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A320261
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Write n in binary, then modify each run of 0's and each run of 1's by prepending a 1. a(n) is the decimal equivalent of the result.
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3
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3, 14, 7, 28, 59, 30, 15, 56, 115, 238, 119, 60, 123, 62, 31, 112, 227, 462, 231, 476, 955, 478, 239, 120, 243, 494, 247, 124, 251, 126, 63, 224, 451, 910, 455, 924, 1851, 926, 463, 952, 1907, 3822, 1911, 956, 1915, 958, 479, 240, 483, 974, 487, 988, 1979, 990
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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) = 3 and f(1) = a(2) + a(3) = 21, and f(2) = a(4)+a(5)+a(6)+a(7) = 132.
Then f(k) = 135*6^(k-2) - 3*2^(k-2) for k >= 0.
Proof: the formula 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)+3, 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) + 6) = 6*f(k+1) + 3*2^(k+1). Solving this first-order recurrence relation with the initial condition f(1) = 21 shows that f(k) = 135*6^(k-2) - 3*2^(k-2) for k > 0.
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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)+3, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018
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EXAMPLE
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6 in binary is 110. Modify each run by prepending a 1 to get 11110, which is 30 in decimal. So a(6) = 30.
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MATHEMATICA
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Table[FromDigits[Flatten[Join[{1}, #]&/@Split[IntegerDigits[n, 2]]], 2], {n, 60}] (* Harvey P. Dale, Apr 26 2019 *)
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PROG
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(Python)
from re import split
return int(''.join('1'+d 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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