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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A002117 Apéry's number or Apéry's constant zeta(3). Decimal expansion of zeta(3) = Sum_{m >= 1} 1/m^3.

Original entry on oeis.org

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

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Author

Keywords

Comments

Sometimes called Apéry's constant.
"A natural question is whether Zeta(3) is a rational multiple of Pi^3. This is not known, though in 1978 R. Apéry succeeded in proving that Zeta(3) is irrational. In Chapter 8 we pointed out that the probability that two random integers are relatively prime is 6/Pi^2, which is 1/Zeta(2). This generalizes to: The probability that k random integers are relatively prime is 1/Zeta(k) ... ." [Stan Wagon]
In 2001 Tanguy Rivoal showed that there are infinitely many odd (positive) integers at which zeta is irrational, including at least one value j in the range 5 <= j <= 21 (refined the same year by Zudilin to 5 <= j <= 11), at which zeta(j) is irrational. See the Rivoal link for further information and references.
The reciprocal of this constant is the probability that three integers chosen randomly using uniform distribution are relatively prime. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 13 2005
Also the value of zeta(1,2), the double zeta-function of arguments 1 and 2. - R. J. Mathar, Oct 10 2011
Also the length of minimal spanning tree for large complete graph with uniform random edge lengths between 0 and 1, cf. link to John Baez's comment. - M. F. Hasler, Sep 26 2017
Sum of the inverses of the cubes (A000578). - Michael B. Porter, Nov 27 2017
This number is the average value of sigma_2(n)/n^2 where sigma_2(n) is the sum of the squares of the divisors of n. - Dimitri Papadopoulos, Jan 07 2022

Examples

			1.2020569031595942853997...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 261.
  • S. R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, pp. 40-53, 500.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead, and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 84.
  • R. William Gosper, Strip Mining in the Abandoned Orefields of Nineteenth Century Mathematics, Computers in Mathematics (Stanford CA, 1986); Lecture Notes in Pure and Appl. Math., Dekker, New York, 125 (1990), 261-284; MR 91h:11154.
  • Xavier Gourdon, Analyse, Les Maths en tête, Ellipses, 1994, Exemple 3, page 224.
  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section F17, Series associated with the zeta-function, p. 391.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press; 6 edition (2008), pp. 47, 268-269.
  • Paul Levrie, The Ubiquitous Apéry Number, Math. Intelligencer, Vol. 45, No. 2, 2023, pp. 118-119.
  • A. A. Markoff, Mémoire sur la transformation de séries peu convergentes en séries très convergentes, Mém. de l'Acad. Imp. Sci. de St. Pétersbourg, XXXVII, 1890.
  • Paul J. Nahin, In Pursuit of Zeta-3, Princeton University Press, 2021.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Stan Wagon, Mathematica In Action, W. H. Freeman and Company, NY, 1991, page 354.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 33.
  • A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with Elementary Solutions, Dover (1987), Ex. 92-93.

Crossrefs

Cf. A197070: 3*zeta(3)/4; A233090: 5*zeta(3)/8; A233091: 7*zeta(3)/8.
Cf. A000578 (cubes).
Cf. sums of inverses: A152623 (tetrahedral numbers), A175577 (octahedral numbers), A295421 (dodecahedral numbers), A175578 (icosahedral numbers).

Programs

  • Magma
    L:=RiemannZeta(: Precision:=100); Evaluate(L,3); // G. C. Greubel, Aug 21 2018
  • Maple
    # Calculates an approximation with n exact decimal places (small deviation
    # in the last digits are possible). Goes back to ideas of A. A. Markoff 1890.
    zeta3 := proc(n) local s, w, v, k; s := 0; w := -1; v := 4;
    for k from 2 by 2 to 7*n/2 do
        w := -w*v/k;
        v := v + 8;
        s := s + 1/(w*k^3);
    od; 20*s; evalf(%, n) end:
    zeta3(10000); # Peter Luschny, Jun 10 2020
  • Mathematica
    RealDigits[ N[ Zeta[3], 100] ] [ [1] ]
    (* Second program (historical interest): *)
    d[n_] := 34*n^3 + 51*n^2 + 27*n + 5; 6/Fold[Function[d[#2-1] - #2^6/#1], 5, Reverse[Range[100]]] // N[#, 108]& // RealDigits // First
    (* Jean-François Alcover, Sep 19 2014, after Apéry's continued fraction *)
  • Maxima
    fpprec : 100$ ev(bfloat(zeta(3)))$ bfloat(%); /* Martin Ettl, Oct 21 2012 */
    
