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A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.

%I A320037 #24 Nov 22 2018 18:53:57
%S A320037 2,9,6,17,38,25,14,33,70,153,78,49,102,57,30,65,134,281,142,305,614,
%T A320037 313,158,97,198,409,206,113,230,121,62,129,262,537,270,561,1126,569,
%U A320037 286,609,1222,2457,1230,625,1254,633,318,193,390,793,398,817,1638,825,414
%N A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.
%C A320037 A variation of A175046. Indices of record values are given by A319423.
%C A320037 From _Chai Wah Wu_, Nov 21 2018: (Start)
%C A320037 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) = 2 and f(1) = a(2) + a(3) = 15.
%C A320037 Then f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
%C A320037 Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. 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) = 15 shows that f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
%C A320037 (End)
%H A320037 Chai Wah Wu, <a href="/A320037/b320037.txt">Table of n, a(n) for n = 1..10000</a>
%H A320037 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 A320037 a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. - _Chai Wah Wu_, Nov 21 2018
%e A320037 6 in binary is 110. Modify each run by appending the opposite digit to get 11001, which is 25 in decimal. So a(6) = 25.
%o A320037 (Python)
%o A320037 from re import split
%o A320037 def A320037(n):
%o A320037     return int(''.join(d+'0' if '1' in d else d+'1' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)
%Y A320037 Cf. A175046, A319423, A320038.
%K A320037 nonn,base
%O A320037 1,1
%A A320037 _Chai Wah Wu_, Oct 04 2018