A165414 a(n) = number of run-lengths that each occur only once in the binary representation of n.

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, 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, 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

Offset

1,4

Links

Gheorghe Coserea, Table of n, a(n) for n = 1..10000

Example

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.

Mathematica

Table[Count[Tally[Length/@Split[IntegerDigits[n, 2]]], _?(#[[2]]==1&)], {n, 120}] (* Harvey P. Dale, May 11 2017 *)

Prog

(PARI)
binruns(n) = {
  if (n == 0, return([1, 0]));
  my(bag = List(), v=0);
  while(n != 0,
        v = valuation(n, 2); listput(bag, v); n >>= v; n++;
        v = valuation(n, 2); listput(bag, v); n >>= v; n--);
  return(Vec(bag));
};
a(n) = {
    my(v = binruns(n), hist = vector(1+logint(n+1, 2)));
    for (i = 1, #v, if (v[i] != 0, hist[v[i]]++));
    #select(k->(k==1), hist)
};
vector(105, i, a(i))  \\ Gheorghe Coserea, Sep 24 2015

Crossrefs

Cf. A005811, A165413.

Sequence in context: A070288 A329643 A352697 * A330868 A178687 A325538

Adjacent sequences: A165411 A165412 A165413 * A165415 A165416 A165417

Keyword

base,nonn

Author

Leroy Quet, Sep 17 2009

Extensions

Extended by Ray Chandler, Mar 13 2010

Status

approved