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

1,1

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

Formula

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

Prog

(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

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.

Sequence in context: A283224 A178834 A331720 * A255457 A146013 A028894

Adjacent sequences: A327970 A327971 A327972 * A327974 A327975 A327976

Keyword

nonn

Author

Antti Karttunen, Oct 03 2019

Status

approved