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%I #22 Jan 24 2017 10:24:59
%S 1,100,101,11000,11001,11010,11011,1110000,1110001,1110010,1110011,
%T 1110100,1110101,1110110,1110111,111100000,111100001,111100010,
%U 111100011,111100100,111100101,111100110,111100111,111101000,111101001,111101010,111101011,111101100,111101101,111101110
%N Elias gamma code (EGC) for n.
%C This sequence is the binary equivalent of A171885 for n>=1 and is also mentioned in the example section of the same.
%C The number of bits of a(n) is equal to A129972(n).
%C Unary(n) = A105279(n-1).
%H Indranil Ghosh, <a href="/A281149/b281149.txt">Table of n, a(n) for n = 1..10000</a>
%H J. Nelson Raja, P. Jaganathan and S. Domnic, <a href="http://dx.doi.org/10.17485/ijst/2015/v8i24/80242">A New Variable-Length Integer Code for Integer Representation and Its Application to Text Compression</a>, Indian Journal of Science and Technology, Vol 8(24), September 2015.
%F For a given integer n, it is stored in two parts. The first part equals 1+floor(log_2 n) and the second part equals n-2^(floor(log_2 n)). The first part is stored in unary and the second part is stored in binary using floor(log_2 n) bits. Now the first and the second parts are concatenated to give the answer.
%e For n = 9, first part is "1110" and the second part is "001". So, a(9) = 1110001.
%o (Python)
%o def unary(n):
%o ....return "1"*(n-1)+"0"
%o def elias_gamma(n):
%o ....if n ==1:
%o ........return "1"
%o ....k=int(math.log(n,2))
%o ....fp=unary(1+k) #fp is the first part
%o ....sp=n-2**(k) #sp is the second part
%o ....nb=k #nb is the number of bits used to store sp in binary
%o ....sp=bin(sp)[2:]
%o ....if len(sp)<nb:
%o ........sp=("0"*(nb-len(sp)))+sp
%o ....return fp+sp
%Y Cf. A105279 (unary code for n), A129972, A171885.
%K nonn,base
%O 1,2
%A _Indranil Ghosh_, Jan 16 2017
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