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A261300
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Concatenate successive run lengths of 0's in the binary expansion of n, each increased by 1.
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4
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0, 1, 2, 11, 3, 21, 12, 111, 4, 31, 22, 211, 13, 121, 112, 1111, 5, 41, 32, 311, 23, 221, 212, 2111, 14, 131, 122, 1211, 113, 1121, 1112, 11111, 6, 51, 42, 411, 33, 321, 312, 3111, 24, 231, 222, 2211, 213, 2121, 2112, 21111, 15, 141, 132, 1311, 123, 1221, 1212, 12111, 114, 1131, 1122, 11211, 1113, 11121, 11112, 111111, 7, 61, 52
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OFFSET
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0,3
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COMMENTS
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Any positive number written in binary has its first bit equal to 1. From there on we count the 0 bits up to the next 1 bit or the end of the number. Each count is increased by one because of the impossibility of representing leading 0's in this database.
The sequence is prefixed by a conventional a(0)=0, which represents an empty sum or concatenation.
The positive integer n written as a binary digit string of length m uniquely decomposes into substrings of '1' followed by a maximal run of '0's where the lengths of these substrings forms a composition of m. a(n) is the concatenation of the parts of the composition of m when written in decimal. See A066099 for the table of composition parts. - Michael Somos, Aug 20 2015
Suggested by Armands Strazds's sequence A258055.
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LINKS
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FORMULA
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a(n) = Sum_{k=0..f(n)-1} T(n,k)*10^g(n,k) for n > 0 with a(0)=0 where f(n) = A000120(n), T(n,k) = T(floor(n/2),k - n mod 2) for k > 0 with T(n,0) = A001511(n), and g(n,k) = Sum_{j=0..k-1} (1 + floor(log_10(T(n,j)))). - Mikhail Kurkov, Nov 25 2019 [verification needed]
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EXAMPLE
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n=2 is written "10" in binary, so following the initial digit '1', there is one (= 1) bit zero; this 1 becomes increased to yield a(2) = 2.
n=3 is written "11" in binary, so following the initial digit '1', there is no (= 0) bit zero; after the next digit '1', there follow again 0 bits '0'. These two 0 are increased to yield two 1's, whence a(3) = 11.
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PROG
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(PARI) A261300(n, s="", c=0)={for(i=2, #n=binary(n), c++; n[i]&&s=concat(s, c+c=0)); eval(concat(s, c++))}
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CROSSREFS
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KEYWORD
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nonn,base,changed
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AUTHOR
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STATUS
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approved
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