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 A165413 a(n) = number of distinct lengths of runs in the binary representation of n. 5
 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 1, 2, 2, 2, 3, 1, 3, 2, 3, 2, 2, 2, 1, 2, 2, 2, 3, 3, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Least k whose value is n: 1, 4, 35, 536, 16775, 1060976, ..., . - Robert G. Wilson v, Sep 30 2009 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA a(n) = 1 for n in A140690. - Robert G. Wilson v, Sep 30 2009 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 distinct values of these run lengths are (1,3,2). Since there are 3 distinct values, then a(92) = 3. MATHEMATICA f[n_] := Length@ Union@ Map[ Length, Split@ IntegerDigits[n, 2]]; Array[f, 105] (* Robert G. Wilson v, Sep 30 2009 *) PROG (Haskell) import Data.List (group, nub) a165413 = length . nub . map length . group . a030308_row -- Reinhard Zumkeller, Mar 02 2013 (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) = #Set(select(k->k, binruns(n))); vector(105, i, a(i))  \\ Gheorghe Coserea, Sep 17 2015 CROSSREFS Cf. A005811, A165414 Cf. A030308, A044813. Cf. A140690 Sequence in context: A022921 A080763 A245920 * A172155 A080573 A186440 Adjacent sequences:  A165410 A165411 A165412 * A165414 A165415 A165416 KEYWORD base,nonn AUTHOR Leroy Quet, Sep 17 2009 EXTENSIONS More terms from Robert G. Wilson v, Sep 30 2009 STATUS approved

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Last modified September 19 11:06 EDT 2019. Contains 327192 sequences. (Running on oeis4.)