A138276
Total number of active nodes of the Rule 150 cellular automaton on an infinite Bethe lattice with coordination number 3 (with a single 1 as initial condition).
Original entry on oeis.org
1, 4, 6, 18, 30, 90, 102, 306, 510, 1530, 1542, 4626, 7110
Offset: 0
Jens Christian Claussen (claussen(AT)theo-physik.uni-kiel.de), Mar 11 2008
Let x_0 be the state (0 or 1) of the focal node and x_i the state of every node that is i steps away from the focal node. In time step n=0, all x_i=0 except x_0=1 (start with a single seed). In the next step, x_1=1 as they have 1 neighbor being 1. For n=2, the x_1 nodes have 1 neighbor being 1 (x_0) and
themselves being 1; the sum being 2, modulo 2, resulting in x_1=0. The focal node itself is 1 and has 3 neighbors being 1, sum being 4, modulo 2, resulting in x_0=0. The outmost nodes x_n are always 1.
Thus one has the patterns
x_0, x_1, x_2, ...
1
1 1
0 0 1
0 0 1 1
0 0 1 0 1
0 0 1 1 1 1
0 0 1 0 0 0 1
0 0 1 1 0 0 1 1
0 0 1 0 1 0 1 0 1
0 0 1 1 1 1 1 1 1 1
0 0 1 0 0 0 0 0 0 0 1
After 2 time steps, the x_0 and x_1 stay frozen at zero and the remaining x_i are generated by Rule 60 (or Rule 90 on half lattice spacing).
These nodes have multiplicities 1,3,6,12,24,48,96,192,384,768,...
The sequence then is obtained by
a(n) = x_0(n) + 3 * Sum_{i=1..n} x_i(n) * 2^(i-1)
A246011
a(n) = Product_{i in row n of A245562} Lucas(i+1), where Lucas = A000204.
Original entry on oeis.org
1, 3, 3, 4, 3, 9, 4, 7, 3, 9, 9, 12, 4, 12, 7, 11, 3, 9, 9, 12, 9, 27, 12, 21, 4, 12, 12, 16, 7, 21, 11, 18, 3, 9, 9, 12, 9, 27, 12, 21, 9, 27, 27, 36, 12, 36, 21, 33, 4, 12, 12, 16, 12, 36, 16, 28, 7, 21, 21, 28, 11, 33, 18, 29, 3, 9, 9, 12, 9, 27, 12, 21, 9, 27, 27, 36, 12, 36, 21, 33, 9, 27, 27, 36, 27
Offset: 0
From _Omar E. Pol_, Feb 15 2015: (Start)
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
3;
3,4;
3,9,4,7;
3,9,9,12,4,12,7,11;
3,9,9,12,9,27,12,21,4,12,12,16,7,21,11,18;
3,9,9,12,9,27,12,21,9,27,27,36,12,36,21,33,4,12,12,16,12,36,16,28,7,21,21,28,11,33,18,29;
...
Right border gives the Lucas numbers (beginning with 1). This is simply a restatement of the theorem that this sequence is the Run Length Transform of A000204.
(End)
-
A000204 := proc(n) option remember; if n <=2 then 2*n-1; else A000204(n-1)+A000204(n-2); fi; end;
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n,base,2); L1:=nops(t1);
out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c,op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c,op(lis)]; fi;
od:
a:=mul(A000204(i+1), i in lis);
ans:=[op(ans),a];
od:
ans;
-
from math import prod
from re import split
from sympy import lucas
def run_length_transform(f): return lambda n: prod(f(len(d)) for d in split('0+', bin(n)[2:]) if d != '') if n > 0 else 1
def A246011(n): return run_length_transform(lambda n:lucas(n+1))(n) # Chai Wah Wu, Oct 24 2024
A247650
Number of terms in expansion of f^n mod 2, where f = (1/x^2+1/x+1+x+x^2)*(1/y^2+1/y+1+y+y^2) mod 2.
Original entry on oeis.org
1, 25, 25, 49, 25, 289, 49, 361, 25, 625, 289, 361, 49, 961, 361, 625, 25, 625, 625, 1225, 289, 3721, 361, 5041, 49, 1225, 961, 1681, 361, 5041, 625, 5929, 25, 625, 625, 1225, 625, 7225, 1225, 9025, 289, 7225, 3721, 5041, 361, 8281, 5041, 5929, 49, 1225
Offset: 0
-
import sympy
from operator import mul
from functools import reduce
x, y = sympy.symbols('x y')
f = ((1/x**2+1/x+1+x+x**2)*(1/y**2+1/y+1+y+y**2)).expand(modulus=2)
A247650_list, g = [1], 1
for n in range(1, 101):
s = [int(d, 2) for d in bin(n)[2:].split('00') if d != '']
g = (g*f).expand(modulus=2)
if len(s) == 1:
A247650_list.append(g.subs([(x, 1), (y, 1)]))
else:
A247650_list.append(reduce(mul, (A247650_list[d] for d in s)))
# Chai Wah Wu, Sep 25 2014
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