A207063 -id:A207063 - cp's OEIS frontend

A206853 a(1)=1, for n>1, a(n) is the least number > a(n-1) such that the Hamming distance D(a(n-1), a(n)) = 2.

1, 2, 4, 7, 11, 13, 14, 22, 26, 28, 31, 47, 55, 59, 61, 62, 94, 110, 118, 122, 124, 127, 191, 223, 239, 247, 251, 253, 254, 382, 446, 478, 494, 502, 506, 508, 511, 767, 895, 959, 991, 1007, 1015, 1019, 1021, 1022, 1534, 1790, 1918, 1982, 2014, 2030, 2038, 2042

Offset: 1

Vladimir Shevelev, Feb 13 2012

For integers a, b, denote by a<+>b the least c>=a, such that D(a,c)=b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>2. Thus this sequence is a Hamming analog of odd numbers 1,3,5,...
A Hamming analog of nonnegative integers is A000225 and a Hamming analog of the triangular numbers is A000975.
All terms are odious (A000069).

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A182187 (n<+>2), A207063 (starting from 0).

Cf. A000225, A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A000069.

read("transforms");
Hamming := proc(a,b)
        XORnos(a,b) ;
        wt(%) ;
end proc:
Dplus := proc(a,b)
        for c from a to 1000000 do
                if Hamming(a,c)=b then
                        return c;
                end if;
        end do:
        return -1 ;
end proc:
A206853 := proc(n)
        option remember;
        if n = 1 then
                1;
        else
                Dplus(procname(n-1),2) ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]]; t = {1}; Do[If[myHammingDistance[t[[-1]], n] == 2, AppendTo[t, n]], {n, 2, 2042}]; t (* T. D. Noe, Mar 07 2012 *)
t={x=1}; Do[i=x+1; While[Count[IntegerDigits[BitXor[x,i],2],1]!=2,i++]; AppendTo[t,x=i],{n,53}]; t (* Jayanta Basu, May 26 2013 *)

next_A206853(n)={my(b=binary(n)); until(norml2(binary(n)-b)==2, n++>=2^#b & b=concat(0,b)); n}
print1(n=1); for(i=1,99,print1(","n=next_A206853(n)))  \\ M. F. Hasler, Apr 07 2012

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

Original entry on oeis.org

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

Vladimir Shevelev, Apr 17 2012

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

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

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.

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

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 *)

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

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

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

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).

More terms from Alois P. Heinz, Apr 17 2012

A182336 a(n) is the least m>=n, such that the Hamming distance D(n,m)=4.

Original entry on oeis.org

15, 14, 13, 12, 11, 10, 9, 8, 19, 18, 17, 16, 17, 16, 16, 17, 31, 30, 29, 28, 27, 26, 25, 24, 33, 32, 32, 33, 32, 33, 34, 35, 47, 46, 45, 44, 43, 42, 41, 40, 51, 50, 49, 48, 49, 48, 48, 49, 63, 62, 61, 60, 59, 58, 57, 56, 64, 65, 66, 67, 68, 69, 70, 71, 79, 78, 77, 76, 75, 74

Offset: 0

Vladimir Shevelev, Apr 25 2012

Or (see comment in A206853) a(n)=n<+>4.

Conjecture: for n > 96, n + 1 <= a(n) <= 9n/8 + 1. - Charles R Greathouse IV, Apr 25 2012

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206853, A206960, A207063, A209544, A209554, A007814, A086799, A182187, A182209.

hamming(n)=my(v=binary(n));sum(i=1,#v,v[i])
a(n)=my(k=n);while(hamming(bitxor(n,k++))!=4,);k \\ Charles R Greathouse IV, Apr 25 2012

Terms corrected by Charles R Greathouse IV, Apr 25 2012

A346261 Lexicographically earliest sequence of decimal words starting with 10 such that each term has Hamming distance at least 2 from all earlier terms.

Original entry on oeis.org

10, 21, 32, 43, 54, 65, 76, 87, 98, 100, 111, 122, 133, 144, 155, 166, 177, 188, 199, 201, 212, 220, 234, 245, 253, 267, 278, 286, 302, 313, 324, 330, 341, 356, 368, 375, 389, 397, 403, 414, 425, 431, 440, 452, 469, 496, 504, 515, 523, 536, 542, 550, 561, 579, 605, 616, 627, 638, 649, 651

Offset: 1

Peter Woodward, Jul 11 2021

			a(1) = 10 by definition.
a(2) = 21 because '2' has not yet appeared in the tens place and '1' has not yet appeared in the ones place.

Michael S. Branicky, Table of n, a(n) for n = 1..10000
J. H. Conway and N. J. A. Sloane, Lexicographic codes: error-correcting codes from game theory, IEEE Transactions on Information Theory, 32:337-348, 1986.

Lexicodes of minimal distance 1,2,3,... over alphabets of size 2: A001477, A001969, A075926, A075928, A075931, A075934, ...; size 3: A001477, A346002, A346003; size 10: A001477, A343444, A333568, A346000, A346001.

Cf. A207063.

def ham(m, n):
    s, t = str(min(m, n))[::-1], str(max(m, n))[::-1]
    return len(t) - len(s) + sum(s[i] != t[i] for i in range(len(s)))
def aupton(terms):
    alst = [10]
    for n in range(2, terms+1):
        an = alst[-1] + 1
        while any(ham(an, aprev) < 2 for aprev in alst[::-1]):  an += 1
        alst.append(an)
    return alst
print(aupton(60)) # Michael S. Branicky, Jul 22 2021

Edited and corrected by N. J. A. Sloane, Jul 20 2021

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A206853 a(1)=1, for n>1, a(n) is the least number > a(n-1) such that the Hamming distance D(a(n-1), a(n)) = 2.

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

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A182336 a(n) is the least m>=n, such that the Hamming distance D(n,m)=4.

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A346261 Lexicographically earliest sequence of decimal words starting with 10 such that each term has Hamming distance at least 2 from all earlier terms.

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Python

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