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 1-3 of 3 results.

A211338 Numbers k for which the number of divisors, tau(k), is congruent to 2 modulo 3.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 16, 17, 19, 23, 24, 29, 30, 31, 37, 40, 41, 42, 43, 47, 53, 54, 56, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 88, 89, 97, 101, 102, 103, 104, 105, 107, 109, 110, 113, 114, 127, 128, 130, 131, 135, 136, 137, 138, 139, 149, 151, 152, 154
Offset: 1

Views

Author

Douglas Latimer, Apr 07 2012

Keywords

Comments

The product of any 2 terms a(i)*a(j) is not a member of the sequence.
Any term a(n) can be expressed as 1 term (required to be greater than 1) from A211485 times 1 nonzero term from A000578. - Douglas Latimer, Apr 20 2012
The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 4, 37, 368, 3681, 36596, 365336, 3653499, 36537962, 365381169, 3653826361, ... . Conjecture: the asymptotic density of this sequence exists and equals 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602), so, a(n) ~ k*n with k = Pi^2/(3*zeta(3)) = 2.73686555524... . This conjecture is true if this sequence and A211337 have the same density (see A059269). - Amiram Eldar, Jan 06 2024

Examples

			The divisors of 16 are: 1, 2, 4, 8, 16 (5 divisors). 5 is congruent to 2 modulo 3. Thus 16 is a member of this sequence.

Links

Crossrefs

This is an extension of A000040 (the prime numbers, which each have 2 divisors).
The union of A059269 and A211337 is the complementary sequence to this one.
The definition of this sequence uses A000005 (the number of divisors of n).
Cf. A000578, A074796, A211485, A346602.

Programs

  • Mathematica
    Select[Range[154], Mod[DivisorSigma[0, #], 3] == 2 &] (* T. D. Noe, Apr 21 2012 *)
  • PARI
    {plnt=1 ; mxind=100 ;for(k=1, 10^6,
if(numdiv(k) % 3 == 2, print(k); plnt++; if(mxind+1 == plnt, break() )))}

Formula

Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012

A074794 Number of numbers k <= n such that tau(k) == 1 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 11, 12, 13, 13, 13, 14, 15, 15, 15, 15, 15, 15, 15, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 20, 21, 21, 21, 21, 22, 22, 23, 24, 24, 24, 24, 25, 25, 25, 25, 25, 26, 26, 26, 27, 27
Offset: 1

Views

Author

Benoit Cloitre, Sep 07 2002

Keywords

Links

Crossrefs

Cf. A000005, A074795, A074796, A211337, A346602.

Programs

  • Mathematica
    Accumulate[Table[Boole[Mod[DivisorSigma[0, n], 3] == 1], {n, 1, 100}]] (* Amiram Eldar, Aug 29 2020 *)
  • PARI
    a(n)=sum(k=1,n,if(numdiv(k)%3-1,0,1))

Formula

a(n) is asymptotic to c*n with c = 0.36....
The constant is conjecturally 3*zeta(3)/Pi^2 = 0.3653814847007... (A346602). See A211337 for more details. - Amiram Eldar, Feb 01 2025

A074795 Number of numbers k <= n such that tau(k) == 0 (mod 3) where tau(k) = A000005(k) is the number of divisors of k.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 11, 11, 11, 11, 12, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21
Offset: 1

Views

Author

Benoit Cloitre, Sep 07 2002

Keywords

Links

Crossrefs

Cf. A000005, A059269, A074794, A074796.

Programs

  • Mathematica
    Accumulate[Boole[Divisible[DivisorSigma[0,Range[90]],3]]] (* Harvey P. Dale, Jan 11 2015 *)
  • PARI
    a(n)=sum(k=1,n,if(numdiv(k)%3,0,1))

Formula

a(n) is asymptotic to c*n with c = 0.26....
The constant is c = 1 - zeta(3)/zeta(2) = 1 - 6*zeta(3)/Pi^2 = 0.2692370305 ... (Sathe, 1945). - Amiram Eldar, Aug 29 2020
Showing 1-3 of 3 results.

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