A333922 a(n) = 1 - (floor(2^n*A002487(n)/A002487(n+1)) mod 2).

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

Offset

0

Comments

The real number Sum_{n >= 0} a(n)/2^n (A346989) is irrational by Cantor's diagonal argument.

Links

Table of n, a(n) for n=0..86.

Wikipedia, Cantor's diagonal argument

Formula

a(2^k-1) = 1 for any k >= 0.

a(7*2^n-1) = 0 for any k > 0.

Example

The first terms, alongside A002487(n)/A002487(n+1) as a fraction and in binary, are:
  n   a(n)  s(n)/s(n+1)  bin(s(n)/s(n+1))
  --  ----  -----------  -----------------------------
   0     1            0    (0).00000000000000000000...
   1     1            1    1.(0)0000000000000000000...
   2     1          1/2    0.1(0)000000000000000000...
   3     1            2   10.00(0)00000000000000000...
   4     0          1/3    0.010(1)0101010101010101...
   5     1          3/2    1.1000(0)000000000000000...
   6     1          2/3    0.10101(0)10101010101010...
   7     1            3   11.000000(0)0000000000000...
   8     1          1/4    0.0100000(0)000000000000...
   9     1          4/3    1.01010101(0)10101010101...
  10     1          3/5    0.100110011(0)0110011001...
  11     1          5/2   10.1000000000(0)000000000...
  12     1          2/5    0.01100110011(0)01100110...
  13     0          5/3    1.101010101010(1)0101010...
  14     1          3/4    0.1100000000000(0)000000...
  15     1            4  100.00000000000000(0)00000...

Prog

(PARI) A002487(n, a=1, b=0)=for(i=0, #binary(n), if(bittest(n, i), b+=a, a+=b)); b
a(n) = 1 - floor(2^n*A002487(n)/A002487(n+1))%2
(Python)
from functools import reduce
def A333922(n): return (sum(reduce(lambda x, y:(x[0], x[0]+x[1]) if int(y) else (x[0]+x[1], x[1]), bin(n)[-1:2:-1], (1, 0)))<<n)//sum(reduce(lambda x, y:(x[0], x[0]+x[1]) if int(y) else (x[0]+x[1], x[1]), bin(n+1)[-1:2:-1], (1, 0)))&1^1 if n else 1 # Chai Wah Wu, May 07 2023

Crossrefs

Cf. A002487, A346989.

Sequence in context: A187944 A105597 A188470 * A071026 A259022 A153490

Adjacent sequences: A333919 A333920 A333921 * A333923 A333924 A333925

Keyword

nonn,base

Author

Rémy Sigrist, Apr 10 2020

Status

approved