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 177 results. Next

A000583 Fourth powers: a(n) = n^4.

Original entry on oeis.org

0, 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625, 65536, 83521, 104976, 130321, 160000, 194481, 234256, 279841, 331776, 390625, 456976, 531441, 614656, 707281, 810000, 923521, 1048576, 1185921
Offset: 0

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Comments

Figurate numbers based on 4-dimensional regular convex polytope called the 4-measure polytope, 4-hypercube or tesseract with Schlaefli symbol {4,3,3}. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004
Totally multiplicative sequence with a(p) = p^4 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A058649. The inverse binomial transforms yields the (finite) 0, 1, 14, 36, 24, the 4th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Generate Pythagorean triangles with parameters a and b to get sides of lengths x = b^2-a^2, y = 2*a*b, and z = a^2 + b^2. In particular use a=n-1 and b=n for a triangle with sides (x1,y1,z1) and a=n and b=n+1 for another triangle with sides (x2,y2,z2). Then x1*x2 + y1*y2 + z1*z2 = 8*a(n). - J. M. Bergot, Jul 22 2013
For n > 0, a(n) is the largest integer k such that k^4 + n is a multiple of k + n. Also, for n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n^2. - Derek Orr, Sep 04 2014
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
a(n+2)/2 is the area of a trapezoid with vertices at (T(n), T(n+1)), (T(n+1), T(n)), (T(n+1), T(n+2)), and (T(n+2), T(n+1)) with T(n)=A000292(n) for n >= 0. - J. M. Bergot, Feb 16 2018

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
  • 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).
  • Dov Juzuk, Curiosa 56: An interesting observation, Scripta Mathematica 6 (1939), 218.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 47.

Crossrefs

Programs

Formula

a(n) = A123865(n)+1 = A002523(n)-1.
Multiplicative with a(p^e) = p^(4e). - David W. Wilson, Aug 01 2001
G.f.: x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5. More generally, g.f. for n^m is Euler(m, x)/(1-x)^(m+1), where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292).
Dirichlet generating function: zeta(s-4). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: (x + 7*x^2 + 6*x^3 + x^4)*e^x. More generally, the general form for the e.g.f. for n^m is phi_m(x)*e^x, where phi_m is the exponential polynomial of order n. - Franklin T. Adams-Watters, Sep 11 2005
Sum_{k>0} 1/a(k) = Pi^4/90 = A013662. - Jaume Oliver Lafont, Sep 20 2009
a(n) = C(n+3,4) + 11*C(n+2,4) + 11*C(n+1,4) + C(n,4). [Worpitzky's identity for powers of 4. See, e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n*A177342(n) - Sum_{i=1..n-1} A177342(i) - (n - 1), with n > 1. - Bruno Berselli, May 07 2010
a(n) + a(n+1) + 1 = 2*A002061(n+1)^2. - Charlie Marion, Jun 13 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 24. - Ant King, Sep 23 2013
From Amiram Eldar, Jan 20 2021: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/720 (A267315).
Product_{n>=2} (1 - 1/a(n)) = sinh(Pi)/(4*Pi). (End)

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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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)

A085964 Decimal expansion of the prime zeta function at 4.

Original entry on oeis.org

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

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Author

Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Jul 06 2003

Keywords

Comments

Mathar's Table 1 (cited below) lists expansions of the prime zeta function at integers s in 10..39. - Jason Kimberley, Jan 05 2017

Examples

			0.0769931397642468449426...
		

References

  • Henri Cohen, Number Theory, Volume II: Analytic and Modern Tools, GTM Vol. 240, Springer, 2007; see pp. 208-209.
  • J. W. L. Glaisher, On the Sums of Inverse Powers of the Prime Numbers, Quart. J. Math. 25, 347-362, 1891.

Crossrefs

Decimal expansion of the prime zeta function: A085548 (at 2), A085541 (at 3), this sequence (at 4), A085965 (at 5) to A085969 (at 9).

Programs

  • Magma
    R := RealField(106);
    PrimeZeta := func;
    [0]cat Reverse(IntegerToSequence(Floor(PrimeZeta(4,87)*10^105)));
    // Jason Kimberley, Dec 30 2016
    
  • Mathematica
    s[n_] := s[n] = Sum[ MoebiusMu[k]*Log[Zeta[4*k]]/k, {k, 1, n}] // RealDigits[#, 10, 104]& // First // Prepend[#, 0]&; s[100]; s[n = 200]; While[s[n] != s[n - 100], n = n + 100]; s[n] (* Jean-François Alcover, Feb 14 2013 *)
    RealDigits[ PrimeZetaP[ 4], 10, 111][[1]] (* Robert G. Wilson v, Sep 03 2014 *)
  • PARI
    sumeulerrat(1/p,4) \\ Hugo Pfoertner, Feb 03 2020

Formula

P(4) = Sum_{p prime} 1/p^4 = Sum_{n>=1} mobius(n)*log(zeta(4*n))/n
Equals A086034 + A085993 + 1/16. - R. J. Mathar, Jul 22 2010
Equals Sum_{k>=1} 1/A030514(k). - Amiram Eldar, Jul 27 2020

A059376 Jordan function J_3(n).

Original entry on oeis.org

1, 7, 26, 56, 124, 182, 342, 448, 702, 868, 1330, 1456, 2196, 2394, 3224, 3584, 4912, 4914, 6858, 6944, 8892, 9310, 12166, 11648, 15500, 15372, 18954, 19152, 24388, 22568, 29790, 28672, 34580, 34384, 42408, 39312, 50652, 48006, 57096
Offset: 1

Views

Author

N. J. A. Sloane, Jan 28 2001

Keywords

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 199, #3.
  • R. Sivaramakrishnan, "The many facets of Euler's totient. II. Generalizations and analogues", Nieuw Arch. Wisk. (4) 8 (1990), no. 2, 169-187.

