A047217 Numbers that are congruent to {0, 1, 2} mod 5.

0, 1, 2, 5, 6, 7, 10, 11, 12, 15, 16, 17, 20, 21, 22, 25, 26, 27, 30, 31, 32, 35, 36, 37, 40, 41, 42, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 62, 65, 66, 67, 70, 71, 72, 75, 76, 77, 80, 81, 82, 85, 86, 87, 90, 91, 92, 95, 96, 97, 100, 101, 102, 105, 106, 107, 110, 111

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Also, the only numbers that are eligible to be the sum of two 4th powers (A004831). - Cino Hilliard, Nov 23 2003

Nonnegative m such that floor(2*m/5) = 2*floor(m/5). - Bruno Berselli, Dec 09 2015

The sequence lists the indices of the multiples of 5 in A007531. - Bruno Berselli, Jan 05 2018

Links

• Mohammed Yaseen, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A007531, A030341, A004831 (two 4th powers).

Cf. similar sequences with formula n+i*floor(n/3) listed in A281899.

Programs

Magma

I:=[0, 1, 2, 5]; [n le 4 select I[n] else Self(n-1)+Self(n-3)-Self(n-4): n in [1..70]]; // Vincenzo Librandi, Apr 25 2012

Magma

&cat [[5*n,5*n+1,5*n+2]: n in [0..30]]; // Bruno Berselli, Dec 09 2015

Maple

seq(op([5*i,5*i+1,5*i+2]),i=0..100); # Robert Israel, Sep 02 2014

Mathematica

Select[Range[0,120], MemberQ[{0,1,2}, Mod[#,5]]&] (* Harvey P. Dale, Jan 20 2012 *)

PARI

a(n)=n--\3*5+n%3 \\ Charles R Greathouse IV, Oct 22 2011

PARI

concat(0, Vec(x^2*(1+x+3*x^2)/(1-x)^2/(1+x+x^2) + O(x^100))) \\ Altug Alkan, Dec 09 2015

PARI

is(n) = n%5 < 3 \\ Felix Fröhlich, Jan 05 2018

Formula

a(n+1) = Sum_{k>=0} A030341(n,k)*b(k) with b(0)=1 and b(k)=5*3^(k-1) for k>0. - Philippe Deléham, Oct 22 2011

G.f.: x^2*(1+x+3*x^2)/(1-x)^2/(1+x+x^2). - Colin Barker, Feb 17 2012

a(n) = 5 + a(n-3) for n>3. - Robert Israel, Sep 02 2014

a(n) = floor((5/4)*floor(4*(n-1)/3)). - Bruno Berselli, May 03 2016

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

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

a(n) = (15*n-21-6*cos(2*n*Pi/3)+2*sqrt(3)*sin(2*n*Pi/3))/9.

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

a(n) = n - 1 + 2*floor((n-1)/3). - Bruno Berselli, Feb 06 2017

Sum_{n>=2} (-1)^n/a(n) = sqrt(1-2/sqrt(5))*Pi/5 + 3*log(2)/5. - Amiram Eldar, Dec 10 2021