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%I #28 May 12 2024 10:03:51
%S 1,2,0,1,3,1,0,1,2,4,2,0,2,1,0,1,2,3,1,3,5,3,2,0,1,3,1,0,2,1,0,1,2,3,
%T 1,2,4,2,1,3,4,6,4,2,4,3,2,0,1,2,0,2,4,2,1,0,1,3,1,0,2,1,0,1,2,3,1,2,
%U 4,2,1,2,3,5,3,1,3,2,1,3,4,5,3,5,7,5,4,2,3,5,3,2,4,3,2,0,1,2,0,1,3,1,0,2,3
%N a(n) = the number of digits in the binary representation of n that each do not precede or follow a similarly valued digit.
%C A165316(n) + a(n) = the number of digits in the binary representation of n.
%C Also number of parts equal to 1 in the composition having index n. For the definition of the index of a composition see A298644. For example, a(18) = 3 since the binary form of 18 is (1)00(1)(0) which has 3 runs of length 1 (each placed between parentheses). The command c(n) from the Maple program yields the composition having index n. - _Emeric Deutsch_, Jan 29 2018
%H John Tyler Rascoe, <a href="/A165317/b165317.txt">Table of n, a(n) for n = 1..8191</a>
%e 184 in binary is 10111001. There are exactly three binary digits (the first and last 1's, and the leftmost 0) that are each not adjacent to a similar digit. So a(184) = 3.
%p Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: a := proc (n) local ct, j: ct := 0: for j to nops(c(n)) do if c(n)[j] = 1 then ct := ct+1 else end if end do: ct end proc: seq(a(n), n = 1 .. 105); # most of the Maple program is due to _W. Edwin Clark_. # _Emeric Deutsch_, Jan 29 2018
%o (Python)
%o def a(n): return ((n^(n<<1))&(n^(n>>1))).bit_count() + ((n&3)==2)
%o print([a(n) for n in range(1, 106)]) # _Michael S. Branicky_, May 12 2024
%Y Cf. A165316, A298644, A101211.
%K base,nonn,easy
%O 1,2
%A _Leroy Quet_, Sep 14 2009
%E Extended by _Ray Chandler_, Mar 13 2010
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