%I #55 Feb 10 2023 12:02:27

%S 1,5,5,7,5,17,7,19,5,25,17,19,7,31,19,25,5,25,25,35,17,61,19,71,7,35,

%T 31,41,19,71,25,77,5,25,25,35,25,85,35,95,17,85,61,71,19,91,71,77,7,

%U 35,35,49,31,107,41,121,19,95,71,85,25,113,77,103

%N Number of terms in expansion of f^n mod 2, where f = 1/x^2 + 1/x + 1 + x + x^2 mod 2.

%C This is the number of cells that are ON after n generations in a one-dimensional cellular automaton defined by the odd-neighbor rule where the neighborhood consists of 5 contiguous cells.

%C a(n) is also the number of odd entries in row n of A035343. - _Leon Rische_, Feb 02 2023

%H Chai Wah Wu, <a href="/A247649/b247649.txt">Table of n, a(n) for n = 0..10000</a>

%H N. J. A. Sloane, <a href="http://arxiv.org/abs/1503.01168">On the Number of ON Cells in Cellular Automata</a>, arXiv:1503.01168 [math.CO], 2015.

%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>

%F The values of a(n) for n in A247647 (or A247648) determine all the values, as follows. Parse the binary expansion of n into terms from A247647 separated by at least two zeros: m_1 0...0 m_2 0...0 m_3 ... m_r 0...0. Ignore any number (one or more) of trailing zeros. Then a(n) = a(m_1)*a(m_2)*...*a(m_r). For example, n = 37_10 = 100101_2 is parsed into 1.00.101, and so a(37) = a(1)*a(5) = 5*17 = 85. This is a generalization of the Run Length Transform.

%e The first few generations are:

%e ..........X..........

%e ........XXXXX........

%e ......X.X.X.X.X......

%e ....XX..X.X.X..XX.... (f^3)

%e ..X...X...X...X...X..

%e XXXX.XXX.XXX.XXX.XXXX

%e ...

%e f^3 mod 2 = x^6 + x^5 + x^2 + 1/x^2 + 1/x^5 + 1/x^6 + 1 has 7 terms, so a(3) = 7.

%e From _Omar E. Pol_, Mar 02 2015: (Start)

%e Also, written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:

%e 1;

%e 5;

%e 5, 7;

%e 5,17, 7,19;

%e 5,25,17,19, 7,31,19,25;

%e 5,25,25,35,17,61,19,71, 7,35,31,41,19,71,25,77;

%e 5,25,25,35,25,85,35,95,17,85,61,71,19,91,71,77,7,35,35,49,31,107,41,121,19, ...

%e (End)

%e It follows from the Generalized Run Length Transform result mentioned in the comments that in each row the first quarter of the terms (and no more) are equal to 5 times the beginning of the sequence itself. It cannot be said that the rows converge (in any meaningful sense) to five times the sequence. - _N. J. A. Sloane_, Mar 03 2015

%o (Python)

%o import sympy

%o from functools import reduce

%o from operator import mul

%o x = sympy.symbols('x')

%o f = 1/x**2+1/x+1+x+x**2

%o A247649_list, g = [1], 1

%o for n in range(1,1001):

%o s = [int(d,2) for d in bin(n)[2:].split('00') if d != '']

%o g = (g*f).expand(modulus=2)

%o if len(s) == 1:

%o A247649_list.append(g.subs(x,1))

%o else:

%o A247649_list.append(reduce(mul,(A247649_list[d] for d in s)))

%o # _Chai Wah Wu_, Sep 25 2014

%Y Cf. A071053, A247647, A247648, A253085, A255490.

%Y Partial sums are in A255654.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Sep 25 2014