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 1-9 of 9 results.

A059269 Numbers m for which the number of divisors, tau(m), is divisible by 3.

Original entry on oeis.org

4, 9, 12, 18, 20, 25, 28, 32, 36, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 100, 108, 116, 117, 121, 124, 126, 132, 140, 144, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 180, 188, 196, 198, 200, 204, 207, 212, 220, 224, 225, 228
Offset: 1

Views

Author

Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001

Keywords

Comments

tau(n) is divisible by 3 iff at least one prime in the prime factorization of n has exponent of the form 3*m + 2. This sequence is an extension of the sequence A038109 in which the numbers has at least one prime with exponent 2 (the case of m = 0 here ) in their prime factorization.
The union of A211337 and A211338 is the complementary sequence to this one. - Douglas Latimer, Apr 12 2012
Numbers whose cubefree part (A050985) is not squarefree (A005117). - Amiram Eldar, Mar 09 2021

Examples

			a(7) = 28 is a term because the number of divisors of 28, d(28) = 6, is divisible by 3.
		

Crossrefs

Characteristic function: A353470.

Programs

  • Maple
    with(numtheory): for n from 1 to 1000 do if tau(n) mod 3 = 0 then printf(`%d,`,n) fi: od:
  • Mathematica
    Select[Range[230], Divisible[DivisorSigma[0, #], 3] &] (* Amiram Eldar, Jul 26 2020 *)
  • PARI
    is(n)=vecmax(factor(n)[,2]%3)==2 \\ Charles R Greathouse IV, Apr 10 2012
    
  • PARI
    is(n)=numdiv(n)%3==0 \\ Charles R Greathouse IV, Sep 18 2015

Formula

Conjecture: a(n) ~ k*n where k = 1/(1 - Product(1 - (p-1)/(p^(3*i)))) = 3.743455... where p ranges over the primes and i ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012
The asymptotic density of this sequence is 1 - zeta(3)/zeta(2) = 1 - 6*zeta(3)/Pi^2 = 0.2692370305... (Sathe, 1945). Therefore, the above conjecture, a(n) ~ k*n, is true, but k = 1/(1-6*zeta(3)/Pi^2) = 3.7141993349... - Amiram Eldar, Jul 26 2020
A001248 UNION A030515 UNION A030627 UNION A030630 UNION A030633 UNION A030636 UNION ... - R. J. Mathar, May 05 2023

Extensions

More terms from James Sellers, Jan 24 2001

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.
		

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).

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

A336590 Numbers k such that k/A008834(k) is squarefree, where A008834(k) is the largest cube dividing k.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88
Offset: 1

Views

Author

Amiram Eldar, Jul 26 2020

Keywords

Comments

Numbers such that none of the exponents in their prime factorization is of the form 3*m + 2.
Cohen (1962) proved that for a given number k >= 2 the asymptotic density of numbers whose exponents in their prime factorization are of the forms k*m or k*m + 1 only is zeta(k)/zeta(2). In this sequence k = 3, and therefore its asymptotic density is zeta(3)/zeta(2) = 6*zeta(3)/Pi^2 = 0.7307629694... (A253905).
Also, numbers k whose number of divisors, A000005(k), is not divisible by 3, i.e., complement of A059269.

Examples

			6 is a term since 6 = 2^1 * 3^1 and 1 is not of the form 3*m + 2.
9 is not a term since 9 = 3^2 and 2 is of the form 3*m + 2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[100], Max[Mod[FactorInteger[#][[;;,2]], 3]] < 2 &]

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

Crossrefs

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

A353446 Let g be the inverse Möbius transform of the Eisenstein integer-valued function f defined in A353445. a(n) is twice the real part of g(n).

