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 117 results. Next

A001014 Sixth powers: a(n) = n^6.

Original entry on oeis.org

0, 1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441, 1000000, 1771561, 2985984, 4826809, 7529536, 11390625, 16777216, 24137569, 34012224, 47045881, 64000000, 85766121, 113379904, 148035889, 191102976, 244140625, 308915776, 387420489, 481890304
Offset: 0

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Comments

Numbers both square and cubic. - Patrick De Geest
Totally multiplicative sequence with a(p) = p^6 for prime p. - Jaroslav Krizek, Nov 01 2009
Numbers n for which the order of the torsion subgroup of the elliptic curve y^2 = x^3 + n is t = 6, cf. Gebel link. - Artur Jasinski, Jun 30 2010
Note that Sum_{n>=1} 1/a(n) = Pi^6 / 945. - Mohammad K. Azarian, Nov 01 2011
The binomial transform yields A056468. The inverse binomial transform yields the (finite) 0, 1, 62, 540, ..., 720, the 6th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
For n > 0, a(n) is the largest number k such that k + n^3 divides k^2 + n^3. - Derek Orr, Oct 01 2014

Examples

			The 6th powers of the first few integers are: 0^6 = 0 = a(0), 1^6 = 1 = a(1), 2^6 = 64 = a(2), 3^6 = 9^3 = 729 = a(3), 4^6 = 2^12 = 4096 = a(4), 5^6 = 25^3 = 15625 = a(5), etc.
		

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, eq. (6.37).
  • Granino A. Korn and Theresa M.Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
  • 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

Subsequence of A201217.
Cf. A000540 (partial sums), A022522 (first differences), A008292.
Intersection of A000290 (squares) and A000578 (cubes).
Cf. A002604 (n^6+1), A123866 (n^6-1), A013664 (zeta(6)), A275703 (eta(6)).
Cf. A003358 - A003368 (sums of 2, ..., 12 positive sixth powers).

Programs

Formula

a(n) = A123866(n) + 1 = A002604(n) - 1.
G.f.: -x*(1+x)*(x^4+56*x^3+246*x^2+56*x+1) / (x-1)^7. - Simon Plouffe in his 1992 dissertation
Multiplicative with a(p^e) = p^(6e). - David W. Wilson, Aug 01 2001
E.g.f.: (x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6)*exp(x). Generally, the e.g.f. for n^m is Sum_{k=1..m} A008277(m,k)*x^k*exp(x). - Geoffrey Critzer, Aug 25 2013
From Ant King, Sep 23 2013: (Start)
Signature {7, -21, 35, -35, 21, -7, 1}.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + 720. (End)
a(n) == 1 (mod 7) if gcd(n, 7) = 1, otherwise a(n) == 0 (mod 7). See A109720. - Jake Lawrence, May 28 2016
From Ilya Gutkovskiy, Jul 06 2016: (Start)
Dirichlet g.f.: zeta(s-6).
Sum_{n>=1} 1/a(n) = Pi^6/945 = A013664. (End)
a(n) = Sum_{k=1..6} Eulerian(6, k)*binomial(n+6-k, 6), with Eulerian(6, k) = A008292(6, k) (the numbers are 1, 57, 302, 302, 57, 1) for n >= 0. Worpitzki's identity for powers of 6. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = 31*zeta(6)/32 = 31*Pi^6/30240 (A275703). - Amiram Eldar, Oct 08 2020
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (cosh(Pi)-cos(sqrt(3)*Pi))*sinh(Pi)/(2*Pi^3).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)^2/(6*Pi^2). (End)

Extensions

Comments from 2010 - 2011 edited by M. F. Hasler, Jul 05 2024

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

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

A030516 Numbers with 7 divisors. 6th powers of primes.

