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Bitwise XOR of trajectories (centrally aligned) of rule 30, and its mirror image, rule 86, when both are started from a lone 1-bit, with the latter delayed by one step: a(n) = A110240(n) XOR 2*A265281(n-1).
8

%I #17 Oct 05 2019 18:21:01

%S 5,23,73,359,1233,6143,19225,93495,325729,1518895,4833289,23453735,

%T 81443089,398815039,1271974489,6168932215,21231239841,99197620591,

%U 314863189193,1541326542823,5312985402193,26258203294847,82884499362201,400683454289591,1406328980294113,6532877164215983,20744329255918985,100303645024039591

%N Bitwise XOR of trajectories (centrally aligned) of rule 30, and its mirror image, rule 86, when both are started from a lone 1-bit, with the latter delayed by one step: a(n) = A110240(n) XOR 2*A265281(n-1).

%H Antti Karttunen, <a href="/A327976/b327976.txt">Table of n, a(n) for n = 1..1024</a>

%H Antti Karttunen, <a href="/A327976/a327976.png">Terms a(1)-a(256) drawn as binary strings, with 1 bit = 3x3 pixels resolution</a>

%H Antti Karttunen, <a href="/A327976/a327976_1.png">Terms a(1)-a(1024) drawn as binary strings, with 1 bit = 1 pixel resolution</a>

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

%F a(n) = A110240(n) XOR 2*A265281(n-1) = A110240(n) XOR 2*A030101(A110240(n-1)).

%o (PARI)

%o A269160(n) = bitxor(n, bitor(2*n, 4*n)); \\ From A269160.

%o A110240(n) = if(!n,1,A269160(A110240(n-1)));

%o A327973(n) = bitxor(A110240(n), 2*A110240(n-1));

%o A269161(n) = bitxor(4*n, bitor(2*n, n));

%o A265281(n) = if(!n,1,A269161(A265281(n-1)));

%o A327976(n) = bitxor(A110240(n), 2*A265281(n-1));

%o \\ Use this one for writing b-files:

%o A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));

%o A327976write(up_to) = { my(s=1, t, n=0); for(n=1,up_to, t = A269160(s); write("b327976.txt", n, " ", bitxor(2*A030101(s), t)); s = t); };

%o (Python)

%o def A269160(n): return(n^((n<<1)|(n<<2)))

%o def A269161(n): return((n<<2)^((n<<1)|n))

%o def genA327976():

%o '''Yield successive terms of A327976.'''

%o s1 = 1

%o s2 = 1

%o while True:

%o s1 = A269160(s1)

%o yield (s1^(s2<<1))

%o s2 = A269161(s2)

%Y Cf. A110240, A265281, A269160, A269161, A030101, A327974 (gives the middle bit), A328108 (binary weight).

%Y Cf. also A327971, A327972, A327973, A328103, A328104 for other such combinations.

%K nonn

%O 1,1

%A _Antti Karttunen_, Oct 04 2019