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 33 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

Views

Author

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

Views

Author

Keywords

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

Views

Author

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)

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

Views

Author

Keywords

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

Views

Author

Keywords

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

Views

Author

Keywords

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)

A013668 Decimal expansion of zeta(10).

Original entry on oeis.org

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

Views

Author

Keywords

Examples

			1.0009945751278180853371459589003190170060195315644775172577889946362914...
		

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

Equals Pi^10/93555.
zeta(10) = 4/3*2^10/(2^10 - 1)*( Sum_{n even} n^2*p(n)/(n^2 - 1)^11 ), where p(n) = 3*n^10 + 55*n^8 + 198*n^6 + 198*n^4 + 55*n^2 + 3 is a row polynomial of A091043. - Peter Bala, Dec 05 2013
zeta(10) = Sum_{n >= 1} (A010052(n)/n^5) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^5 ). - Mikael Aaltonen, Feb 20 2015
zeta(10) = Product_{k>=1} 1/(1 - 1/prime(k)^10). - Vaclav Kotesovec, May 02 2020
From Wolfdieter Lang, Sep 16 2020: (Start)
zeta(10) = (1/9!)*Integral_{0..infinity} x^9/(exp(x) - 1). See Abramowitz-Stegun, 23.2.7., for s=10, p. 807. The value of the integral is (128/33)*Pi^10 = (3.6324091...)*10^5.
zeta(10) = (4/1448685)*Integral_{0..infinity} x^9/(exp(x) + 1). See Abramowitz-Stegun, 23.2.8., for s=10, p. 807. The value of the integral is (511/132)*Pi^10 = (3.625314565...)*10^5. (End)

A030631 Numbers with 13 divisors.

Original entry on oeis.org

4096, 531441, 244140625, 13841287201, 3138428376721, 23298085122481, 582622237229761, 2213314919066161, 21914624432020321, 353814783205469041, 787662783788549761, 6582952005840035281, 22563490300366186081, 39959630797262576401, 116191483108948578241
Offset: 1

Views

Author

Keywords

Comments

12th powers of primes. The n-th number with p divisors is equal to the n-th prime raised to power p-1, where p is prime. - Omar E. Pol, May 06 2008

Crossrefs

Subsequence of A008456.

Programs

Formula

a(n) = A000040(n)^(13-1) = A000040(n)^(12). - Omar E. Pol, May 06 2008
From Amiram Eldar, Jan 24 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(12)/zeta(24) = 218517792968475/(236364091*Pi^12).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(12) = 638512875/(691*Pi^12) = 1/A013670. (End)

A008456 12th powers: a(n) = n^12.

Original entry on oeis.org

0, 1, 4096, 531441, 16777216, 244140625, 2176782336, 13841287201, 68719476736, 282429536481, 1000000000000, 3138428376721, 8916100448256, 23298085122481, 56693912375296, 129746337890625, 281474976710656, 582622237229761
Offset: 0

Views

Author

Keywords

Comments

Numbers which are square, cubic and quartic. - Doug Bell, Jun 03 2017

Crossrefs

a(n) = A123868(n) + 1.
Cf. A000290 (squares), A000578 (cubes), A000583 (4th powers), A001014 (6th powers), A008454 (10th powers), A008455 (11th powers), A010801 (13th powers).
Cf. A013670 (zeta(12)).

Programs

Formula

Multiplicative with a(p^e) = p^(12*e). - David W. Wilson, Aug 01 2001
a(n) = A000290(n)^6 = A000578(n)^4 = A000583(n)^3 = A001014(n)^2. - Doug Bell, Jun 03 2017
From Amiram Eldar, Oct 08 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(12) = 691*Pi^12/638512875 (A013670).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2047*zeta(12)/2048 = 1414477*Pi^12/1307674368000. (End)
a(n) = 13*a(n-1)-78*a(n-2)+286*a(n-3)-715*a(n-4)+1287*a(n-5)-1716*a(n-6)+1716*a(n-7)-1287*a(n-8)+715*a(n-9)-286*a(n-10)+78*a(n-11)-13*a(n-12)+a(n-13). - Wesley Ivan Hurt, Dec 02 2021
Intersection of A000578 and A000583; i.e., cubes and 4th powers. - M. F. Hasler, Jul 03 2025

A013959 a(n) = sigma_11(n), the sum of the 11th powers of the divisors of n.

Original entry on oeis.org

1, 2049, 177148, 4196353, 48828126, 362976252, 1977326744, 8594130945, 31381236757, 100048830174, 285311670612, 743375541244, 1792160394038, 4051542498456, 8649804864648, 17600780175361, 34271896307634, 64300154115093, 116490258898220, 204900053024478
Offset: 1

Views

Author

Keywords

Comments

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
Related to congruence properties of the Ramanujan tau function since A000594(n) == a(n) (mod 691) = A046694(n). - Benoit Cloitre, Aug 28 2002

Crossrefs

Programs

Formula

G.f.: Sum_{k>=1} k^11*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
Dirichlet g.f.: zeta(s-11)*zeta(s). - Ilya Gutkovskiy, Sep 10 2016
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(11*e+11)-1)/(p^11-1).
Sum_{k=1..n} a(k) = zeta(12) * n^12 / 12 + O(n^13). (End)
Showing 1-10 of 33 results. Next