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%I #16 May 11 2017 11:41:33
%S 1,0,1,2,0,2,1,2,1,0,1,0,1,2,1,2,1,1,1,1,0,1,1,2,1,1,1,2,1,2,1,2,1,1,
%T 3,0,1,0,3,1,1,0,1,0,1,1,1,2,3,0,0,0,1,0,3,0,3,1,3,2,1,2,1,2,1,1,3,2,
%U 1,2,1,2,0,1,0,1,0,2,3,1,1,1,0,1,0,1,1,2,0,1,0,2,1,1,1,2,3,2,1,1,0,1,1,2,0
%N a(n) = number of run-lengths that each occur only once in the binary representation of n.
%H Gheorghe Coserea, <a href="/A165414/b165414.txt">Table of n, a(n) for n = 1..10000</a>
%e 92 in binary is 1011100. There is a run of one 1, followed by a run of one 0, then a run of three 1's, then finally a run of two 0's. The run lengths are therefore (1,1,3,2). The values of these run lengths that each only occur once are (3,2). Since there are 2 values that occur once, then a(92) = 2.
%t Table[Count[Tally[Length/@Split[IntegerDigits[n,2]]],_?(#[[2]]==1&)],{n,120}] (* _Harvey P. Dale_, May 11 2017 *)
%o (PARI)
%o binruns(n) = {
%o if (n == 0, return([1, 0]));
%o my(bag = List(), v=0);
%o while(n != 0,
%o v = valuation(n,2); listput(bag, v); n >>= v; n++;
%o v = valuation(n,2); listput(bag, v); n >>= v; n--);
%o return(Vec(bag));
%o };
%o a(n) = {
%o my(v = binruns(n), hist = vector(1+logint(n+1, 2)));
%o for (i = 1, #v, if (v[i] != 0, hist[v[i]]++));
%o #select(k->(k==1), hist)
%o };
%o vector(105, i, a(i)) \\ _Gheorghe Coserea_, Sep 24 2015
%Y Cf. A005811, A165413.
%K base,nonn
%O 1,4
%A _Leroy Quet_, Sep 17 2009
%E Extended by _Ray Chandler_, Mar 13 2010
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