A327973 Bitwise XOR of two successive generations (centrally aligned) in the trajectory of rule 30 started from a lone 1 cell: a(n) = A110240(n) XOR 2*A110240(n-1).

5, 23, 93, 335, 1493, 5351, 23853, 85951, 382405, 1369943, 6103965, 21996687, 97906325, 350709671, 1562619373, 5631262591, 25064000389, 89782414999, 400033474525, 1441615751887, 6416397448021, 22984338788455, 102408232210605, 369052763468095, 1642598765228869, 5883986891577303, 26216498605021469, 94477513773305103

Offset: 1

Author

Antti Karttunen, Oct 03 2019

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1024
• Antti Karttunen, Terms a(1)-a(256) drawn as binary strings, with 1 bit = 3x3 pixels resolution
• Antti Karttunen, Terms a(1)-a(1024) drawn as binary strings, with 1 bit = 1 pixel resolution
• Index entries for sequences related to cellular automata

Crossrefs

Cf. A110240, A269160, A327974 (gives the middle bit), A328107 (binary weight of terms).

Cf. also A327971, A327972, 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)));
A327973(n) = bitxor(A110240(n), 2*A110240(n-1));
\\ Use this one for writing b-files:
A327973write(up_to) = { my(s=1, t, n=0); for(n=1,up_to, t = A269160(s); write("b327973.txt", n, " ", bitxor(2*s, t)); s = t); };

Python

def A269160(n): return(n^((n<<1)|(n<<2)))
def genA327973():
    '''Yield successive terms of A327973.'''
    s = 1
    while True:
       t = A269160(s)
       yield (t^(s<<1))
       s = t

Formula

a(n) = A110240(n) XOR 2*A110240(n-1).