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A182187
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a(n) is the least m >= n such that the Hamming distance D(n,m) = 2.
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7
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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
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listen;
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OFFSET
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0,1
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COMMENTS
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a(n) = n<+>2 (see comment in A206853).
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LINKS
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FORMULA
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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.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(Sage)
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)
(Python)
def a(n):
m = n + 1
while bin(n^m).count('1') != 2: m += 1
return m
(PARI) a(n) = bitxor(n, 3<<valuation(n>>1+1, 2)); \\ Kevin Ryde, Jul 09 2021
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CROSSREFS
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Cf. A209544 (primes which are not terms), A209554 (and also not n<+>3).
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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