A083039 Number of divisors of n that are <= 3.
1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3
Offset: 1
Examples
The divisors of 6 are 1, 2, 3 and 6. Of those divisors, 1, 2 and 3 are <= 3. That's three divisors, therefore, a(6) = 3. - _David A. Corneth_, Sep 30 2017
Links
- Antti Karttunen, Table of n, a(n) for n = 1..12000
- D. A. Gewurz and F. Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seq., 6 (2003), 03.1.6
- M. D. Hirschhorn, J. A. Sellers, Enumeration of unigraphical partitions, JIS 11 (2008) 08.4.6
- Index entries for two-way infinite sequences
- Index entries for linear recurrences with constant coefficients, signature (-1, 0, 1, 1).
Programs
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Mathematica
LinearRecurrence[{-1, 0, 1, 1},{1, 2, 2, 2},90] (* Ray Chandler, Aug 26 2015 *)
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PARI
a(n)=[3,1,2,2,2,1][n%6+1];
Formula
G.f.: x/(1-x) + x^2/(1-x^2) + x^3/(1-x^3).
a(n) = a(n-6) = a(-n).
a(n) = 11/6 - (1/2)*(-1)^n - (1/3)*cos(2*Pi*n/3) - (1/3)*3^(1/2)*sin(2*Pi*n/3). - Richard Choulet, Dec 12 2008
a(n) = Sum_{k=1..1} cos(n*(k - 1)/1*2*Pi)/1 + Sum_{k=1..2} cos(n*(k - 1)/2*2*Pi)/2 + Sum_{k=1..3} cos(n*(k - 1)/3*2*Pi)/3. - Mats Granvik, Sep 09 2012
a(n) = log_2(gcd(n,2) + gcd(n,6)). - Gary Detlefs, Feb 15 2014
a(n) = Sum_{d|n, d<=3} 1. - Wesley Ivan Hurt, Oct 30 2023
Comments