A047240 Numbers that are congruent to {0, 1, 2} mod 6.

0, 1, 2, 6, 7, 8, 12, 13, 14, 18, 19, 20, 24, 25, 26, 30, 31, 32, 36, 37, 38, 42, 43, 44, 48, 49, 50, 54, 55, 56, 60, 61, 62, 66, 67, 68, 72, 73, 74, 78, 79, 80, 84, 85, 86, 90, 91, 92, 96, 97, 98, 102, 103, 104, 108, 109, 110, 114, 115, 116, 120, 121, 122

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Partial sums of 0,1,1,4,1,1,4,... - Paul Barry, Feb 19 2007

Numbers k such that floor(k/3) = 2*floor(k/6). - Bruno Berselli, Oct 05 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A010872, A030341, A047234, A047242.

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

Programs

Magma

[0] cat [6*Floor(n/3) + (n mod 3): n in [1..65]]; // Vincenzo Librandi, Oct 23 2011

Maple

A047240:=n->2*n-3-cos(2*n*Pi/3)+sin(2*n*Pi/3)/sqrt(3): seq(A047240(n), n=1..100); # Wesley Ivan Hurt, Jun 14 2016
select(k -> modp(iquo(k,3), 2) = 0, [$0..122]); # Peter Luschny, Oct 05 2017

Mathematica

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 2 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)
a047240[n_] := Flatten[Map[6 # + {0, 1, 2} &, Range[0, n]]]; a047240[20] (* data *) (* Hartmut F. W. Hoft, Mar 06 2017 *)

PARI

a(n)=n\3*6 + n%3 \\ Charles R Greathouse IV, Oct 07 2015

Python

[k for k in range(123) if (k//3) % 2 == 0] # Peter Luschny, Oct 05 2017

Formula

From Paul Barry, Feb 19 2007: (Start)

G.f.: x*(1 + x + 4*x^2)/((1 - x)*(1 - x^3)).

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

a(n) = n-1 + 3*floor((n-1)/3). - Philippe Deléham, Apr 21 2009

a(n) = 6*floor(n/3) + (n mod 3). - Gary Detlefs, Mar 09 2010

a(n+1) = Sum_{k>=0} A030341(n,k)*b(k) with b(0)=1 and b(k)=2*3^k for k>0. - Philippe Deléham, Oct 22 2011.

a(n) = 2*n - 2 - A010872(n-1). - Wesley Ivan Hurt, Jul 07 2013

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

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

a(3*k) = 6*k-4, a(3*k-1) = 6*k-5, a(3*k-2) = 6*k-6. (End)

Sum_{n>=2} (-1)^n/a(n) = (3-sqrt(3))*Pi/18 + log(2+sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 14 2021

E.g.f.: 4 + exp(x)*(2*x - 3) - exp(-x/2)*(3*cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2))/3. - Stefano Spezia, Jul 26 2024

Extensions

Paul Barry formula adapted for offset 1 by Wesley Ivan Hurt, Jun 14 2016