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.

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

Author

Colin Mallows, Apr 13 2008

Keywords

• nonn

Comments

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)

Examples

			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................
		

Links

• 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

Crossrefs

Cf. A136290.

Programs

Maple

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

Formula

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

Extensions

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

More terms from John W. Layman, Jun 26 2008