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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. 6
 1, 6, 3, 12, 25, 14, 7, 24, 49, 102, 51, 28, 57, 30, 15, 48, 97, 198, 99, 204, 409, 206, 103, 56, 113, 230, 115, 60, 121, 62, 31, 96, 193, 390, 195, 396, 793, 398, 199, 408, 817, 1638, 819, 412, 825, 414, 207, 112, 225, 454, 227, 460, 921, 462, 231, 120, 241 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS A variation of A175046. Indices of record values are given by A319423. From Chai Wah Wu, Nov 25 2018: (Start) 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. Then f(k) = 10*6^(k-1) - 2^(k-1) for k > 0. 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. (End) LINKS Chai Wah Wu, Table of n, a(n) for n = 1..10000 Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018. FORMULA a(n) = floor(A175046(n)/2). 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 EXAMPLE 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. MATHEMATICA a[n_] := Split[IntegerDigits[n, 2]] /. {a0:{(0)...} :> Prepend[a0, 1], a1:{(1)...} :> Prepend[a1, 0]} // Flatten // FromDigits[#, 2]&; Array[a, 60] (* Jean-François Alcover, Dec 02 2018 *) PROG (Python) from re import split def A320038(n):     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) CROSSREFS Cf. A175046, A319423, A320037. Sequence in context: A322091 A329583 A050132 * A214498 A128756 A049784 Adjacent sequences:  A320035 A320036 A320037 * A320039 A320040 A320041 KEYWORD nonn AUTHOR Chai Wah Wu, Oct 04 2018 STATUS approved

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Last modified May 18 19:58 EDT 2021. Contains 344002 sequences. (Running on oeis4.)