  • PARI
    default(realprecision, 20080); x=zeta(3); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002117.txt", n, " ", d)); \\ Harry J. Smith, Apr 19 2009
    
  • Python
    from mpmath import mp, apery
    mp.dps=109
    print([int(z) for z in list(str(apery).replace('.', ''))[:-1]]) # Indranil Ghosh, Jul 08 2017
    

Formula

Lima gives an approximation to zeta(3) as (236*log(2)^3)/197 - 283/394*Pi*log(2)^2 + 11/394*Pi^2*log(2) + 209/394*log(sqrt(2) + 1)^3 - 5/197 + (93*Catalan*Pi)/197. - Jonathan Vos Post, Oct 14 2009 [Corrected by Wouter Meeussen, Apr 04 2010]
zeta(3) = 5/2*Integral_(x=0..2*log((1+sqrt(5))/2), x^2/(exp(x)-1)) + 10/3*(log((1+sqrt(5))/2))^3. - Seiichi Kirikami, Aug 12 2011
zeta(3) = -4/3*Integral_{x=0..1} log(x)/x*log(1+x) = Integral_{x=0..1} log(x)/x*log(1-x) = -4/7*Integral_{x=0..1} log(x)/x*log((1+x)/(1-x)) = 4*Integral_{x=0..1} 1/x*log(1+x)^2 = 1/2*Integral_{x=0..1} 1/x*log(1-x)^2 = -16/7*Integral_{x=0..Pi/2} x*log(2*cos(x)) = -4/Pi*Integral_{x=0..Pi/2} x^2*log(2*cos(x)). - Jean-François Alcover, Apr 02 2013, after R. J. Mathar
From Peter Bala, Dec 04 2013: (Start)
zeta(3) = (16/7)*Sum_{k even} (k^3 + k^5)/(k^2 - 1)^4.
zeta(3) - 1 = Sum_{k >= 1} 1/(k^3 + 4*k^7) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n - 1)^6/((2*n - 1)*(n^2 - n + 5) - ...))))) (continued fraction).
More generally, there is a sequence of polynomials P(n,x) (of degree 2*n) such that
zeta(3) - Sum_{k = 1..n} 1/k^3 = Sum_{k >= 1} 1/( k^3*P(n,k-1)*P(n,k) ) = 1/((2*n^2 + 2*n + 1) - 1^6/(3*(2*n^2 + 2*n + 3) - 2^6/(5*(2*n^2 + 2*n + 7) - 3^6/(7*(2*n^2 + 2*n + 13) - ...)))) (continued fraction). See A143003 and A143007 for details.
Series acceleration formulas:
zeta(3) = (5/2)*Sum_{n >= 1} (-1)^(n+1)/( n^3*binomial(2*n,n) )
= (5/2)*Sum_{n >= 1} P(n)/( (2*n(2*n - 1))^3*binomial(4*n,2*n) )
= (5/2)*Sum_{n >= 1} (-1)^(n+1)*Q(n)/( (3*n(3*n - 1)*(3*n - 2))^3*binomial(6*n,3*n) ), where P(n) = 24*n^3 + 4*n^2 - 6*n + 1 and Q(n) = 9477*n^6 - 11421*n^5 + 5265*n^4 - 1701*n^3 + 558*n^2 - 108*n + 8 (Bala, section 7). (End)
zeta(3) = Sum_{n >= 1} (A010052(n)/n^(3/2)) = Sum_{n >= 1} ( (floor(sqrt(n)) - floor(sqrt(n-1)))/n^(3/2) ). - Mikael Aaltonen, Feb 22 2015
zeta(3) = Product_{k>=1} 1/(1 - 1/prime(k)^3). - Vaclav Kotesovec, Apr 30 2020
zeta(3) = 4*(2*log(2) - 1 - 2*Sum_{k>=2} zeta(2*k+1)/2^(2*k+1)). - Jorge Coveiro, Jun 21 2020
zeta(3) = (4*zeta'''(1/2)*(zeta(1/2))^2-12*zeta(1/2)*zeta'(1/2)*zeta''(1/2)+8*(zeta'(1/2))^3-Pi^3*(zeta(1/2))^3)/(28*(zeta(1/2))^3). - Artur Jasinski, Jun 27 2020
zeta(3) = Sum_{k>=1} H(k)/(k+1)^2, where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Jul 31 2020
From Artur Jasinski, Sep 30 2020: (Start)
zeta(3) = (5/4)*Li_3(1/f^2) + Pi^2*log(f)/6 - 5*log(f)^3/6,
zeta(3) = (8/7)*Li_3(1/2) + (2/21)*Pi^2 log(2) - (4/21) log(2)^3, where f is golden ratio (A001622) and Li_3 is the polylogarithm function, formulas published by John Landen in 1780, p. 118. (End)
zeta(3) = (1/2)*Integral_{x=0..oo} x^2/(e^x-1) dx (Gourdon). - Bernard Schott, Apr 28 2021
From Peter Bala, Jan 18 2022: (Start)
