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A343852 a(n) is the least k > 0 such that the binary expansions of k and of n + k have the same numbers of 0's and of 1's.

%I A343852 #36 Dec 19 2024 19:47:52
%S A343852 5,10,9,20,21,18,17,40,19,42,38,36,37,34,33,80,35,38,37,84,35,76,74,
%T A343852 72,73,74,70,68,69,66,65,160,67,70,69,76,67,74,73,168,75,70,69,152,67,
%U A343852 148,146,144,71,146,145,148,140,140,138,136,137,138,134,132,133
%N A343852 a(n) is the least k > 0 such that the binary expansions of k and of n + k have the same numbers of 0's and of 1's.
%C A343852 This is the binary analog of A343888.
%C A343852 The required comparisons are (1) that the number of ones in the binary expansion of k must equal the number of ones in the binary expansion of k+n, and (2) that the number of zeroes in the binary expansion of k must equal the number of zeroes in the binary expansion of k+n. See the example below. - _Harvey P. Dale_, Dec 19 2024
%H A343852 Rémy Sigrist, <a href="/A343852/b343852.txt">Table of n, a(n) for n = 1..8191</a>
%F A343852 a(n) <= A004757(n).
%e A343852 The first terms, alongside the binary expansions of a(n) and of n + a(n), are:
%e A343852   n   a(n)  bin(a(n))  bin(n+a(n))
%e A343852   --  ----  ---------  -----------
%e A343852    1     5        101          110
%e A343852    2    10       1010         1100
%e A343852    3     9       1001         1100
%e A343852    4    20      10100        11000
%e A343852    5    21      10101        11010
%e A343852    6    18      10010        11000
%e A343852    7    17      10001        11000
%e A343852    8    40     101000       110000
%e A343852    9    19      10011        11100
%e A343852   10    42     101010       110100
%e A343852   11    38     100110       110001
%e A343852   12    36     100100       110000
%e A343852   13    37     100101       110010
%e A343852   14    34     100010       110000
%e A343852   15    33     100001       110000
%t A343852 lk[n_]:=Module[{k=1},While[DigitCount[k,2,0]!=DigitCount[n+k,2,0]||DigitCount[k,2,1]!=DigitCount[n+k,2,1],k++];k]; Array[lk,70] (* _Harvey P. Dale_, Dec 19 2024 *)
%o A343852 (PARI) a(n) = { for (k=1, oo, if (#binary(k)==#binary(n+k) && hammingweight(k)==hammingweight(n+k), return (k))) }
%o A343852 (Python)
%o A343852 def a(n):
%o A343852   k = 1
%o A343852   while k.bit_length() != (n+k).bit_length() or bin(k).count('1') != bin(n+k).count('1'): k += 1
%o A343852   return k
%o A343852 print([a(n) for n in range(1, 62)]) # _Michael S. Branicky_, May 04 2021
%Y A343852 Cf. A000120, A004757, A023416, A293198, A343888.
%K A343852 nonn,base
%O A343852 1,1
%A A343852 _Rémy Sigrist_, May 03 2021