A182417 -id:A182417 - cp's OEIS frontend

A182310 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/2)) + 1, where XOR is the bitwise exclusive-or operator.

0, 1, 2, 4, 7, 5, 8, 13, 12, 11, 15, 9, 14, 10, 16, 25, 22, 30, 18, 28, 19, 27, 23, 29, 20, 31, 17, 26, 24, 21, 32, 49, 42, 64, 97, 82, 124, 67, 99, 83, 123, 71, 101, 88, 117, 80, 121, 70, 102, 86, 126, 66, 100, 87, 125, 68, 103, 85, 128, 193, 162, 244, 143

Offset: 0

Alex Ratushnyak, Apr 24 2012

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

			a(5) = ( a(4) XOR floor(a(4)/2) ) + 1 = (7 XOR 3) + 1 = 4+1 = 5.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, XOR
Wikipedia, Bitwise operation XOR

Cf. A105081, A182417.

#include 
#include 
typedef unsigned long long ULL;
int main(int argc, char **argv) {
  ULL a=0, i=0, p, prev;
  while(1) {
    prev = a,   a = (a^(a/2)) + 1;
  #if 0  // "if 1" to print indices of 2^x
    ++i;
    if ( (i & ((1<<30)-1))==0 )  printf(".");
    if ((a^prev) > prev)
       printf("\n%llu at %llu,  log2: %llu ", a, i, (ULL)(100.0*log2(i)));
  #else
    printf("%llu, ", prev);
  #endif
    // Test reversion:
    p=a-1;
    ULL t=p/2;
    while (t)  p^=t, t/=2;
    if (p!=prev) printf("Reversion failed!"), exit(1);
  }
  return 0;
}  // from Alex Ratushnyak, Apr 26 2012

import Data.Bits (xor)
a182310 n = a182310_list !! n
a182310_list = 0 : map (+ 1)
   (zipWith xor a182310_list $ map (`div` 2) a182310_list) :: [Integer]
-- Reinhard Zumkeller, Apr 25 2012

a:= proc(n) option remember; `if`(n=0, 0, 1+
     (t-> Bits[Xor](t, iquo(t, 2)))(a(n-1)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 29 2021

NestList[BitXor[#,Floor[#/2]]+1&,0,70] (* Harvey P. Dale, Apr 18 2015 *)

terms(n) = my(x=0, i=0); while(i < n, print1(x, ", "); x=bitxor(x, floor(x/2)) + 1; i++)
/* Print initial 200 terms as follows: */
terms(200) \\ Felix Fröhlich, Jun 29 2021

a(n) = A105081(a(n-1)+1). - Jon Maiga, Jun 27 2021

A182418 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/6)) + 1, where XOR is the bitwise exclusive-or operator.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 14, 13, 16, 19, 17, 20, 24, 29, 26, 31, 27, 32, 38, 33, 37, 36, 35, 39, 34, 40, 47, 41, 48, 57, 49, 58, 52, 61, 56, 50, 59, 51, 60, 55, 63, 54, 64, 75, 72, 69, 79, 67, 73, 70, 78, 68, 80, 94, 82, 96, 113, 100, 117

Offset: 0

Alex Ratushnyak, Apr 27 2012

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, 280, 370, 445, 555, 614, 704, 761, 896, 1000, 1088, 1210, 1284, 1389, 1505, 1591, 1677, 1765, 1839, 1968, 2060, 2170, 2250, 2366, 2478, 2606, 2713, 2806, 2857, 2997, 3096, 3193, 3254, 3393, 3487, 3583, 3664, 3741, 3858, 3925, 4073, 4192, 4313, 4394, 4520
Conjecture:
As n -> infinity, a(n) -> infinity.

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

Cf. A182310, A182417.

#include 
int main(int argc, char **argv) {
  unsigned long long a=0, prev;
  while(1) {
    prev = a,   a = (a^(a/6)) + 1;
    printf("%llu, ", prev);
  }
  return 0;     // indices of 2^x: see C program of A182310.
}               // from Alex Ratushnyak, Apr 27 2012

NestList[BitXor[#, Quotient[#, 6]] + 1 &, 0, 63] (* Ivan Neretin, Sep 03 2015 *)

N=100; v=vector(N);
for (n=1, N-1, v[n+1] = bitxor( v[n], v[n] \ 6 ) + 1 );
v /* show terms */
/* Joerg Arndt, Apr 28 2012 */

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

Original entry on oeis.org

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

Alex Ratushnyak, Apr 27 2012

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.

Cf. A182310, A182417, A182418

#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

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 */

cp's OEIS Frontend

A182310 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/2)) + 1, where XOR is the bitwise exclusive-or operator.

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A182418 a(0)=0, a(n+1) = (a(n) XOR floor(a(n)/6)) + 1, where XOR is the bitwise exclusive-or operator.

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

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