zeta(3) = 1 + Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)) = 25/24 + (2!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*(4*n^4 + 2^4)) = 28333/27000 + (3!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*(4*n^4 + 2^4)*(4*n^4 + 3^4)). In general, for k >= 1, we have zeta(3) = r(k) + (k!)^4*Sum_{n >= 1} 1/(n^3*(4*n^4 + 1)*...*(4*n^4 + k^4)), where r(k) is rational.
zeta(3) = (6/7) + (64/7)*Sum_{n >= 1} n/(4*n^2 - 1)^3.
More generally, for k >= 0, it appears that zeta(3) = a(k) + b(k)*Sum_{n >= 1} n/( (4*n^2 - 1)*(4*n^2 - 9)*...*(4*n^2 - (2*k+1)^2) )^3, where a(k) and b(k) are rational.
zeta(3) = (10/7) - (128/7)*Sum_{n >= 1} n/(4*n^2 - 1)^4.
More generally, for k >= 0, it appears that zeta(3) = c(k) + d(k)*Sum_{n >= 1} n/( (4*n^2 - 1)*(4*n^2 - 9)*...*(4*n^2 - (2*k+1)^2) )^4, where c(k) and d(k) are rational. [added Nov 27 2023: for the values of a(k), b(k), c(k) and d(k) see the Bala 2023 link, Sections 8 and 9.]
zeta(3) = 2/3 + (2^13)/(3*7)*Sum_{n >= 1} n^3/(4*n^2 - 1)^6. (End)
zeta(3) = -Psi(2)(1/2)/14 (the second derivative of digamma function evaluated at 1/2). - Artur Jasinski, Mar 18 2022
zeta(3) = -(8*Pi^2/9) * Sum_{k>=0} zeta(2*k)/((2*k+1)*(2*k+3)*4^k) = (2*Pi^2/9) * (log(2) + 2 * Sum_{k>=0} zeta(2*k)/((2*k+3)*4^k)) (Scheufens, 2011, Glasser Math. Comp. 22 1968). - Amiram Eldar, May 28 2022
zeta(3) = Sum_{k>=1} (30*k-11) / (4*(2k-1)*k^3*(binomial(2k,k))^2) (Gosper, 1986 and Richard K. Guy reference). - Bernard Schott, Jul 20 2022
zeta(3) = (4/3)*Integral_{x >= 1} x*log(x)*(1 + log(x))*log(1 + 1/x^x) dx = (2/3)*Integral_{x >= 1} x^2*log(x)^2*(1 + log(x))/(1 + x^x) dx. - Peter Bala, Nov 27 2023
zeta_3(n) = 1/180*(-360*n^3*f(-3, n/4) + Pi^3*(n^4 + 20*n^2 + 16))/(n*(n^2 + 4)), where f(-3, n) = Sum_{k>=1} 1/(k^3*(exp(Pi*k/n) - 1)). Will give at least 1 digit of precision/term, example: zeta_3(5) = 1.202056944732.... - Simon Plouffe, Dec 21 2023
zeta(3) = 1 + (1/2)*Sum_{n >= 1} (2*n + 1)/(n^3*(n + 1)^3) = 5/4 - (1/4)*Sum_{n >= 1} (2*n + 1)/(n^4*(n + 1)^4) = 147/120 + (2/15)*Sum_{n >= 1} (2*n + 1)/(n^5*(n + 1)^5) - (64/15)*Sum_{n >= 1} (n + 1)/(n^5*(n + 2)^5) = 19/16 + (128/21)*Sum_{n >= 1} (n + 1)/(n^6*(n + 2)^6) - (1/21)*Sum_{n >= 1} (2*n + 1)/(n^6*(n + 1)^6). - Peter Bala, Apr 15 2024
Equals 7*Pi^3/180 - 2*Sum_{k>=1} 1/(k^3*(exp(2*Pi*k) - 1)) [Grosswald] (see Finch). - Stefano Spezia, Nov 01 2024
Equals 10*Integral_{x=0..1/2} arcsinh(x)^2/x dx = -5*Integral_{x=0..2*log(phi)} x*log(2*sinh(x/2))dx [Munthe Hjortnaes] (see Finch). - Stefano Spezia, Nov 03 2024
Equals Li_3(1) = Integral_{x=0..1} Li_2(x)/x dx = Integral_{x=0..1} Integral_{y=0..1} Li_1(xy)/xy dydx = Integral_{x=0..1} Integral_{y=0..1} Integral_{z=0..1} Li_0(xyz)/xyz dzdydx (see Beukers), in general Integral_{x_1,...,x_k=0..1} Li_{3-k}(Product_{n=1..k} x_n)/(Product_{n=1..k} x_n) dx_k...dx_1 = zeta(3), for any k > 0. - Miko Labalan, Dec 23 2024
zeta(3) = (1/2)*Sum_{m >= 1}(Sum_{n >= 1} 1/(m*n*(m+n))). - Ricardo Bittencourt, Feb 24 2025
zeta(3) = Integral_{x=0..1} Integral_{y=0..1} Integral_{z=0..1} 1/(1 - x*y*z) dz dy dx. - Kritsada Moomuang, May 22 2025
zeta(3) = Sum_{i, j >= 1} 1/(i^2*j*binomial(i+j, i)) = Sum_{k >= 1} 1/(k + 1)^2 * Sum_{j = 1..k} 1/j = zeta(2, 1) (multiple zeta value due to Euler). - Peter Bala, Aug 05 2025

