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%I #13 Oct 02 2021 11:06:27
%S 1,3,2,6,7,5,3,7,13,15,14,10,11,7,4,12,15,27,26,30,31,29,15,11,21,23,
%T 22,14,15,9,5,13,25,31,30,54,55,53,27,31,61,63,62,58,59,31,28,20,23,
%U 43,42,46,47,45,23,15,29,31,30,18,19,11,6,14,27,51,50,62,63,61,31,55,109,111
%N Concatenation of run lengths in binary expansion of n, written in base 2, then converted to base 10.
%H Rémy Sigrist, <a href="/A175930/b175930.txt">Table of n, a(n) for n = 1..8192</a>
%F From _Rémy Sigrist_, Jul 02 2019: (Begin)
%F a(2^k-1) = k for any k > 0.
%F a(2^k) = A004755(k) for any k > 0.
%F (End)
%e 6 = 110, two runs, lengths 2 and 1, so we write down 101 and convert it to base 10, getting 5. So a(6) = 5.
%t f[n_] := FromDigits[Flatten[IntegerDigits[Length /@ Split[IntegerDigits[n, \ 2]], 2]], 2]
%t Array[f, NUMBER OF TERMS]
%o (PARI) a(n) = my (b=[]); while (n, my (x=valuation(n+(n%2), 2)); b = concat(binary(x), b); n \= 2^x); fromdigits(b, 2) \\ _Rémy Sigrist_, Jul 02 2019
%o (Python)
%o from itertools import groupby
%o def a(n):
%o c = "".join(bin(len(list(g)))[2:] for k, g in groupby(bin(n)[2:]))
%o return int(c, 2)
%o print([a(n) for n in range(1, 75)]) # _Michael S. Branicky_, Oct 02 2021
%Y Cf. A004755, A101211, A321226.
%K base,easy,look,nonn
%O 1,2
%A _Dylan Hamilton_, Oct 23 2010
%E Edited by _N. J. A. Sloane_, Oct 23 2010
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