Crossrefs

See A059379 and A059380 (triangle of values of J_k(n)), A000010 (J_1), A007434 (J_2), A059377 (J_4), A059378 (J_5), A069091 - A069095 (J_6 through J_10).

Programs

  • Maple
    J := proc(n,k) local i,p,t1,t2; t1 := n^k; for p from 1 to n do if isprime(p) and n mod p = 0 then t1 := t1*(1-p^(-k)); fi; od; t1; end; # (with k = 3)
    A059376 := proc(n)
        add(d^3*numtheory[mobius](n/d),d=numtheory[divisors](n)) ;
    end proc: # R. J. Mathar, Nov 03 2015
  • Mathematica
    JordanJ[n_, k_: 1] := DivisorSum[n, #^k*MoebiusMu[n/#] &]; f[n_] := JordanJ[n, 3]; Array[f, 39]
    f[p_, e_] := p^(3*e) - p^(3*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 12 2020 *)
  • PARI
    for(n=1,120,print1(sumdiv(n,d,d^3*moebius(n/d)),","))
    
  • PARI
    for (n = 1, 1000, write("b059376.txt", n, " ", sumdiv(n, d, d^3*moebius(n/d))); ) \\ Harry J. Smith, Jun 26 2009
    
  • PARI
    seq(n) = dirmul(vector(n,k,k^3), vector(n,k,moebius(k)));
    seq(39)  \\ Gheorghe Coserea, May 11 2016
    
  • Python
    from math import prod
    from sympy import factorint
    def A059376(n): return prod(p**(3*(e-1))*(p**3-1) for p, e in factorint(n).items()) # Chai Wah Wu, Jan 29 2024

Formula

Multiplicative with a(p^e) = p^(3e) - p^(3e-3). - Vladeta Jovovic, Jul 26 2001
a(n) = Sum_{d|n} d^3*mu(n/d). - Benoit Cloitre, Apr 05 2002
Dirichlet generating function: zeta(s-3)/zeta(s). - Franklin T. Adams-Watters, Sep 11 2005
A063453(n) divides a(n). - R. J. Mathar, Mar 30 2011
a(n) = Sum_{k=1..n} gcd(k,n)^3 * cos(2*Pi*k/n). - Enrique Pérez Herrero, Jan 18 2013
a(n) = n^3*Product_{distinct primes p dividing n} (1-1/p^3). - Tom Edgar, Jan 09 2015
G.f.: Sum_{n>=1} a(n)*x^n/(1 - x^n) = x*(1 + 4*x + x^2)/(1 - x)^4. - Ilya Gutkovskiy, Apr 25 2017
Sum_{d|n} a(d) = n^3. - Werner Schulte, Jan 12 2018
Sum_{k=1..n} a(k) ~ 45*n^4 / (2*Pi^4). - Vaclav Kotesovec, Feb 07 2019
From Amiram Eldar, Oct 12 2020: (Start)
lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k^3 = 1/zeta(4) (A215267).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + p^3/(p^3-1)^2) = 1.2253556451... (End)
O.g.f.: Sum_{n >= 1} mu(n)*x^n*(1 + 4*x^n + x^(2*n))/(1 - x^n)^4 = x + 7*x^2 + 26*x^3 + 56*x^4 + 124*x^5 + .... - Peter Bala, Jan 31 2022
From Peter Bala, Jan 01 2024: (Start)
a(n) = Sum_{d divides n} d * J_2(d) * phi(n/d) = Sum_{d divides n} d^2 * phi(d) * J_2(n/d), where J_2(n) = A007434(n).
a(n) = Sum_{k = 1..n} gcd(k, n) * J_2(gcd(k, n)) = Sum_{1 <= j, k <= n} gcd(j, k, n)^2 * J_1(gcd(j, k, n)). (End)
a(n) = Sum_{1 <= i, j <= n, lcm(i, j) = n} phi(i)*J_2(j) = Sum_{1 <= i, j, k <= n, lcm(i, j, k) = n} phi(i)*phi(j)*phi(k), where J_2(n) = A007434(n). - Peter Bala, Jan 29 2024

A082020 Decimal expansion of 15/Pi^2.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, May 09 2003

Keywords

Comments

3/(2*Pi^2) (the same decimal expansion with an offset 0) is the probability that the greatest common divisor of two numbers selected at random is 2 (Christopher, 1956). - Amiram Eldar, May 23 2020
Sum of 1/n^2 over all squarefree n, see Penn link. - Charles R Greathouse IV, Jan 01 2022
Equals the asymptotic mean of the abundancy index of the cubefree numbers (A004709) (Jakimczuk and Lalín, 2022). - Amiram Eldar, May 12 2023

Examples

			1.51981775463506657...
		

Crossrefs

Programs

Formula

Product_{n >= 1} (1+1/prime(n)^2) = 15/Pi^2 (Ramanujan).
Equals zeta(2)/zeta(4) = A013661/A013662 = Sum_{n>=1} mu(n)^2/n^2 = Sum_{n>=1} |mu(n)|/n^2 . - Enrique Pérez Herrero, Jan 15 2012
Equals Sum_{n>=1} 1/A005117(n)^2 . - Enrique Pérez Herrero, Mar 30 2012
Equals lim_{n->oo} (1/n) * Sum_{k=1..n} psi(k)/k, where psi(k) is the Dedekind psi function (A001615). - Amiram Eldar, May 12 2019.
Equals Sum_{k>=1} A007434(k)/k^4. - Amiram Eldar, Jan 25 2024

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)
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