Extensions

More terms from David W. Wilson
Additional comments from Robert G. Wilson v, Dec 08 2000
Quotation from Stan Wagon corrected by N. J. A. Sloane on Dec 24 2005. Thanks to Jose Brox for noticing this error.
Edited by M. F. Hasler, Sep 26 2017

A000584 Fifth powers: a(n) = n^5.

Original entry on oeis.org

0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049, 100000, 161051, 248832, 371293, 537824, 759375, 1048576, 1419857, 1889568, 2476099, 3200000, 4084101, 5153632, 6436343, 7962624, 9765625, 11881376, 14348907, 17210368, 20511149
Offset: 0

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Keywords

Comments

Totally multiplicative sequence with a(p) = p^5 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A059338. The inverse binomial transform yields the (finite) 0, 1, 30, 150, 240, 120, the 5th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Equals sum of odd numbers from n^2*(n-1)+1 (A100104) to n^2*(n+1)-1 (A003777). - Bruno Berselli, Mar 14 2014
a(n) mod 10 = n mod 10. - Reinhard Zumkeller, May 10 2014
Numbers of the form a(n) + a(n+1) + ... + a(n+k) are nonprime for all n, k>=0; this can be proved by the method indicated in the comment in A256581. - Vladimir Shevelev and Peter J. C. Moses, Apr 04 2015

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums give A000539.

Programs

Formula

G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (x-1)^6. [Simon Plouffe in his 1992 dissertation]
Multiplicative with a(p^e) = p^(5e). - David W. Wilson, Aug 01 2001
E.g.f.: exp(x)*(x+15*x^2+25*x^3+10*x^4+x^5). - Geoffrey Critzer, Jun 12 2013
a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 120. - Ant King, Sep 23 2013
a(n) = n + Sum_{j=0..n-1}{k=1..4}binomial(5,k)*j^(5-k). - Patrick J. McNab, Mar 28 2016
From Kolosov Petro, Oct 22 2018: (Start)
a(n) = Sum_{k=1..n} A300656(n,k).
a(n) = Sum_{k=0..n-1} A300656(n,k). (End)
a(n) = Sum_{k=1..5} Eulerian(5, k)*binomial(n+5-k, 5), with Eulerian(5, k) = A008292(5, k), the numbers 1, 26, 66, 26, 1, for n >= 0. Worpitzki's identity for powers of 5. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(5) (A013663).
Sum_{n>=1} (-1)^(n+1)/a(n) = 15*zeta(5)/16 (A267316). (End)

Extensions

More terms from Henry Bottomley, Jun 21 2001

A013662 Decimal expansion of zeta(4).

Original entry on oeis.org

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

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Examples

			1.082323233711138191516003696541167...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 89, Exercise.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section F17, Series associated with the zeta-function, p. 391.
  • L. D. Landau and E. M. Lifschitz, Band V, Statistische Physik, Akademie Verlag, 1966, pp. 172 and 180-181.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 162.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 33.

