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A327971
Bitwise XOR of trajectories of rule 30 and its mirror image, rule 86, when both are started from a lone 1 cell: a(n) = A110240(n) XOR A265281(n).
8
0, 0, 10, 20, 130, 396, 2842, 4420, 38610, 124220, 684490, 1385044, 8891330, 26281036, 192525274, 269101060, 2454365330, 8588410876, 43860512138, 89059958420, 551714970626, 1663794165260, 12235920695450, 19683098342340, 164315052318034, 538162708968636, 2894532467106378, 6192136868790228, 37503903254935874, 114926395086966988, 814341599153559130
OFFSET
0,3
COMMENTS
Each term is a binary palindrome when its trailing zeros (in base 2) are omitted, that is, a term of A057890.
Compare the binary string illustrations drawn for the first 1024 terms of this sequence and for A327976, which has almost the same definition.
FORMULA
a(n) = A110240(n) XOR A265281(n).
a(n) = A280508(A110240(n)) = A110240(n) XOR A030101(A110240(n)).
a(n) = A280508(A265281(n)) = A265281(n) XOR A030101(A265281(n)).
For n >= 1, a(n) = (1/2) * (A327973(n-1) XOR A327976(n-1)).
PROG
(PARI)
A269160(n) = bitxor(n, bitor(2*n, 4*n)); \\ From A269160.
A110240(n) = if(!n, 1, A269160(A110240(n-1)));
A269161(n) = bitxor(4*n, bitor(2*n, n));
A265281(n) = if(!n, 1, A269161(A265281(n-1)));
A327971(n) = bitxor(A110240(n), A265281(n));
(PARI)
A030101(n) = if(n<1, 0, subst(Polrev(binary(n)), x, 2));
A327971write(up_to) = { my(s=1, n=0); for(n=0, up_to, write("b327971.txt", n, " ", bitxor(s, A030101(s))); s = A269160(s)); };
(Python)
def A269160(n): return(n^((n<<1)|(n<<2)))
def A269161(n): return((n<<2)^((n<<1)|n))
def genA327971():
'''Yield successive terms of A327971.'''
s1 = 1
s2 = 1
while True:
yield (s1^s2)
s1 = A269160(s1)
s2 = A269161(s2)
CROSSREFS
Cf. A003987, A030101, A057890, A110240, A265281, A280508, A328106 (binary weight of terms).
Cf. also A327972, A327973, A327976, A328103, A328104 for other such combinations.
Sequence in context: A220012 A154330 A006993 * A250107 A038693 A217318
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 03 2019
STATUS
approved