Original entry on oeis.org

2, 1, 1, 0, 1, 2, 1, 2, 0, -1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 0, -1, -1, 1, 1, 0, 2, 2, 0, 1, 1, 1, 0, 2, -1, 2, 0, 1, 2, -1, 1, 1, 1, 1, 0, 0, -1, 1, 2, 0, 0, 2, 0, 1, 1, -1, 1, -1, 2, 1, 0, 1, -1, 0, 2, 2, 1, 1, 0, 2, 1, 1, 0, 1, 2, 0, 0, 2, 1, 1, -1, 1, -1, 1, 0, -1, 2, -1, 1, 1, 0, -1, 0, 2, -1, 2, 0, 1, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen and Peter Munn, Apr 19 2022

Keywords

Comments

The imaginary part of g(n) is A353354(n)*(sqrt(3)/2)*i.
f(n), g(n), and so also a(n), are determined by the cubefree part of n, A050985(n). If the cubefree part is not squarefree, g(n) is 0; otherwise g(n) = x^(A195017(A050985(n))), where x = (1 + sqrt(3)*i)/2, the primitive 6th root of unity with positive imaginary part.
The above formula arises from g being multiplicative (because f is multiplicative). g(prime(m)^k) is 1 for k == 0 (mod 3), 0 for k == 2 (mod 3), and for k == 1 (mod 3) the result depends on the parity of m. g(prime(m)^(3k+1)) is 1+w for odd m, -w for even m, where w is the cube root of unity with positive imaginary part. 1+w and -w are the primitive 6th roots of unity.
So the range of g is the 6 sixth roots of unity and 0 itself: these are the 7 Eisenstein integers closest to 0, and they are clearly closed under multiplication. The range of (a(n)) is [-2..2]. g(n) and a(n) are 0 if and only if the cubefree part of n is not squarefree. (Compare with the Moebius function being 0 when its argument is not squarefree.) Otherwise a(n) is even if and only if n is in A332820.

Crossrefs

Sequences used in a definition of this sequence: A008966, A050985, A090882, A087204, A195017, A353445.
See A353354 for the imaginary part.
Positions of particular values depend on A059269, A211337, A211338, A332820 as shown in the formula section.
Positions of even numbers: A353355.

Programs

  • PARI
    A050985(n) = { my(f=factor(n)); f[, 2] = apply(x->(x % 3), f[, 2]); factorback(f); }; \\ From A050985
    A087204(n) = ([2, 1, -1, -2, -1, 1][1+(n%6)]);
    A195017(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i,2] * (-1)^(1+primepi(f[i,1])))); };
    A353446(n) = { my(u=A050985(n)); issquarefree(u) * A087204(abs(A195017(u))); };

Formula

a(n) = A008966(m) * A087204(A090882(m)) = A008966(m) * A087204(|A195017(m)|), where m = A050985(n), the cubefree part of n, and A008966(.) is the characteristic function of squarefree numbers.
For n >= 1, -2 <= a(n) <= 2.
{n : a(n) = -2} = {A211338} INTERSECT {A332820}.
{n : a(n) = -1} = {A211337} \ {A332820}.
{n : a(n) = 0} = {A059269}.
{n : a(n) = 1} = {A211338} \ {A332820}.
{n : a(n) = 2} = {A211337} INTERSECT {A332820}.

A074943 a(n) = tau(n) mod 3.

Original entry on oeis.org

1, 2, 2, 0, 2, 1, 2, 1, 0, 1, 2, 0, 2, 1, 1, 2, 2, 0, 2, 0, 1, 1, 2, 2, 0, 1, 1, 0, 2, 2, 2, 0, 1, 1, 1, 0, 2, 1, 1, 2, 2, 2, 2, 0, 0, 1, 2, 1, 0, 0, 1, 0, 2, 2, 1, 2, 1, 1, 2, 0, 2, 1, 0, 1, 1, 2, 2, 0, 1, 2, 2, 0, 2, 1, 0, 0, 1, 2, 2, 1, 2, 1, 2, 0, 1, 1, 1, 2, 2, 0, 1, 0, 1, 1, 1, 0, 2, 0, 0, 0, 2, 2, 2, 2, 2
Offset: 1

Views

Author

Benoit Cloitre, Oct 04 2002

Keywords

Crossrefs

Programs

Formula

From Amiram Eldar, Apr 16 2024: (Start)
a(n) = A010872(A000005(n)).
a(A059269(n)) = 0; a(A211337(n)) = 1; a(A211338(n)) = 2.
Conjecture: Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 9*zeta(3)/Pi^2 = 1.0961... . The conjecture is true if A211337 and A211338 have the same asymptotic density (see also A059269). (End)

