A047220 Numbers that are congruent to {0, 1, 3} mod 5.

0, 1, 3, 5, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 25, 26, 28, 30, 31, 33, 35, 36, 38, 40, 41, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 60, 61, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 80, 81, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 100, 101, 103, 105, 106

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

First differences are (1,2,2), repeat, with period 3 (A130196). - N. J. A. Sloane, Dec 03 2015

Also numbers k such that k*(k+2)*(k+4) is divisible by 5. - Bruno Berselli, Dec 28 2017

Maximum sum of degeneracies over all decompositions of the complete graph of order n into four factors. The extremal decompositions are characterized in the Bickle link below. - Allan Bickle, Dec 21 2021

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Allan Bickle, Nordhaus-Gaddum Theorems for k-Decompositions, Congr. Num. 211 (2012) 171-183.
• Z. Füredi, A. Kostochka, M. Stiebitz, R. Skrekovski, and D. West, Nordhaus-Gaddum-type theorems for decompositions into many parts, J. Graph Theory 50 (2005), 273-292.
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A001622, A011655, A130196 (first differences).

Programs

Magma

I:=[0, 1, 3, 5]; [n le 4 select I[n] else Self(n-1)+Self(n-3)-Self(n-4): n in [1..70]]; // Vincenzo Librandi, Apr 26 2012

Maple

seq(floor(5*(n-1)/3), n=1..56); # Gary Detlefs, Feb 20 2010
seq(2*n-floor(n/3)-(n^2 mod 3), n=0..55); # Gary Detlefs, Mar 19 2010

Mathematica

Table[Floor[5*(n-1)/3], {n,100}] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)

PARI

a(n)=n + 2*(n-1)\3 - 1 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = floor(5*(n-1)/3). - Gary Detlefs, Feb 20 2010

a(n) = 2*n - floor(n/3) - (n^2 mod 3), with offset 0. - Gary Detlefs, Mar 19 2010

G.f.: x^2*(1 + 2*x + 2*x^2)/(1 - x)^2/(1 + x + x^2). - Colin Barker, Feb 17 2012

a(n) = n + floor(2*(n-1)/3) - 1. - Arkadiusz Wesolowski, Sep 18 2012

From Wesley Ivan Hurt, Jun 14 2016: (Start)

a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.

a(n) = 5*n/3 - 2 + 2*sin(2*n*Pi/3)/(3*sqrt(3)).

a(3*k) = 5*k-2, a(3*k-1) = 5*k-4, a(3*k-2) = 5*k-5. (End)

E.g.f.: 2 + (5*x - 6)*exp(x)/3 + 2*sin(sqrt(3)*x/2)*(cosh(x/2) - sinh(x/2))/(3*sqrt(3)). - Ilya Gutkovskiy, Jun 14 2016

Sum_{n>=2} (-1)^n/a(n) = sqrt(1-2/sqrt(5))*Pi/5 + 2*log(phi)/sqrt(5) + log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 16 2023