Crossrefs

Programs

  • Magma
    SetDefaultRealField(RealField(110)); L:=RiemannZeta(); Evaluate(L,4); // G. C. Greubel, May 30 2019
    
  • Maple
    evalf(Pi^4/90,120); # Muniru A Asiru, Sep 19 2018
  • Mathematica
    RealDigits[Zeta[4],10,120][[1]] (* Harvey P. Dale, Dec 18 2012 *)
  • Maxima
    ev(zeta(4),numer) ; /* R. J. Mathar, Feb 27 2012 */
    
  • PARI
    default(realprecision, 20080); x=Pi^4/90; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b013662.txt", n, " ", d)); \\ Harry J. Smith, Apr 29 2009
    
  • Sage
    numerical_approx(zeta(4), digits=100) # G. C. Greubel, May 30 2019

Formula

zeta(4) = Pi^4/90 = A092425/90. - Harry J. Smith, Apr 29 2009
From Peter Bala, Dec 03 2013: (Start)
Definition: zeta(4) := Sum_{n >= 1} 1/n^4.
zeta(4) = (4/17)*Sum_{n >= 1} ( (1 + 1/2 + ... + 1/n)/n )^2 and
zeta(4) = (16/45)*Sum_{n >= 1} ( (1 + 1/3 + ... + 1/(2*n-1))/n )^2 (see Borwein and Borwein).
zeta(4) = (256/90)*Sum_{n >= 1} n^2*(4*n^2 + 3)*(12*n^2 + 1)/(4*n^2 - 1)^5.
Series acceleration formulas:
zeta(4) = (36/17)*Sum_{n >= 1} 1/( n^4*binomial(2*n,n) ) (Comtet)
= (36/17)*Sum_{n >= 1} P(n)/( (2*n*(2*n - 1))^4*binomial(4*n,2*n) )
= (36/17)*Sum_{n >= 1} Q(n)/( (3*n*(3*n - 1)*(3*n - 2))^4*binomial(6*n,3*n) ),
where P(n) = 80*n^4 - 48*n^3 + 24*n^2 - 8*n + 1 and Q(n) = 137781*n^8 - 275562*n^7 + 240570*n^6 - 122472*n^5 + 41877*n^4 - 10908*n^3 + 2232*n^2 - 288*n + 16 (see section 8 in the Bala link). (End)
zeta(4) = 2/3*2^4/(2^4 - 1)*( Sum_{n even} n^2*p(n)/(n^2 - 1)^5 ), where p(n) = 3*n^4 + 10*n^2 + 3 is a row polynomial of A091043. See A013664, A013666, A013668 and A013670. - Peter Bala, Dec 05 2013
zeta(4) = Sum_{n >= 1} ((floor(sqrt(n))-floor(sqrt(n-1)))/n^2). - Mikael Aaltonen, Jan 18 2015
zeta(4) = Product_{k>=1} 1/(1 - 1/prime(k)^4). - Vaclav Kotesovec, May 02 2020
From Wolfdieter Lang, Sep 16 2020: (Start)
zeta(4) = (1/3!)*Integral_{x=0..oo} x^3/(exp(x) - 1) dx. See Abramowitz-Stegun, 23.2.7., for s=2, p. 807, and Landau-Lifschitz, Band V, p. 172, eq. (4), for x=4. See also A231535.
zeta(4) = (4/21)*Integral_{x=0..oo} x^3/(exp(x) + 1) dx. See Abramowitz-Stegun, 23.2.8., for s=2, p. 807, and Landau-Lifschitz, Band V, p. 172, eq. (1), for x=4. See also A337711. (End)
zeta(4) = (72/17) * Integral_{x=0..Pi/3} x*(log(2*sin(x/2)))^2. See Richard K. Guy reference. - Bernard Schott, Jul 20 2022
From Peter Bala, Nov 12 2023: (Start)
zeta(4) = 1 + (4/3)*Sum_{k >= 1} (1 - 2*(-1)^k)/(k*(k + 1)^4*(k + 2)) = 35053/32400 + 48*Sum_{k >= 1} (1 - 2*(-1)^k)/(k*(k + 1)*(k + 2)*(k + 3)^4*(k + 4)*(k + 5)*(k + 6)).
More generally, it appears that for n >= 0, zeta(4) = c(n) + (4/3)*(2*n + 1)!^2 * Sum_{k >= 1} (1 - 2*(-1)^k)/( (k + 2*n + 1)^3*Product_{i = 0..4*n+2} (k + i) ), where {c(n) : n >= 0} is a sequence of rational approximations to zeta(4) beginning [1, 35053/32400, 2061943067/ 1905120000, 18594731931460103/ 17180389306080000, 257946156103293544441/ 238326360453941760000, ...]. (End)
From Peter Bala, Apr 27 2025: (Start)
zeta(4) = 1/4! * Integral_{x >= 0} x^4 * exp(x)/(exp(x) - 1)^2 dx = 8/7 * 1/4! * Integral_{x >= 0} x^4 * exp(x)/(exp(x) + 1)^2 dx.
zeta(4) = 1/5! * Integral_{x >= 0} x^5 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3*5*7) * Integral_{x >= 0} x^5 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
10*zeta(4) = Sum_{k>=1} H(k)^3/(k*(k+1)), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number (Ramachandra, 1981). - Amiram Eldar, May 30 2025
zeta(4) = Integral_{x=0..1} Li(3,x)/x dx, where Li(n,x) is the polylogarithm function. - Kritsada Moomuang, Jun 14 2025
zeta(4) = Sum_{i, j >= 1} 1/(i^3*j*binomial(i+j, i)) = 4/3 * Sum_{i, j >= 1} 1/(i^2*j^2*binomial(i+j, i)). - Peter Bala, Aug 03 2025

