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%I #13 Nov 25 2022 22:13:18
%S 1,5,20,50,110,200,340,525,780,1095,1500,1980,2570,3250,4060,4975,
%T 6040,7225,8580,10070,11750
%N Coordination sequence for the faces of the uniform infinite surface that is formed from congruent regular pentagons and from which there is a continuous function that maps the faces 1:1 to regular pentagons in the plane.
%C Each pentagon shares each of its edges with a congruent pentagon. Obviously, the images of many pentagonal faces overlap.
%C Equivalently, at step 0, embed the edges of a regular pentagon in the plane. At step n >= 1, embed the edges to form all regular pentagons that include an edge that was present after step n-1. a(n) is the number of pentagons whose set of embedded edges is completed in step n.
%e Position the seed pentagon in the complex plane centered at 0, scaled so that the centers of adjacent pentagons are at unit distance. Let the centers of the pentagons completed in step 1 be at 1, exp(2 Pi i / 5), exp(4 Pi i / 5), exp(6 Pi i / 5) and exp(8 Pi i / 5). Call the seed pentagon an "even" pentagon, for reasons that become obvious.
%e The step-1 pentagons have orientation opposite to that of the seed pentagon, so have neighbors whose centers are displaced by exp(Pi i / 5), exp(3 Pi i / 5), -1, exp (7 Pi i / 5) and exp(9 Pi i / 5). So call these pentagons "odd".
%e This arrangement is clearly like a checkerboard in that even pentagons are adjacent to odd pentagons and vice versa.
%e We give an "address" to every pentagon in the embedding by starting from the seed pentagon and giving the coefficient of Pi i / 5 in the argument of each displacement exponential. These coefficients range from 0 to 9. They alternate in parity, starting with an even coefficient. To illustrate, the seed pentagon has address [], an empty string of coefficients. Its neighbors are [0], [2], [4], [6], and [8]. [0]'s neighbors are [01], [03], [05], [07], and [09] -- of these, all but [05] are completed in step 2, as [05] is another name for the seed pentagon. In general, an occurrence of (n) cancels with an occurrence of (n+5) anywhere in the string, since the corresponding exponentials negate each other.
%e There are no other "coincidences" at step 2. So there are 20 new pentagons:
%e [01], [03], [07], [09], [21], [23], [25], [29], [41], [43], [45], [47], [63], [65], [67], [69], [81], [85], [87], [89]
%e -- after deleting the five "revisits" to the seed pentagon: [05], [27], [49], [61], and [83].
%e At step 3, the addresses beginning [01] are [010], [012], [014], [018] (with (1) and (6) canceling from [016] to give [0]). But every pentagon that is addressed as [abc] can be seen to be addressed as [cba] just by commutativity of addition. So we may allocate to [01] a half share of each of [012], [014] and [018], plus all of [010], counting 1 + 3/2 = 2.5 pentagons. Each of the 20 step-2 addresses generates step-3 addresses in a similar way, so giving 20 * 2.5 = 50 new pentagons.
%Y Bisections (conjectured): A008383 (even), A063490 (odd, divided by 5).
%Y Partial sums: A175898 (apparently).
%K nonn,more
%O 0,2
%A _Peter Munn_ and _Allan C. Wechsler_, Nov 24 2022
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