102327162179548832876169...
		

Crossrefs

Cf. similar sums:
A233091 for Sum_{i>=0} 1/(2i+1)^3;
A153071 for Sum_{i>=0} (-1)^i/(2i+1)^3;
A251809 for Sum_{i>=0} (-1)^floor(i/2)/(2i+1)^3.

Programs

Formula

Equals Sum_{i>=0} (-1)^floor(i/3)/(2i+1)^3 = +1 +1/3^3 +1/5^3 -1/7^3 -1/9^3 -1/11^3 +1/13^3 +1/15^3 +1/17^3 - ...
Showing 1-10 of 10 results.