A393076 a(n) is the last bit in the bit string s_n, where s_0 is 10 and s_n is the concatenation of s_{n-1} with itself with the last bit removed and then reversed.

0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0

Offset

0,1

Comments

For odd n, s_n is palindromic because if s_n = xPx then s_{n+2} = PxxPxPxxP.

a(n) is also the central bit in s_{n+1}.

For any given n, the next 2^n terms after a(n) can be obtained by alternatingly reading the bits from the end and from the start of s_n.

Links

Andrei Zabolotskii, Table of n, a(n) for n = 0..20000

Formula

For n > 0, a(2*n) = a(2*n-1).

Example

s_1 = reverse(10..1) = 101 (where .. is concatenation), so a(1) = 1.
s_2 = reverse(101..10) = 01101, so a(2) = 1.
s_3 = reverse(01101..0110) = 011010110, so a(3) = 0.
Eight more terms can be obtained directly from s_3, without constructing s_4:
   a(n)   0 1 1  0  1 0 1 1 0 <- this is s_3
     n    4 6 8 10 11 9 7 5 3

Prog

(Python)
s, sr, a = 0b10, 0b01, []
for i in range(11):
    a += [s&1]
    s, sr = ((sr&((1<<(2**i))-1))<<(2**i+1))+sr, (s<<(2**i))+(s>>1)
for i in range(100):
    a += [s&1, s&1]
    s >>= 1
print(a)

Crossrefs

Related to A282390.

A334820 is defined by the same rule without reversion.

Sequence in context: A189135 A369961 A219189 * A029691 A209229 A365089

Adjacent sequences: A393072 A393074 A393075 * A393077 A393078 A393079

Keyword

nonn

Author

Andrei Zabolotskii, Jan 31 2026

Status

approved