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 22 results. Next

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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Keywords

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

A013663 Decimal expansion of zeta(5).

Original entry on oeis.org

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

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Comments

In a widely distributed May 2011 email, Wadim Zudilin gave a rebuttal to v1 of Kim's 2011 preprint: "The mistake (unfixable) is on p. 6, line after eq. (3.3). 'Without loss of generality' can be shown to work only for a finite set of n_k's; as the n_k are sufficiently large (and N is fixed), the inequality for epsilon is false." In a May 2013 email, Zudilin extended his rebuttal to cover v2, concluding that Kim's argument "implies that at least one of zeta(2), zeta(3), zeta(4) and zeta(5) is irrational, which is trivial." - Jonathan Sondow, May 06 2013
General: zeta(2*s + 1) = (A000364(s)/A331839(s)) * Pi^(2*s + 1) * Product_{k >= 1} (A002145(k)^(2*s + 1) + 1)/(A002145(k)^(2*s + 1) - 1), for s >= 1. - Dimitris Valianatos, Apr 27 2020

Examples

			1/1^5 + 1/2^5 + 1/3^5 + 1/4^5 + 1/5^5 + 1/6^5 + 1/7^5 + ... =
1 + 1/32 + 1/243 + 1/1024 + 1/3125 + 1/7776 + 1/16807 + ... = 1.036927755143369926331365486457...
		

References

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

Crossrefs

Programs

Formula

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(5) = Sum_{n >= 1} 1/n^5.
zeta(5) = 2^5/(2^5 - 1)*(Sum_{n even} n^5*p(n)*p(1/n)/(n^2 - 1)^6 ), where p(n) = n^2 + 3. See A013667, A013671 and A013675. (End)
zeta(5) = Sum_{n >= 1} (A010052(n)/n^(5/2)) = Sum_{n >= 1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n^(5/2)). - Mikael Aaltonen, Feb 22 2015
zeta(5) = Product_{k>=1} 1/(1 - 1/prime(k)^5). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(5) = (-1/30)*Integral_{x=0..1} log(1-x^4)^5/x^5.
zeta(5) = (1/24)*Integral_{x=0..infinity} x^4/(exp(x)-1).
zeta(5) = (2/45)*Integral_{x=0..infinity} x^4/(exp(x)+1).
zeta(5) = (1/(1488*zeta(1/2)^5))*(-5*Pi^5*zeta(1/2)^5 + 96*zeta'(1/2)^5 - 240*zeta(1/2)*zeta'(1/2)^3*zeta''(1/2) + 120*zeta(1/2)^2*zeta'(1/2)*zeta''(1/2)^2 + 80*zeta(1/2)^2*zeta'(1/2)^2*zeta'''(1/2)- 40*zeta(1/2)^3*zeta''(1/2)*zeta'''(1/2) - 20*zeta(1/2)^3*zeta'(1/2)*zeta''''(1/2)+4*zeta(1/2)^4*zeta'''''(1/2)). (End).
From Peter Bala, Oct 29 2023: (Start)
zeta(3) = (8/45)*Integral_{x >= 1} x^3*log(x)^3*(1 + log(x))*log(1 + 1/x^x) dx = (2/45)*Integral_{x >= 1} x^4*log(x)^4*(1 + log(x))/(1 + x^x) dx.
zeta(5) = 131/128 + 26*Sum_{n >= 1} (n^2 + 2*n + 40/39)/(n*(n + 1)*(n + 2))^5.
zeta(5) = 5162893/4976640 - 1323520*Sum_{n >= 1} (n^2 + 4*n + 56288/12925)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^5. Taking 10 terms of the series gives a value for zeta(5) correct to 20 decimal places.
Conjecture: for k >= 1, there exist rational numbers A(k), B(k) and c(k) such that zeta(5) = A(k) + B(k)*Sum_{n >= 1} (n^2 + 2*k*n + c(k))/(n*(n + 1)*...*(n + 2*k))^5. A similar conjecture can be made for the constant zeta(3). (End)
zeta(5) = (694/204813)*Pi^5 - Sum_{n >= 1} (6280/3251)*(1/(n^5*(exp(4*Pi*n)-1))) + Sum_{n >= 1} (296/3251)*(1/(n^5*(exp(5*Pi*n)-1))) - Sum_{n >= 1} (1073/6502)*(1/(n^5*(exp(10*Pi*n)-1))) + Sum_{n >= 1} (37/6502)*(1/(n^5*(exp(20*Pi*n)-1))). - Simon Plouffe, Jan 06 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(5) = 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) - 1)^2 dx = (16/15) * 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) + 1)^2 dx.
zeta(5) = 1/6! * Integral_{x >= 0} x^6 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^3 * 5^2) * Integral_{x >= 0} x^6 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
zeta(5) = Sum_{i, j >= 1} 1/((i^4)*j*binomial(i+j, i)). More generally, zeta(n+1) = Sum_{i, j >= 1} 1/((i^n)*j*binomial(i+j, i)) for n >= 1. - Peter Bala, Aug 07 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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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)

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)

A179645 a(n) = prime(n)^8.

