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.

Previous Showing 41-46 of 46 results.

A383057 Decimal expansion of the asymptotic mean of A056671(k)/A034444(k), the ratio between the number of squarefree unitary divisors and the number of unitary divisors over the positive integers.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Apr 15 2025

Keywords

Comments

The asymptotic mean of the inverse ratio A034444(k)/A056671(k) is 15/Pi^2 (A082020).

Examples

			0.78936260126098902910370862925139689276851676052691...
		

Crossrefs

The unitary analog of A308043.

Programs

  • Mathematica
    $MaxExtraPrecision = 300; m = 300; f[p_] := 1 - 1/(2*p^2); c = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, n]*(PrimeZetaP[n]), {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    prodeulerrat(1 - 1/(2*p^2))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A056671(k)/A034444(k).
Equals Product_{p prime} (1 - 1/(2*p^2)).

A384558 The sum of the exponential divisors of n that are exponentially odd numbers (A268335).

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 10, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 30, 5, 26, 30, 14, 29, 30, 31, 34, 33, 34, 35, 6, 37, 38, 39, 50, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 60, 55, 70, 57, 58, 59, 30, 61, 62, 21, 10, 65, 66, 67, 34, 69
Offset: 1

Views

Author

Amiram Eldar, Jun 03 2025

Keywords

Comments

First differs from A384559 at n = 512: a(512) = 522, while A384559(512) = 514.
The number of these divisors is A368979(n), and the largest of them is A331737(n).
The indices of records of a(n)/n are the primorial numbers (A002110) cubed, i.e., 1 and the terms of A115964.

Crossrefs

Programs

  • Maple
    A384558:=proc(n)
        local a, pe,p,e,af,d;
        a := 1;
        for pe in ifactors(n)[2] do
            p := op(1,pe) ;
            e := op(2,pe) ;
            af := 0 ;
            for d in numtheory[divisors](e) do
                if type(d,'odd') then
                    af := af+p^d ;
                end if;
            end do:
            a := a*af ;
        end do;
        a
    end proc:
    seq(A384558(n), n=1..100); # R. J. Mathar, Jun 04 2025
  • Mathematica
    f[p_, e_] := DivisorSum[e, p^# &, OddQ[#] &]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, sumdiv(f[i,2], d, (d % 2) * f[i,1]^d));}

Formula

Multiplicative with a(p^e) = Sum_{d|e, d odd} p^d.
a(n) = n if and only if n is squarefree (A005117).
a(n) < n if and only if n is in A072587.
a(n) > n if and only if n is in A374459.
limsup_{n->oo} a(n)/n = Product_{p prime} (1 + 1/p^2) = 15/Pi^2 (A082020).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 + 1/(p*(p^2-1)) - 1/(p^2-1) + (1-1/p) * Sum_{k>=1} p^(2*k+1)/(p^(4*k+2)-1)) = 0.80824764393216997768... .

A059479 Number of 3 X 3 matrices with elements from {0,...,n-1} such that the middle element of each of the eight lines of three (rows, columns and diagonals) is the square (mod n) of the difference of the end elements.

Original entry on oeis.org

1, 4, 9, 64, 25, 36, 49, 256, 729, 100, 121, 576, 169, 196, 225, 4096, 289, 2916, 361, 1600, 441, 484, 529, 2304, 15625, 676, 6561, 3136, 841, 900, 961, 16384, 1089, 1156, 1225, 46656, 1369, 1444, 1521, 6400, 1681, 1764, 1849, 7744, 18225, 2116, 2209
Offset: 1

Views

Author

John W. Layman, Feb 15 2001

Keywords

Comments

This sequence is multiplicative. - Mitch Harris, Apr 19 2005
The sequence enumerates the solutions of a system of polynomials equations modulo n, hence is multiplicative by the Chinese Remainder Theorem. The middle entry of the 3 X 3 is zero modulo n. - Michael Somos, Apr 30 2005

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(3*e - (Mod[e, 2])); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)
  • PARI
    a(n)=if(n<1,0,n^3/core(n)) /* Michael Somos, Apr 30 2005 */

