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).

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

Author

Antti Karttunen, Oct 03 2019

Keywords

• nonn

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..1023
• Antti Karttunen, Terms up to a(255) drawn as binary strings, with 1 bit = 3x3 pixels resolution
• Antti Karttunen, Terms up to a(1023) drawn as binary strings, with 1 bit = 1 pixel resolution
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to cellular automata

Crossrefs

Cf. A003987, A030101, A057890, A110240, A265281, A280508, A328106 (binary weight of terms).

Cf. also A327972, A327973, A327976, A328103, A328104 for other such combinations.

Programs

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)

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)).