A289840 Complex cross sequence (see Comments lines for definition).

0, 1, 3, 11, 19, 27, 35, 67, 83, 99, 115, 163, 179, 195, 211, 275, 323, 355, 387, 467, 483, 499, 515, 579, 627, 675, 707, 787, 803, 819, 835, 899, 947, 995, 1027, 1107, 1123, 1139, 1155, 1219, 1267, 1315, 1347, 1427, 1443, 1459, 1475, 1539, 1587, 1635, 1667, 1747, 1763, 1779, 1795, 1859, 1907, 1955, 1987, 2067

Offset: 0

Author

Omar E. Pol, Jul 14 2017

Keywords

• nonn
• easy

Comments

The sequence arises from a "hybrid" cellular automaton on the infinite square grid, which consist of two successive generations using toothpicks of length 2 (cf. A139250) followed by two successive generations using the rules of the D-toothpick sequence A220500.

In other words (and more precisely) we have that:

1) If n is congruent to 1 or 2 mod 4 (cf. A042963), for example: 1, 2, 5, 6, 9, 10, ..., the elements added to the structure at n-th stage must be toothpicks of length 2. These toothpicks are connected to the structure by their midpoints.

2) If n is a positive integer of the form 4*k-1 (cf. A004767), for example: 3, 7, 11, 15, ..., the elements added to the structure at n-th stage must be D-toothpicks of length sqrt(2) and eventually D-toothpicks of length sqrt(2)/2, in both cases the D-toothpicks are connected to the structure by their endpoints, in the same way as in the even-indexed stages of A220500.

3) If n is a positive multiple of 4 (cf. A008586) the elements added to the structure at n-th stage must be toothpicks of length 1 connected by their endpoints, in the same way as in the odd-indexed stages of A220500.

a(n) is the total number of elements in the structure after n generations.

A289841 (the first differences) gives the number of elements added at n-th stage.

Note that after 19 generations the structure is a 72-gon which essentially looks like a diamond (as a square that has been rotated 45 degrees).

The surprising fact is that from n = 20 up to 27 the structure is gradually transformed into a square cross.

The diamond mentioned above can be interpreted as the center of the cross. The diamond has an area equal to 384 and it contains 222 polygonal regions (or enclosures) of 11 distinct shapes. Missing two heptagonal shapes which are in the arms of the square cross only.

In total the complex cross contains 13 distinct shapes of polygonal regions. There are ten polygonal shapes that have an infinite number of copies. On the other hand, three of these polygonal shapes have a finite number of copies because they are in the center of the cross only. For example: there are only four copies of the concave 14-gon, which is also the largest polygon in the structure.

For n => 27 the shape of the square cross remains forever because its four arms grow indefinitely.

Every arm has a minimum width equal to 8, and a maximum width equal to 12.

Every arm also has a periodic structure which can be dissected in infinitely many clusters of area equal to 64. Every cluster is a 30-gon that contains 40 polygonal regions of nine distinct shapes.

If n is a number of the form 8*k-3 (cf. A017101) and greater than 19, for example: 27, 35, 43, 51, ..., then at n-th stage a new cluster is finished in every arm of the cross.

The behavior is similar to A290220 and A294020 in the sense that these three cellular automata have the property of self-limiting their growth only in some directions of the square grid. - Omar E. Pol, Oct 29 2017

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
• Index entries for sequences related to cellular automata
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,1,-1).

Crossrefs

Cf. A004767, A008586, A017101, A042963, A139250, A220500, A289841, A290220 (a simpler cross), A294020.

Programs

PARI

concat(0, Vec(x*(1 + 2*x + 8*x^2 + 8*x^3 + 8*x^4 + 8*x^5 + 32*x^6 + 16*x^7 + 15*x^8 + 14*x^9 + 40*x^10 + 8*x^11 + 8*x^12 + 8*x^13 + 32*x^14 + 32*x^15 + 16*x^16 + 16*x^17 + 32*x^18 + 16*x^24) / ((1 - x)^2*(1 + x)*(1 + x^2)*(1 + x^4))+ O(x^50))) \\ Colin Barker, Nov 12 2017

Formula

From Colin Barker, Nov 11 2017: (Start)

G.f.: x*(1 + 2*x + 8*x^2 + 8*x^3 + 8*x^4 + 8*x^5 + 32*x^6 + 16*x^7 + 15*x^8 + 14*x^9 + 40*x^10 + 8*x^11 + 8*x^12 + 8*x^13 + 32*x^14 + 32*x^15 + 16*x^16 + 16*x^17 + 32*x^18 + 16*x^24) / ((1 - x)^2*(1 + x)*(1 + x^2)*(1 + x^4)).

a(n) = a(n-1) + a(n-8) - a(n-9) for n>19.

(End)