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A244643
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Number of cells with state 1 at n-th stage in some 2D reversible second order cellular automata (see comments for precise definition).
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7
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0, 1, 4, 5, 16, 9, 20, 21, 64, 25, 36, 29, 80, 41, 84, 85, 256, 89, 100, 61, 144, 65, 116, 109, 320, 121, 164, 125, 336, 169, 340, 341, 1024, 345, 356, 189, 400, 161, 244, 205, 576, 209, 260, 181, 464, 225, 436, 429, 1280, 441, 484, 285, 656, 289, 500, 461, 1344, 505, 676, 509, 1360, 681, 1364, 1365, 4096, 1369
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OFFSET
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-1,3
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COMMENTS
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Consider cellular automata with two states defined in A244642 and second order cellular automata with four states generated from them. If we start with a single cell with state 1 and all the others 0, then number cells with state 1 in subsequent steps will be the terms in the sequence a(n). Number of cells with state 2 is a(n-1), where a(-1) = 0.
Equals row 4 in the array shown in A178568.
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LINKS
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FORMULA
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a(-1) = 0, a(0) = 1, a(2^k + j) = 4*a(j) + a(2^k - j - 2).
a(2n+1) = 4a(n), a(2n+2) = a(n+1) + a(n).
G.f.: Product_{k>=0} (1 + 4*x^(2^k) + x^(2^(k+1))). - Ilya Gutkovskiy, Mar 16 2021
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EXAMPLE
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For n=4, a(4) = 16 and a(n-1) = a(3) = 5.
....1
...121
..1.1.1
.1212121
..1.1.1
...121
....1
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MATHEMATICA
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msb[1]=1; msb[n_] := 2 msb[Quotient[n, 2]];
a[-1] = 0; a[0] = 1; a[n_] := 4 a[n-msb[n]] + a[2 msb[n]-n-2];
Table[a[n], {n, -1, 64}] (* recursion *)
Table[PolynomialMod[Fibonacci[n, f]/.f->(x+1/x+y+1/y), 2], {n, 0, 24}]/.{x->1, y->1} (* calculation using GF(2) polynomial representation of CA *)
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PROG
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(Axiom)
msb n == if n=1 then 1 else 2*msb(quo(n, 2))
a n == if n<1 then n+1 else 4*a(n-msb(n))+a(2*msb(n)-n-2)
[a(n) for n in -1..64]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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