

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.


5



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
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OFFSET

0,1


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 nth 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
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)


LINKS



FORMULA

Conjecture: a(n) = a(n3) + 6, implying g.f. 3*(1+x^2)/((1x)^2*(1+x+x^2)).  R. J. Mathar, Apr 15 2008
Conjecture: a(n) = 2*(n+1+sin(2*(n+1)*Pi/3)/sqrt(3)).  Wesley Ivan Hurt, Sep 27 2017


EXAMPLE

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


MAPLE

isAdjac := proc(a, b, c) abs(ba) = 1 and abs(cb)=1 and abs(ac)=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


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