Original entry on oeis.org

64, 729, 15625, 117649, 1771561, 4826809, 24137569, 47045881, 148035889, 594823321, 887503681, 2565726409, 4750104241, 6321363049, 10779215329, 22164361129, 42180533641, 51520374361, 90458382169, 128100283921
Offset: 1

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Comments

These are the numbers p^6 with p prime. - Jorge Coveiro, Apr 13 2004
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

Programs

Formula

a(n) = A000040(n)^(7-1) = A000040(n)^6. - Omar E. Pol, May 06 2008
A056595(a(n)) = 3. - Reinhard Zumkeller, Aug 15 2011
Sum_{n>=1} 1/a(n) = P(6) = 0.0170700868... (A085966). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(6)/zeta(12) = 675675/(691*Pi^6) (A269404).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(6) = 945/Pi^6 = 1/A013664. (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)

A082695 Decimal expansion of zeta(2)*zeta(3)/zeta(6).

Original entry on oeis.org

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

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Author

Benoit Cloitre, Apr 12 2003

Keywords

Comments

Equals the Dirichlet zeta-function Sum_{n>=1} A001615(n)/n^s at s=3. - R. J. Mathar, Apr 02 2011
Dressler shows that this is the average value of A014197, that is, the number of values m such that phi(m) <= n is asymptotically n times this constant. Erdős had shown earlier that this limit exists. - Charles R Greathouse IV, Nov 26 2013
From Stanislav Sykora, Nov 14 2014: (Start)
Equals lim_{n->infinity} (Sum_{k=1..n} k/phi(k))/n, i.e., the limit mean value of k/phi(k), where phi(k) is Euler's totient function.
Also equals lim_{n->infinity} (Sum_{k=1..n} 1/phi(k))/log(n).
Proofs are trivial using the formulas for Sum_{k=1..n} k/phi(k) and Sum_{k=1..n} 1/phi(k) listed in the Wikipedia link.
For the limit mean value of phi(k)/k, see A059956. (End)
The asymptotic mean of A005361. - Amiram Eldar, Apr 13 2020

Examples

			1.94359643682075920505707036257476343718785850176780571602663568890 ...
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.7, p. 116.
  • Joe Roberts, Lure of the Integers, Mathematical Association of America, 1992. See p. 74.

Crossrefs

Programs

  • Mathematica
    First@RealDigits[ Zeta[2]*Zeta[3]/Zeta[6], 10, 100]
    RealDigits[ 315 Zeta[3]/(2 Pi^4), 10, 111][[1]] (* Robert G. Wilson v, Aug 11 2014 *)
  • PARI
    zeta(3)*315/2/Pi^4

Formula

Decimal expansion of Product_{p prime} (1+1/p/(p-1)) = zeta(2)*zeta(3)/zeta(6) = 1.94359643682075920505707...
The sum of the reciprocals of the powerful numbers, A001694. - T. D. Noe, May 03 2006
Equals A013661 * A002117 / A013664 = 1 / A068468 = zeta(3) * 315/(2*Pi^4) = zeta(3) * A157292.
Equals Sum_{k>=1} mu(k)^2/(k*phi(k)) (the sum of reciprocals of the squarefree numbers multiplied by their Euler totient function values, A000010). - Amiram Eldar, Aug 18 2020

Extensions

New definition from Eric W. Weisstein, May 04 2006

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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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

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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)

A013670 Decimal expansion of zeta(12).

Original entry on oeis.org

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

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Author

Keywords

Examples

			1.0002460865533080482986379980477396709604160884580034045330409521332520...
		

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(12) = 2/3*2^12/(2^12 - 1)*( Sum_{n even} n^2*p(n)/(n^2 - 1)^13 ), where p(n) = 7*n^12 + 182*n^10 + 1001*n^8 + 1716*n^6 + 1001*n^4 + 182*n^2 + 7 is a row polynomial of A091043. - Peter Bala, Dec 05 2013
zeta(12) = Sum_{n >= 1} (A010052(n)/n^6) = Sum {n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^6 ). - Mikael Aaltonen, Feb 20 2015
zeta(12) = 691/638512875*Pi^12 (see A002432). - Rick L. Shepherd, May 30 2016
zeta(12) = Product_{k>=1} 1/(1 - 1/prime(k)^12). - Vaclav Kotesovec, May 02 2020
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