Original entry on oeis.org

256, 6561, 390625, 5764801, 214358881, 815730721, 6975757441, 16983563041, 78310985281, 500246412961, 852891037441, 3512479453921, 7984925229121, 11688200277601, 23811286661761, 62259690411361, 146830437604321
Offset: 1

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Author

Keywords

Examples

			a(1) = 256 since the eighth power of the first prime is 2^7 = 256. - _Wesley Ivan Hurt_, Mar 27 2014
		

Crossrefs

Programs

Formula

a(n) = A000040(n)^8 = A001016(A000040(n)). - Wesley Ivan Hurt, Mar 27 2014
Sum_{n>=1} 1/a(n) = P(8) = 0.0040614053... (A085968). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 24 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(8)/zeta(16) = 34459425/(3617*Pi^8) = A013666/A013674.
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(8) = 9450/Pi^8 = 1/A013666. (End)

A266562 Decimal expansion of the generalized Glaisher-Kinkelin constant A(15).

Original entry on oeis.org

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

Views

Author

G. C. Greubel, Dec 31 2015

Keywords

Comments

Also known as the 15th Bendersky constant.

Examples

			0.342830806132816736571711146340672378141726945483236877251076164....
		

Crossrefs

Cf. A019727 (A(0)), A074962 (A(1)), A243262 (A(2)), A243263 (A(3)), A243264 (A(4)), A243265 (A(5)), A266553 (A(6)), A266554 (A(7)), A266555 (A(8)), A266556 (A(9)), A266557 (A(10)), A266558 (A(11)), A266559 (A(12)), A260662 (A(13)), A266560 (A(14)), A266563 (A(16)), A266564 (A(17)), A266565 (A(18)), A266566 (A(19)), A266567 (A(20)).

Programs

  • Mathematica
    Exp[N[(BernoulliB[16]/16)*(EulerGamma + Log[2*Pi] - Zeta'[16]/Zeta[16]), 200]]

Formula

A(k) = exp(H(k)*B(k+1)/(k+1) - zeta'(-k)), where B(k) is the k-th Bernoulli number, H(k) the k-th harmonic number, and zeta'(x) is the derivative of the Riemann zeta function.
A(15) = exp(H(15)*B(16)/16 - zeta'(-15)) = exp((B(16)/16)*(EulerGamma + log(2*Pi) - zeta'(16)/zeta(16))).
Equals (2*Pi*exp(gamma) * Product_{p prime} p^(1/(p^16-1)))^c, where gamma is Euler's constant (A001620), and c = Bernoulli(16)/16 = -3617/8160 (Van Gorder, 2012). - Amiram Eldar, Feb 08 2024

A266563 Decimal expansion of the generalized Glaisher-Kinkelin constant A(16).

Original entry on oeis.org

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

Views

Author

G. C. Greubel, Dec 31 2015

Keywords

Comments

Also known as the 16th Bendersky constant.

Examples

			0.16981839784277560774730955168312711879515291428637735860...
		

Crossrefs

Cf. A019727 (A(0)), A074962 (A(1)), A243262 (A(2)), A243263 (A(3)), A243264 (A(4)), A243265 (A(5)), A266553 (A(6)), A266554 (A(7)), A266555 (A(8)), A266556 (A(9)), A266557 (A(10)), A266558 (A(11)), A266559 (A(12)), A260662 (A(13)), A266560 (A(14)), A266562 (A(15)), A266564 (A(17)), A266565 (A(18)), A266566 (A(19)), A266567 (A(20)).

Programs

  • Mathematica
    Exp[N[(BernoulliB[16]/4)*(Zeta[17]/Zeta[16]), 200]]

Formula

A(k) = exp(H(k)*B(k+1)/(k+1) - zeta'(-k)), where B(k) is the k-th Bernoulli number, H(k) the k-th harmonic number, and zeta'(x) is the derivative of the Riemann zeta function.
A(16) = exp((B(16)/4)*(zeta(17)/zeta(16))) = exp(-zeta'(-16)).
A(16) = exp(-16! * Zeta(17) / (2^17 * Pi^16)). - Vaclav Kotesovec, Jan 01 2016

A053167 Smallest 4th power divisible by n.

Original entry on oeis.org

1, 16, 81, 16, 625, 1296, 2401, 16, 81, 10000, 14641, 1296, 28561, 38416, 50625, 16, 83521, 1296, 130321, 10000, 194481, 234256, 279841, 1296, 625, 456976, 81, 38416, 707281, 810000, 923521, 256, 1185921, 1336336, 1500625, 1296, 1874161, 2085136
Offset: 1

Views

Author

Henry Bottomley, Feb 29 2000

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(e + Mod[4 - Mod[e, 4], 4]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019*)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(f[i,2] + (4-f[i,2])%4));} \\ Amiram Eldar, Oct 27 2022

Formula

a(n) = (n/A000190(n))^4 = (n*A007913(n))^2/A008835(n*A007913(n)).
From Amiram Eldar, Jul 29 2022: (Start)
Multiplicative with a(p^e) = p^(e + ((4-e) mod 4)).
Sum_{n>=1} 1/a(n) = Product_{p prime} ((p^4+3)/(p^4-1)) = 1.341459051107600424... . (End)
Sum_{k=1..n} a(k) ~ c * n^5, where c = (zeta(16)/(5*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.1230279197... . - Amiram Eldar, Oct 27 2022
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