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A320262
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Write n in binary, then modify each run of 0's and each run of 1's by appending a 0. a(n) is the decimal equivalent of the result.
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2
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2, 8, 6, 16, 34, 24, 14, 32, 66, 136, 70, 48, 98, 56, 30, 64, 130, 264, 134, 272, 546, 280, 142, 96, 194, 392, 198, 112, 226, 120, 62, 128, 258, 520, 262, 528, 1058, 536, 270, 544, 1090, 2184, 1094, 560, 1122, 568, 286, 192, 386, 776, 390, 784, 1570, 792, 398
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OFFSET
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1,1
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COMMENTS
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A variation of A175046. Indices of record values are given by A319423.
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) = 14.
Then f(k) = 15*6^(k-1) - 2^(k-1) for k >= 0.
Proof: the equation for f is true for k = 0. 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)+2, a(4n+2) = 4*a(2n+1) and a(4n+3) = 2*a(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) = 14 shows that f(k) = 15*6^(k-1) - 2^(k-1) for k > 0.
(End)
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LINKS
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FORMULA
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a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+2, a(4n+2) = 4*a(2n+1) and a(4n+3) = 2*a(2n+1)+2. - Chai Wah Wu, Nov 21 2018
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EXAMPLE
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6 in binary is 110. Modify each run by appending a 0 to get 11000, which is 24 in decimal. So a(6) = 24.
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MATHEMATICA
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Array[FromDigits[Flatten@ Map[Append[#, 0] &, Split@ IntegerDigits[#, 2]], 2] &, 55] (* Michael De Vlieger, Nov 23 2018 *)
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PROG
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(Python)
from re import split
return int(''.join(d+'0' for d in split('(0+)|(1+)', bin(n)[2:]) if d != '' and d != None), 2)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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