Formula

a(n) = A008833(n)*n^2, where A008833(n) is the largest square that divides n.
Multiplicative with a(p^e) = p^(3e - (e % 2)). - Mitch Harris, Jun 09 2005
Dirichlet g.f.: zeta(s-2)*zeta(2s-6)/zeta(2s-4). - R. J. Mathar, Mar 30 2011
Sum_{k=1..n} a(k) ~ zeta(3/2) * n^(7/2) / (7*zeta(3)). - Vaclav Kotesovec, Sep 16 2020
Sum_{n>=1} 1/a(n) = 15*zeta(6)/Pi^2 = A082020 * A013664 = 1.546176... . - Amiram Eldar, Nov 03 2022

A380857 Squares of numbers that are neither squarefree nor prime powers.

Original entry on oeis.org

144, 324, 400, 576, 784, 1296, 1600, 1936, 2025, 2304, 2500, 2704, 2916, 3136, 3600, 3969, 4624, 5184, 5625, 5776, 6400, 7056, 7744, 8100, 8464, 9216, 9604, 9801, 10000, 10816, 11664, 12544, 13456, 13689, 14400, 15376, 15876, 17424, 18225, 18496, 19600, 20736
Offset: 1

Views

Author

Michael De Vlieger, Feb 06 2025

Keywords

Comments

Proper subset of A359280 which is a proper subset of A286708 (powerful numbers that are not prime powers, a proper subset of A126706).
Does not intersect A362605.

Crossrefs

Cf. A059404, A126706, A177492 (k^2 for k in A120944), A286708, A359280, A362605, A378768 (k^2 for k in A286708).

Programs

  • Mathematica
    Select[Range[150], Nor[PrimePowerQ[#], SquareFreeQ[#]] &]^2
  • PARI
    isok(k) = !issquarefree(k) && !isprimepower(k); \\ A126706
    apply(sqr, select(isok, [1..200])) \\ Michel Marcus, Feb 07 2025
    
  • Python
    from math import isqrt
    from sympy import primepi, integer_nthroot, mobius
    def A380857(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return int(n+sum(primepi(integer_nthroot(x,k)[0]) for k in range(2,x.bit_length()))+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)))
        return bisection(f,n,n)**2 # Chai Wah Wu, Feb 08 2025

Formula

a(n) = A126706(n)^2.
Sum_{n>=1} 1/a(n) = Pi^2/6 - 15/Pi^2 - Sum_{p prime} 1/(p^2*(p^2-1)) = A013661 - A082020 + A085548 - A154945 = 0.025670434597226178881... . - Amiram Eldar, Feb 08 2025

A383647 Decimal expansion of 15/(2*Pi^4).

Original entry on oeis.org

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

Views

Author

Stefano Spezia, May 03 2025

Keywords

Examples

			0.07699486691013251391864587450339020606370851390...
		

References

  • G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 64.

Crossrefs

Programs

  • Mathematica
    Join[{0},RealDigits[15/(2Pi^4),10,100][[1]]]

Formula

Equals Sum_{n > 0} 1/A030059(n)^4.
Equals 10/A151927. - Hugo Pfoertner, May 03 2025

A335204 Decimal expansion of Product_{p = prime} (1 + 6/p^2).

Original entry on oeis.org

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

Views

Author

Jude Thaddeus Poole Jr., Andrew Hinton, Reid Huntley, May 26 2020

Keywords

Examples

			6.95743589392521746246609006782918530413...
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 10000; Do[Print[5/2*Exp[-N[Sum[(-1)^j*6^j*(PrimeZetaP[2*j] - 1/4^j)/j, {j, 1, t}], 120]]], {t, 100, 1000, 100}] (* Vaclav Kotesovec, May 29 2020 *)
  • PARI
    prodeulerrat(1 + 6/p^2) \\ Amiram Eldar, Mar 17 2021

Extensions

More terms from Vaclav Kotesovec, May 29 2020
Previous Showing 41-46 of 46 results.