A084302 Remainder of tau(n) modulo 6.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 0, 2, 4, 4, 5, 2, 0, 2, 0, 4, 4, 2, 2, 3, 4, 4, 0, 2, 2, 2, 0, 4, 4, 4, 3, 2, 4, 4, 2, 2, 2, 2, 0, 0, 4, 2, 4, 3, 0, 4, 0, 2, 2, 4, 2, 4, 4, 2, 0, 2, 4, 0, 1, 4, 2, 2, 0, 4, 2, 2, 0, 2, 4, 0, 0, 4, 2, 2, 4, 5, 4, 2, 0, 4, 4, 4, 2, 2, 0, 4, 0, 4, 4, 4, 0, 2, 0, 0, 3, 2, 2, 2, 2, 2
Offset: 1

Views

Author

Labos Elemer, Jun 02 2003

Keywords

Comments

The sums of the first 10^k terms, for k = 1, 2, ..., are 27, 236, 2275, 22166, 220070, 2195376, 21933228, 219259514, 2192385128, 21923168052, ... . Conjecture: the asymptotic mean of this sequence is 3*zeta(3)/zeta(2) = 3 * A253905 = 2.192288... . The conjecture is true if A211337 and A211338 have an equal asymptotic density (see also A059269). - Amiram Eldar, Jul 11 2024

Crossrefs

Programs

Formula

a(n) = A000005(n) modulo 6.

A214153 Numbers k for which k and tau(k) are both congruent to 1 modulo 3.

Original entry on oeis.org

1, 10, 22, 34, 46, 55, 58, 64, 82, 85, 91, 94, 106, 112, 115, 118, 133, 142, 145, 166, 178, 187, 202, 205, 208, 214, 217, 226, 235, 247, 253, 259, 262, 265, 274, 280, 295, 298, 301, 304, 319, 334, 343, 346, 355, 358, 382, 391, 394, 403, 415, 427, 445, 451
Offset: 1

Views

Author

Gerasimov Sergey, Jul 05 2012

Keywords

Examples

			The divisors of 10 are: 1, 2, 5, 10 (4 divisors). 10 and 4 are both congruent to 1 modulo 3. Thus 10 is a member of this sequence.
		

Crossrefs

Intersection of A016777 and A211337.
Cf. A000005.

Programs

  • Mathematica
    Select[Range[1, 500, 3], Mod[DivisorSigma[0, #], 3] == 1 &] (* T. D. Noe, Jul 09 2012 *)

A369209 Numbers whose number of divisors has the largest prime factor 3.

Original entry on oeis.org

4, 9, 12, 18, 20, 25, 28, 32, 36, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 100, 108, 116, 117, 121, 124, 126, 132, 140, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 180, 188, 196, 198, 200, 204, 207, 212, 220, 224, 225, 228
Offset: 1

Views

Author

Amiram Eldar, Jan 16 2024

Keywords

Comments

Subsequence of A059269 and first differs from it at n = 36: A059269(136) = 44 has 15 = 3 * 5 divisors and thus is not a term of this sequence.
Numbers k such that A000005(k) is in A065119.
Numbers k such that A071188(k) = 3.
Equals the complement of A354181, without the terms of A036537 (i.e., complement(A354181) \ A036537).
The asymptotic density of this sequence is Product_{p prime} (1-1/p) * (Sum_{k>=1} 1/p^(A003586(k)-1)) - A327839 = 0.26087647470200496716... .

Crossrefs

Programs

  • Mathematica
    gpf[n_] := FactorInteger[n][[-1, 1]]; Select[Range[300], gpf[DivisorSigma[0, #]] == 3 &]
  • PARI
    gpf(n) = if(n == 1, 1, vecmax(factor(n)[, 1]));
    is(n) = gpf(numdiv(n)) == 3;
Showing 1-9 of 9 results.