A182187 a(n) is the least m >= n such that the Hamming distance D(n,m) = 2.

3, 2, 4, 5, 7, 6, 10, 11, 11, 10, 12, 13, 15, 14, 22, 23, 19, 18, 20, 21, 23, 22, 26, 27, 27, 26, 28, 29, 31, 30, 46, 47, 35, 34, 36, 37, 39, 38, 42, 43, 43, 42, 44, 45, 47, 46, 54, 55, 51, 50, 52, 53, 55, 54, 58, 59, 59, 58, 60, 61, 63, 62, 94, 95, 67, 66, 68

Offset: 0

Author

Vladimir Shevelev, Apr 17 2012

Keywords

• nonn
• base
• easy

Comments

a(n) = n<+>2 (see comment in A206853).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A206853 (trajectory of 1), A207063 (trajectory of 0).

Cf. A209544 (primes which are not terms), A209554 (and also not n<+>3).

Cf. A086799 ((n-1)<+>1), A182209 (n<+>3), A182336 (n<+>4).

Cf. A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206960, A007814.

Programs

Maple

HD:= (i, j)-> add(h, h=Bits[Split](Bits[Xor](i, j))):
a:= proc(n) local c;
      for c from n do if HD(n, c)=2 then return c fi od
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 17 2012

Mathematica

t={}; Do[i=n+1; While[Count[IntegerDigits[BitXor[n,i],2],1]!=2,i++]; AppendTo[t,i],{n,0,66}]; t (* Jayanta Basu, May 26 2013 *)

PARI

a(n) = bitxor(n, 3<>1+1,2)); \\ Kevin Ryde, Jul 09 2021

Python

def a(n):
  m = n + 1
  while bin(n^m).count('1') != 2: m += 1
  return m
print([a(n) for n in range(67)]) # Michael S. Branicky, Mar 02 2021

Sage

def A182187(n):
    S = n.bits(); T = S; c = n; L = len(S)
    while true:
         H = sum(a != b for a, b in zip(S, T))
         if H == 2: return c
         c += 1; T = c.bits()
         if len(T) > L: L += 1; S.append(0)
[A182187(n) for n in (0..66)]   # Peter Luschny, May 26 2013

Formula

If n is odd, then a(n) = n+2^(A007814(n+1)-1); if n == 2 (mod 4), then a(n) = n+2^(A007814(n+2)-1); if n == 0 (mod 4), then a(n) = n+3.

Using this formula, we can prove the conjecture formulated in comment in A209554 in the case k=2. Moreover, let us show that if N does not have the form 8*t or 8*t+1, then it can be represented in the form n<+>2. Indeed, in the cases N = 8*t+2, 8*t+4, 8*t+6, 8*t+3, 8*t+5 and 8*t+7 it is sufficient to choose n=N-4, n=N-2, n=N-1, n=N-3, n=N-2 and n = N-3 respectively; in the cases 8*t, 8*t+1, for every choice of n <= N, we do not obtain the equality n<+>2 = N.

In addition, note that n<+>1 = n + 2^A007814(n+1) = A086799(n+1).

Extensions

More terms from Alois P. Heinz, Apr 17 2012