A047266 Numbers that are congruent to {0, 1, 5} mod 6.

0, 1, 5, 6, 7, 11, 12, 13, 17, 18, 19, 23, 24, 25, 29, 30, 31, 35, 36, 37, 41, 42, 43, 47, 48, 49, 53, 54, 55, 59, 60, 61, 65, 66, 67, 71, 72, 73, 77, 78, 79, 83, 84, 85, 89, 90, 91, 95, 96, 97, 101, 102, 103, 107, 108, 109, 113, 114, 115, 119, 120, 121, 125

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n+3) is the Hankel transform of A005773(n+3). - Paul Barry, Nov 04 2008

The numbers m == 0, 2 or 10 mod 12 (the doubles of this sequence, that is, 10, 12, 14, 22, 24, 26, 34, ...) have the property that exactly 1/4 of the 3-part partitions of m form the sides of a triangle. See Mathematics Stack Exchange, 2013, link. - Ed Pegg Jr, Dec 19 2013

Row sum of a triangle where two rules build the triangle. #1 Start with the value "1" at the top of the triangle. #2 Require every "triple" to contain the values 1,2,3 (see link below). Compare with A136289 that has "3" at the apex. - Craig Knecht, Oct 17 2015

Nonnegative m such that floor(k*m^2/6) = k*floor(m^2/6), where k = 2, 3, 4 or 5. - Bruno Berselli, Dec 03 2015

Links

• Craig Knecht, Triangular row sum of A204259.
• Mathematics Stack Exchange, Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle, 2013.
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

Cf. A005773, A047240, A047242, A057078, A136289, A204259.

Programs

Magma

[n : n in [0..150] | n mod 6 in [0, 1, 5]]; // Wesley Ivan Hurt, Jun 13 2016

Maple

seq(seq(6*s+j, j=[0,1,5]), s=0..100); # Robert Israel, Dec 01 2014

Mathematica

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 5 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)

PARI

concat(0, Vec(x^2*(1+4*x+x^2)/((1+x+x^2)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Oct 17 2015

Formula

G.f.: x^2*(1+4*x+x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011

a(n) = 2*(n-1) + A057078(n). - Robert Israel, Dec 01 2014

a(n) = a(n-1) + a(n-3) - a(n-4) for n>4. - Wesley Ivan Hurt, Nov 09 2015

From Wesley Ivan Hurt, Jun 13 2016: (Start)

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

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

Sum_{n>=2} (-1)^n/a(n) = log(2)/6 + log(2 + sqrt(3))/sqrt(3). - Amiram Eldar, Dec 14 2021