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 1-5 of 5 results.

A343978 Number of ordered 6-tuples (a,b,c,d,e,f) with gcd(a,b,c,d,e,f)=1 (1<= {a,b,c,d,e,f} <= n).

Original entry on oeis.org

1, 63, 727, 4031, 15559, 45863, 116855, 257983, 526615, 983583, 1755143, 2935231, 4776055, 7407727, 11256623, 16498719, 23859071, 33434063, 46467719, 62949975, 84644439, 111486599, 146142583, 187854119, 240880239, 303814503, 382049919, 473813703, 586746719
Offset: 1

Views

Author

Karl-Heinz Hofmann, May 06 2021

Keywords

References

  • Joachim von zur Gathen and Jürgen Gerhard, Modern Computer Algebra, Cambridge University Press, Second Edition 2003, pp. 53-54.

Crossrefs

Programs

  • PARI
    a(n)={sum(k=1, n+1, moebius(k)*(n\k)^6)} \\ Andrew Howroyd, May 08 2021
    
  • Python
    from labmath import mobius
    def A343978(n): return sum(mobius(k)*(n//k)**6 for k in range(1, n+1))
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A343978(n):
        if n == 0:
            return 0
        c, j, k1 = 1, 2, n//2
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*A343978(k1)
            j, k1 = j2, n//j2
        return n*(n**5-1)-c+j # Chai Wah Wu, May 17 2021

Formula

a(n) = Sum_{k=1..n} mu(k)*floor(n/k)^6.
Lim_{n->infinity} a(n)/n^6 = 1/zeta(6) = A343359 = 945/Pi^6.
a(n) = n^6 - Sum_{k=2..n} a(floor(n/k)). - Seiichi Manyama, Sep 13 2024

Extensions

Edited by N. J. A. Sloane, Jun 13 2021

A344038 Number of ordered 6-tuples (a,b,c,d,e,f) with gcd(a,b,c,d,e,f)=1 (1<= {a,b,c,d,e,f} <= 10^n).

Original entry on oeis.org

1, 983583, 983029267047, 982960635742968103, 982953384128772770413831, 982952672223441253533233827367, 982952600027678075050509511271466303, 982952593055042000417993486008754893529583, 982952592342881094406730790044111038427637071855
Offset: 0

Views

Author

Karl-Heinz Hofmann, May 07 2021

Keywords

Crossrefs

Related counts of k-tuples:
triples: A071778, A342935, A342841;
quadruples: A082540, A343527, A343193;
5-tuples: A343282;
6-tuples: A343978, A344038. - N. J. A. Sloane, Jun 13 2021

Programs

  • PARI
    a(n)={sum(k=1, 10^n+1, moebius(k)*(10^n\k)^6)} \\ Andrew Howroyd, May 08 2021
  • Python
    from labmath import mobius
    def A344038(n): return sum(mobius(k)*(10**n//k)**6 for k in range(1, 10**n+1))
    

Formula

Lim_{n->infinity} a(n)/10^(6*n) = 1/zeta(6) = A343359 = 945/Pi^4.
a(n) = A343978(10^n).

Extensions

Edited by N. J. A. Sloane, Jun 13 2021

A342683 Decimal expansion of 1/zeta(8).

Original entry on oeis.org

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

Views

Author

Karl-Heinz Hofmann, May 18 2021

Keywords

Comments

1/zeta(8) is the probability that 8 randomly selected numbers will be coprime.

Examples

			0.9959392011255151468348364728055453240050227784589...
		

Crossrefs

Programs

  • Maple
    evalf(9450/Pi^8,100) ; # R. J. Mathar, Jun 04 2021
  • Mathematica
    RealDigits[1/Zeta[8], 10, 100][[1]] (* Amiram Eldar, May 18 2021 *)
  • PARI
    1/zeta(8)

Formula

Equals 1/A013666 = 9450/Pi^8.
From Amiram Eldar, Jun 01 2023: (Start)
Equals Sum_{k>=1} mu(k)/k^8, where mu is the Möbius function (A008683).
Equals Product_{p prime} (1 - 1/p^8). (End)

A355265 Bicubeful numbers.

Original entry on oeis.org

64, 128, 192, 256, 320, 384, 448, 512, 576, 640, 704, 729, 768, 832, 896, 960, 1024, 1088, 1152, 1216, 1280, 1344, 1408, 1458, 1472, 1536, 1600, 1664, 1728, 1792, 1856, 1920, 1984, 2048, 2112, 2176, 2187, 2240, 2304, 2368, 2432, 2496, 2560, 2624, 2688, 2752
Offset: 1

Views

Author

Peter Luschny, Jul 12 2022

Keywords

Comments

Let lp(n, e) denote the largest positive integer b such that b^e divides n. For example for e = 1, 2, 3, 4 the sequences (lp(n, e), n >= 1) are A000027, A000188, A053150, and A053164. Let rad(n) = A007947(n) be the squarefree kernel of n. k is in this sequence if lp(n, 3) does not divide rad(n). The case e = 1 gives A013929, and the case e = 2 is A046101.
The asymptotic density of this sequence is 1 - 1/zeta(6) = 1 - 945/Pi^6 = 0.017047... . - Amiram Eldar, Jul 13 2022

Examples

			n = 512 = 2^9, rad(n) = 2, lp(n, 3) = 8 since n/8^3 = 1. But 8 does not divide 2.
n = 704 = 2^6*11, rad(n) = 22, lp(n, 3) = 4 since n/4^3 = 11. But 4 does not divide 22.
		

Crossrefs

Cf. A007947, A000188, A053150, A053164, A013929, A046101 (biquadrateful).

Programs

  • Maple
    with(NumberTheory):
    isBicubeful := n -> irem(Radical(n), LargestNthPower(n, 3)) <> 0:
    select(isBicubeful, [`$`(1..2752)]);
  • Mathematica
    bicubQ[n_] := AnyTrue[FactorInteger[n][[;; , 2]], # > 5 &]; Select[Range[3000], bicubQ] (* Amiram Eldar, Jul 13 2022 *)
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A355265_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:any(map(lambda m:m>5,factorint(n).values())),count(max(startvalue,1)))
    A355265_list = list(islice(A355265_gen(),30)) # Chai Wah Wu, Jul 12 2022

Formula

A number k is bicubeful iff it is divisible by the 6th power of an integer > 1.

A341901 Decimal expansion of 1/zeta(9).

Original entry on oeis.org

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

Views

Author

Karl-Heinz Hofmann, Jun 04 2021

Keywords

Comments

1/zeta(9) is the probability that 9 randomly selected numbers will be coprime.

Examples

			0.997995632730762156864676132105099996320941848...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[1/Zeta[9], 10, 100][[1]]
  • PARI
    1/zeta(9)

Formula

Equals 1/A013667.
From Amiram Eldar, Jun 01 2023: (Start)
Equals Sum_{k>=1} mu(k)/k^9, where mu is the Möbius function (A008683).
Equals Product_{p prime} (1 - 1/p^9). (End)
Showing 1-5 of 5 results.