A013664 Decimal expansion of zeta(6).

Original entry on oeis.org

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

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Keywords

Examples

			1.01734306198444913...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.

Crossrefs

Programs

Formula

Equals Pi^6/945 = A092732/945. - Mohammad K. Azarian, Mar 03 2008
zeta(6) = 8/3*2^6/(2^6 - 1)*( Sum_{n even} n^2*p(n)/(n^2 - 1)^7 ), where p(n) = n^6 + 7*n^4 + 7*n^2 + 1 is a row polynomial of A091043. See A013662, A013666, A013668 and A013670. - Peter Bala, Dec 05 2013
Definition: zeta(6) = Sum_{n >= 1} 1/n^6. - Bruno Berselli, Dec 05 2013
zeta(6) = Sum_{n >= 1} (A010052(n)/n^3). - Mikael Aaltonen, Feb 20 2015
zeta(6) = Sum_{n >= 1} (A010057(n)/n^2). - A.H.M. Smeets, Sep 19 2018
zeta(6) = Product_{k>=1} 1/(1 - 1/prime(k)^6). - Vaclav Kotesovec, May 02 2020
From Wolfdieter Lang, Sep 16 2020: (Start)
zeta(6) = (1/5!)*Integral_{x=0..infinity} x^5/(exp(x) - 1) dx. See Abramowitz-Stegun, 23.2.7., for s=6, p. 807. See also A337710 for the value of the integral.
zeta(6) = (4/465)*Integral_{x=0..infinity} x^5/(exp(x) + 1) dx. See Abramowitz-Stegun, 23.2.8., for s=6, p. 807. The value of the integral is (31/252)*Pi^6 = 118.2661309... . (End)
From Peter Bala, Apr 27 2025: (Start)
zeta(6) = 1/6! * Integral_{x >= 0} x^6 * exp(x)/(exp(x) - 1)^2 dx = 2^5/(2^5 - 1) * 1/6! * Integral_{x >= 0} x^6 * exp(x)/(exp(x) + 1)^2 dx.
zeta(6) = 1/7! * Integral_{x >= 0} x^7 * exp(x)*(exp(x) + 1) /(exp(x) - 1)^3 dx = 2/(3*7*15*31) * Integral_{x >= 0} x^7 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)

A050997 Fifth powers of primes.

Original entry on oeis.org

32, 243, 3125, 16807, 161051, 371293, 1419857, 2476099, 6436343, 20511149, 28629151, 69343957, 115856201, 147008443, 229345007, 418195493, 714924299, 844596301, 1350125107, 1804229351, 2073071593, 3077056399, 3939040643, 5584059449, 8587340257, 10510100501
Offset: 1

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Comments

Numbers k such that A062799(k) = 5.
Let r(n) = (a(n)+1)/(a(n)-1) if a(n) mod 4 = 3, (a(n)-1)/(a(n)+1) otherwise; then Product_{n>=1} r(n) = (31/33) * (244/242) * (3124/3126) * (16808/16806) * ... = 246016/259875. - Dimitris Valianatos, Mar 09 2020

Crossrefs

Programs

Formula

A056595(a(n)) = 3. - Reinhard Zumkeller, Aug 15 2011
Sum_{n>=1} 1/a(n) = P(5) = 0.0357550174... (A085965). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(5)/zeta(10) (A157291).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(5) = 1/A013663. (End)

A013665 Decimal expansion of zeta(7).

