A204259 -id:A204259 - cp's OEIS frontend

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

N. J. A. Sloane

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

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).

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

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

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

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

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

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

A136289 Start with three pennies touching each other on a tabletop. In each generation, add pennies subject to the rule that a penny can be placed only when (at least) two pennies are already in position to determine its position; sequence gives number of pennies added at generation n.

Original entry on oeis.org

3, 3, 6, 9, 9, 12, 15, 15, 18, 21, 21, 24, 27, 27, 30, 33, 33, 36, 39, 39, 42, 45, 45, 48, 51, 51, 54, 57, 57, 60, 63, 63, 66, 69, 69, 72, 75, 75, 78, 81, 81, 84, 87, 87, 90, 93, 93, 96, 99, 99, 102, 105, 105, 108, 111, 111, 114, 117, 117, 120, 123, 123, 126, 129, 129, 132

Offset: 0

Colin Mallows, Apr 13 2008

nonn

Is there a recurrence or generating function?
Place 3 at the apex of a triangle and require every "triple" to contain the values 1,2,3. See link below. Compare with A204259. - Craig Knecht, Jul 29 2015
a(n) is also the number of the circles added at n-th iteration of the pattern generated by the construction rules: (i) At n = 0, there are three circles of radius s with centers at the vertices of an equilateral triangle of side length s. (ii) At n > 0, draw circles by placing center at the intersection points of the circumferences of circles in the previous iteration, with overlaps forbidden. The pattern seems to be the flower of life. See illustration. - Kival Ngaokrajang, Oct 02 2015
From Charlie Neder, Jun 22 2018: (Start)
The conjectured formulae are correct: Looking at Knecht's diagram one can see that a line of three numbers in any direction must be a permutation of {1,2,3} by applying the triple rule three times to one of the adjacent pairs, like so:
                 o o      c o      c a      c a
                a b o    a b o    a b o    a b c
Applying this rule repeatedly to the rightmost diagonal shows that it repeats {3,1,2} with period 3, so row 3k+1 will always end in 3 and row 3k+2 will always end in 1. Applying the triple rule again using the aforementioned 3 and 1 shows that row 3k+2 ends in {2,1}. Therefore, for row n = 3k+p, k lines of three sum to 6 and the remainder, if present, sums to 3.  (End)

			After four generations we have:
.............4...3...4............
..................................
.......4...3...2...2...3...4......
..................................
.........3...1...0...1...3........
..................................
.......4...2...0...0...2...4......
..................................
.........3...2...1...2...3........
..................................
...........4...3...3...4..........
..................................
.................4................

Craig Knecht, Every triple contains 1,2,3 starting with 3 at the top.
Kival Ngaokrajang, Illustration of the flower of life pattern
Sacred Geometry, Flower of life

Cf. A136290.

isAdjac := proc(a,b,c) abs(b-a) = 1 and abs(c-b)=1 and abs(a-c)=1 ; end: neighbrs := proc(x) local y,phi ; y := {} ; for phi from 0 to 5 do y := y union {x+expand(exp(I*phi*Pi/3)) } ; od: end: doesMatch2 := proc(genLin,x) local p ; for p in combinat[choose](genLin,2) do if isAdjac(x,op(1,p),op(2,p)) then RETURN(true) ; fi ; od: RETURN(false) ; end: A136289 := proc(gen) local newgen,o,candid,x,genLin,g ; newgen := {}; genLin := {} ; for g in gen do genLin := genLin union g ; od: for o in op(-1,gen) do candid := neighbrs(o) ; for x in candid do if not x in newgen then if not x in genLin then if doesMatch2(genLin,x) then newgen := newgen union {x} ; fi ; fi ; fi ; od: od: RETURN( [op(gen),newgen] ) ; end: gen := [{0,1,expand(exp(I*Pi/3))}] : for n from 1 do printf("%d,", nops(op(n,gen)) ) ; gen := A136289(gen) od: # R. J. Mathar, Apr 15 2008

Conjecture: a(n) = a(n-3) + 6, implying g.f. 3*(1+x^2)/((1-x)^2*(1+x+x^2)). - R. J. Mathar, Apr 15 2008
Conjecture: a(n) = 2n + 1 + ((n+2) mod 3). - Wesley Ivan Hurt, Jul 07 2013
Conjecture: a(n) = 3*floor(2*n/3) + 3. - Jon E. Schoenfield, Jul 30 2015
Conjecture: a(n) = 2*(n+1+sin(2*(n+1)*Pi/3)/sqrt(3)). - Wesley Ivan Hurt, Sep 27 2017

More terms from R. J. Mathar, Apr 15 2008

More terms from John W. Layman, Jun 26 2008

cp's OEIS Frontend

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

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