A182419 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/8)) + 1, where XOR is the bitwise exclusive-or operator.

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 19, 18, 17, 20, 23, 22, 21, 24, 28, 32, 37, 34, 39, 36, 33, 38, 35, 40, 46, 44, 42, 48, 55, 50, 53, 52, 51, 54, 49, 56, 64, 73, 65, 74, 68, 77, 69, 78, 72, 66, 75, 67, 76, 70, 79, 71, 80, 91, 81, 92, 88, 84, 95

Offset: 0

Author

Alex Ratushnyak, Apr 27 2012

Keywords

• nonn
• base

Comments

Para-random values on each interval (2^k,2^(k+1)-1), then jump to the next interval (2^(k+1),2^(k+2)-1).

Floor(100*log_2(indices of 2^x)):

0, 100, 200, 300, 358, 445, 542, 642, 708, 752, 898, 985, 1042, 1176, 1245, 1319, 1462, 1521, 1588, 1714, 1809, 1906, 1963, 2049, 2149, 2260, 2402, 2481, 2553, 2657, 2770, 2835, 2929, 3004, 3164, 3281, 3346, 3472, 3557, 3646, 3694, 3801, 3935, 4027, 4135, 4214, 4294, 4415

Given a(n), the previous term a(n-1) can be unambiguously reconstructed as in the C program.

As n -> infinity, a(n) -> infinity.

Crossrefs

Cf. A182310, A182417, A182418

Programs

C

#include 
int main(int argc, char **argv) {
  unsigned long long a=0;
  for (int j=0; j<1000; ++j) {
    printf("%llu, ", a);
    a = (a^(a/8)) + 1;
  }
  return 0;     // indices of 2^x: see C program of A182310.
}               // from Alex Ratushnyak, Apr 27 2012

PARI

N=100; v=vector(N);
for (n=1, N-1, v[n+1] = bitxor( v[n], v[n] \ 8 ) + 1 );
v /* show terms */
/* Joerg Arndt, Apr 28 2012 */