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A047266
Numbers that are congruent to {0, 1, 5} mod 6.
5
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
1,3
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
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
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 *)
PROG
(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
(Magma) [n : n in [0..150] | n mod 6 in [0, 1, 5]]; // Wesley Ivan Hurt, Jun 13 2016
CROSSREFS
KEYWORD
nonn,easy
STATUS
approved