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A182417 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/4)) + 1, where XOR is the bitwise exclusive-or operator. 3
0, 1, 2, 3, 4, 6, 8, 11, 10, 9, 12, 16, 21, 17, 22, 20, 18, 23, 19, 24, 31, 25, 32, 41, 36, 46, 38, 48, 61, 51, 64, 81, 70, 88, 79, 93, 75, 90, 77, 95, 73, 92, 76, 96, 121, 104, 115, 112, 109, 119, 107, 114, 111, 117, 105, 116, 106, 113, 110, 118, 108 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

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*log2(indices of 2^x)):

0, 100, 200, 258, 345, 445, 490, 622, 708, 812, 847, 965, 1074, 1170, 1205, 1331, 1372, 1524, 1578, 1697, 1798, 1923, 2013, 2092, 2205, 2323, 2363, 2479, 2535, 2674, 2785, 2904, 2973, 3083, 3174, 3210, 3343, 3468, 3605, 3653, 3774, 3885, 3934, 4021, 4124, 4265, 4344, 4444

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

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

LINKS

Ivan Neretin, Table of n, a(n) for n = 0..10000

MATHEMATICA

NestList[BitXor[#, Quotient[#, 4]] + 1 &, 0, 60] (* Ivan Neretin, Sep 03 2015 *)

PROG

(C)

#include <stdio.h>

int main(int argc, char **argv) {

  unsigned long long a=0, p, t, prev;

  while(1) {

    prev = a,   a = (a^(a/4)) + 1;

    printf("%llu, ", prev);

    // Test reversion:

    p=a-1;

    t=p/4;

    while (t)  p^=t, t/=4;

    if (p!=prev) printf("Reversion failed!"), exit(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 */

CROSSREFS

Cf. A182310.

Sequence in context: A247334 A237450 A165514 * A189704 A223543 A306802

Adjacent sequences:  A182414 A182415 A182416 * A182418 A182419 A182420

KEYWORD

nonn,base,look

AUTHOR

Alex Ratushnyak, Apr 27 2012

STATUS

approved

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Last modified June 14 21:27 EDT 2021. Contains 345041 sequences. (Running on oeis4.)