Original entry on oeis.org

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

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Comments

From Dimitris Valianatos, Apr 29 2020: (Start)
Let p_n = Product_{k >= 1, 4*k-1 is prime} (((4*k - 1)^n + 1) / ((4*k - 1)^n - 1)).
Then (2^(n + 1) / (2^n - 1)) * Sum_{k >= 1} 1 / (4*k - 3)^n = ((p_n + 1) / p_n) * Sum_{k >= 1} 1 / k^n = ((p_n + 1) / p_n) * zeta(n), n >= 3 odd number.
For n = 7, p_7 = 1.00091744947834007403796003463414...
The product (256 / 127) * Sum_{k >= 1} 1 / (4*k - 3)^7 = 2.01577429320860871987548541116538... is equal to the product ((p_7 + 1) / p_7) * Sum_{k >= 1} 1 / k^7 = 1.9990833914636834116748... * zeta(7) = 2.01577429320860871987548541116538... (End)

Examples

			1.0083492773819228268397975498497967595998635605652387064172831365716014...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3, p. 43.

Crossrefs

Programs

Formula

zeta(7) = Sum_{n >= 1} (A010052(n)/n^(7/2)) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^(7/2) ). - Mikael Aaltonen, Feb 22 2015
zeta(7) = Product_{k>=1} 1/(1 - 1/prime(k)^7). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(7) = (-1/840)*Integral_{x=0..1} log(1-x^6)^7/x^7.
zeta(7) = (1/720)*Integral_{x=0..oo} x^6/(exp(x)-1).
zeta(7) = (4/2835)*Integral_{x=0..oo} x^6/(exp(x)+1).
zeta(7) = (1/(182880*Zeta(1/2)^7))*(-61*Pi^7*zeta(1/2)^7 + 2880* zeta'(1/2)^7 - 10080*zeta(1/2)*zeta'(1/2)^5*zeta''(1/2) + 10080* zeta(1/2)^2*zeta'(1/2)^3*zeta''(1/2)^2 - 2520*zeta(1/2)^3*zeta'(1/2)* zeta''(1/2)^3 + 3360*zeta(1/2)^2*zeta'(1/2)^4*zeta'''(1/2) - 5040 zeta(1/2)^3*zeta'(1/2)^2*zeta''(1/2)*zeta'''(1/2) + 840*zeta(1/2)^4* zeta''(1/2)^2*zeta'''(1/2) + 560*zeta(1/2)^4*zeta'(1/2)*zeta'''(1/2)^3 - 840*zeta(1/2)^3*zeta'(1/2)^3*zeta''''(1/2) + 840*zeta(1/2)^4*zeta'(1/2)* zeta''(1/2)*zeta''''(1/2) - 140*zeta(1/2)^5*zeta'''(1/2)*zeta''''(1/2) + 168*zeta(1/2)^4*zeta'(1/2)^2*zeta'''''(1/2) - 84*zeta(1/2)^5*zeta''(1/2)* zeta'''''(1/2) - 28*zeta(1/2)^5*zeta'(1/2)*zeta''''''(1/2) + 4* zeta(1/2)^6*zeta'''''''(1/2)). (End)
Equals 19*Pi^7/56700 - 2*Sum_{k>=1} 1/(k^7*(exp(2*Pi*k) - 1)) [Grosswald] (see Finch). - Stefano Spezia, Nov 01 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(7) = 1/7! * Integral_{x >= 0} x^7 * exp(x)/(exp(x) - 1)^2 dx = 2^6/(2^6 - 1) * 1/7! * Integral_{x >= 0} x^7 * exp(x)/(exp(x) + 1)^2 dx.
zeta(7) = 1/8! * Integral_{x >= 0} x^8 * exp(x)*(exp(x) + 1) /(exp(x) - 1)^3 dx = 1/ (2*3*7*15*63) * Integral_{x >= 0} x^8 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)

A013667 Decimal expansion of zeta(9).

Original entry on oeis.org

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

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Examples

			1.0020083928260822...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Crossrefs

Programs

  • Maple
    evalf(Zeta(9)) ; # R. J. Mathar, Oct 16 2015
  • Mathematica
    RealDigits[Zeta[9],10,100][[1]] (* Harvey P. Dale, Aug 27 2014 *)

Formula

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(9) = Sum_{n >= 1} 1/n^9.
zeta(9) = 2^9/(2^9 - 1)*( Sum_{n even} n^7*p(n)*p(1/n)/(n^2 - 1)^10 ), where p(n) = n^4 + 10*n^2 + 5. See A013663, A013671 and A013675. (End)
zeta(9) = Sum_{n >= 1} (A010052(n)/n^(9/2)) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^(9/2) ). - Mikael Aaltonen, Feb 22 2015
zeta(9) = Product_{k>=1} 1/(1 - 1/prime(k)^9). - Vaclav Kotesovec, May 02 2020
From Peter Bala, Apr 27 2025: (Start)
zeta(9) = 1/9! * Integral_{x >= 0} x^9 * exp(x)/(exp(x) - 1)^2 dx = 2^9/(2^9 - 1) * 1/9! * Integral_{x >= 0} x^9 * exp(x)/(exp(x) + 1)^2 dx.
zeta(9) = 1/10! * Integral_{x >= 0} x^10 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^5 * 5^3 * 7 * 17) * Integral_{x >= 0} x^10 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)

A013666 Decimal expansion of zeta(8).

