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A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.

%I A320038 #23 Dec 02 2018 07:44:33
%S A320038 1,6,3,12,25,14,7,24,49,102,51,28,57,30,15,48,97,198,99,204,409,206,
%T A320038 103,56,113,230,115,60,121,62,31,96,193,390,195,396,793,398,199,408,
%U A320038 817,1638,819,412,825,414,207,112,225,454,227,460,921,462,231,120,241
%N A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.
%C A320038 A variation of A175046. Indices of record values are given by A319423.
%C A320038 From _Chai Wah Wu_, Nov 25 2018: (Start)
%C A320038 Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 1 and f(1) = a(2) + a(3) = 9.
%C A320038 Then f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
%C A320038 Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) =  Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 9 shows that f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
%C A320038 (End)
%H A320038 Chai Wah Wu, <a href="/A320038/b320038.txt">Table of n, a(n) for n = 1..10000</a>
%H A320038 Chai Wah Wu, <a href="https://arxiv.org/abs/1810.02293">Record values in appending and prepending bitstrings to runs of binary digits</a>, arXiv:1810.02293 [math.NT], 2018.
%F A320038 a(n) = floor(A175046(n)/2).
%F A320038 a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - _Chai Wah Wu_, Nov 25 2018
%e A320038 6 in binary is 110. Modify each run by prepending the opposite digit to get 01110, which is 14 in decimal. So a(6) = 14.
%t A320038 a[n_] := Split[IntegerDigits[n, 2]] /. {a0:{(0)...} :> Prepend[a0, 1], a1:{(1)...} :> Prepend[a1, 0]} // Flatten // FromDigits[#, 2]&;
%t A320038 Array[a, 60] (* _Jean-François Alcover_, Dec 02 2018 *)
%o A320038 (Python)
%o A320038 from re import split
%o A320038 def A320038(n):
%o A320038     return int(''.join('0'+d if '1' in d else '1'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)
%Y A320038 Cf. A175046, A319423, A320037.
%K A320038 nonn
%O A320038 1,2
%A A320038 _Chai Wah Wu_, Oct 04 2018