A327972 Bitwise XOR of trajectories of rule 30 and rule 150, when both are started from a lone 1 cell: a(n) = A110240(n) XOR A038184(n).

0, 0, 12, 4, 128, 384, 3404, 740, 37056, 127296, 794316, 286532, 8510656, 25560896, 224057484, 42076324, 2446214016, 8430013568, 51732969356, 18062215300, 553213409792, 1655549411840, 14630859361996, 3227756349540, 159219183713088, 546944274202816, 3411332163636556, 1231354981057220, 36554500089286208, 109782277571646400, 962314238681316620

Offset: 0

Author

Antti Karttunen, Oct 03 2019

Keywords

• nonn

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
• Index entries for sequences operating on GF(2)[X]-polynomials

Crossrefs

Cf. A003987, A038184, A048727, A110240, A269160.

Cf. also A327971, A327973, A327976, A328103, A328104 for other such combinations.

Programs

PARI

A048727(n) = bitxor(n, bitxor(2*n, 4*n)); \\ From A048727
A038184(n) = if(!n,1,A048727(A038184(n-1)));
A269160(n) = bitxor(n, bitor(2*n, 4*n)); \\ From A269160.
A110240(n) = if(!n,1,A269160(A110240(n-1)));
A327972(n) = bitxor(A038184(n), A110240(n));
\\ Use this one for writing b-files:
A327972write(up_to) = { my(s1=1, s2=1); for(n=0,up_to, write("b327972.txt", n, " ", bitxor(s1, s2)); s1 = A048727(s1); s2 = A269160(s2)); };

Python

def A048727(n): return(n^(n<<1)^(n<<2))
def A269160(n): return(n^((n<<1)|(n<<2)))
def genA327972():
    '''Yield successive terms of A327972.'''
    s1 = 1
    s2 = 1
    while True:
       yield (s1^s2)
       s1 = A269160(s1)
       s2 = A048727(s2)

Formula

a(n) = A038184(n) XOR A110240(n).

Conjecture: for n > 1, floor(log_2(a(n))) = 2*n - (1,2,1,4,1,2,1,5 according as n == 0..7 (mod 8), respectively). - Alan Michael Gómez Calderón, Mar 02 2023