Original entry on oeis.org

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

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Comments

This sequence is also the decimal expansion of Pi^8/9450. - Mohammad K. Azarian, Mar 03 2008

Examples

			1.00407735619794433937868523850865246525896079064985002032911020265...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Crossrefs

Programs

  • Maple
    Digits := 100 : evalf(Pi^8/9450) ; # R. J. Mathar, Jan 07 2021
  • Mathematica
    RealDigits[Zeta[8], 10, 100][[1]] (* Vincenzo Librandi, Feb 15 2015 *)

Formula

zeta(8) = 2/3*2^8/(2^8 - 1)*( Sum_{n even} n^2*p(n)/(n^2 - 1)^9 ), where p(n) = 5*n^8 + 60*n^6 + 126*n^4 + 60*n^2 + 5 is a row polynomial of A091043. See A013662, A013664, A013668 and A013670. - Peter Bala, Dec 05 2013
zeta(8) = Sum_{n >= 1} (A010052(n)/n^4). - Mikael Aaltonen, Feb 20 2015
zeta(8) = Product_{k>=1} 1/(1 - 1/prime(k)^8). - Vaclav Kotesovec, May 02 2020
From Wolfdieter Lang, Sep 16 2020 (Start):
zeta(8) = (1/7!)*Integral_{0..infinity} x^7/(exp(x) - 1) dx. See Abramowitz-Stegun, 23.2.7., for s=8, p. 807. The value of the integral is 8*Pi^8/15 = 5060.54987... .
zeta(8) = (2^7/(127*7!))*Integral_{0..infinity} x^7/(exp(x) + 1) dx. See Abramowitz-Stegun, 23.2.8., for s=8, p. 807. The prefactor is 8/40005. The value of the integral is (127/240)*Pi^8 = 5021.014329... .(End)
Equals A092736/9450. - R. J. Mathar, Jan 07 2021
From Peter Bala, Apr 27 2025: (Start)
zeta(8) = 1/8! * Integral_{x >= 0} x^8 * exp(x)/(exp(x) - 1)^2 dx = 2^7/(2^7 - 1) * 1/8! * Integral_{x >= 0} x^8 * exp(x)/(exp(x) + 1)^2 dx.
zeta(8) = 1/9! * Integral_{x >= 0} x^9 * exp(x)*(exp(x) + 1) /(exp(x) - 1)^3 dx = 1/(3*15*63*127) * Integral_{x >= 0} x^9 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)

A013669 Decimal expansion of zeta(11).

Original entry on oeis.org

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

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Author

Keywords

Examples

			1.0004941886041194645587022825264699364686064357582...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Crossrefs

Programs

Formula

zeta(11) = Sum_{n >= 1} (A010052(n)/n^(11/2)) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^(11/2) ). - Mikael Aaltonen, Feb 22 2015
zeta(11) = Product_{k>=1} 1/(1 - 1/prime(k)^11). - Vaclav Kotesovec, May 02 2020

Extensions

a(99) corrected by Sean A. Irvine, Sep 05 2018

A262177 Decimal expansion of Q_5 = zeta(5) / (Sum_{k>=1} (-1)^(k+1) / (k^5 * binomial(2k, k))), a conjecturally irrational constant defined by an Apéry-like formula.

Original entry on oeis.org

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

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Author

Jean-François Alcover, Sep 14 2015

Keywords

Comments

The similar constant Q_3 = zeta(3) / (Sum_{k>=1} (-1)^(k+1) / (k^3 * binomial(2k, k))) evaluates to 5/2.

Examples

			2.09486862201003699385024929373294163029675874856778182740127587837438...
		

Crossrefs

Cf. A013663.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692, A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Mathematica
    Q5 = Zeta[5]/Sum[(-1)^(k+1)/(k^5*Binomial[2k, k]), {k, 1, Infinity}]; RealDigits[Q5, 10, 103] // First
  • PARI
    zeta(5)/suminf(k=1, (-1)^(k+1)/(k^5*binomial(2*k,k))) \\ Michel Marcus, Sep 14 2015

Formula

Equals 2*zeta(5)/6F5(1,1,1,1,1,1; 3/2,2,2,2,2; -1/4).
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