A136289 - OEIS

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.

%I #46 Nov 11 2018 22:40:54

%S 3,3,6,9,9,12,15,15,18,21,21,24,27,27,30,33,33,36,39,39,42,45,45,48,

%T 51,51,54,57,57,60,63,63,66,69,69,72,75,75,78,81,81,84,87,87,90,93,93,

%U 96,99,99,102,105,105,108,111,111,114,117,117,120,123,123,126,129,129,132

%N 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.

%C Is there a recurrence or generating function?

%C 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

%C 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

%C From _Charlie Neder_, Jun 22 2018: (Start)

%C 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:

%C o o c o c a c a

%C a b o a b o a b o a b c

%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)

%H Craig Knecht, <a href="/A136289/a136289.jpg">Every triple contains 1,2,3 starting with 3 at the top.</a>

%H Kival Ngaokrajang, <a href="/A136289/a136289.pdf">Illustration of the flower of life pattern</a>

%H Sacred Geometry, <a href="http://www.bibliotecapleyades.net/geometria_sagrada/esp_geometria_sagrada_6.htm">Flower of life</a>

%F 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

%F Conjecture: a(n) = 2n + 1 + ((n+2) mod 3). - _Wesley Ivan Hurt_, Jul 07 2013

%F Conjecture: a(n) = 3*floor(2*n/3) + 3. - _Jon E. Schoenfield_, Jul 30 2015

%F Conjecture: a(n) = 2*(n+1+sin(2*(n+1)*Pi/3)/sqrt(3)). - _Wesley Ivan Hurt_, Sep 27 2017

%e After four generations we have:

%e .............4...3...4............