|
%I #30 Jan 04 2021 17:03:55
%S 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,
%T 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,
%U 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
%N a(n) is the number of distinct lengths of runs in the binary representation of n.
%C Least k whose value is n: 1, 4, 35, 536, 16775, 1060976, ..., = A165933. - _Robert G. Wilson v_, Sep 30 2009
%H Reinhard Zumkeller, <a href="/A165413/b165413.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = 1 for n in A140690. - _Robert G. Wilson v_, Sep 30 2009
%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 distinct values of these run lengths are (1,3,2). Since there are 3 distinct values, then a(92) = 3.
%t f[n_] := Length@ Union@ Map[ Length, Split@ IntegerDigits[n, 2]]; Array[f, 105] (* _Robert G. Wilson v_, Sep 30 2009 *)
%o (Haskell)
%o import Data.List (group, nub)
%o a165413 = length . nub . map length . group . a030308_row
%o -- _Reinhard Zumkeller_, Mar 02 2013
%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) = #Set(select(k->k, binruns(n)));
%o vector(105, i, a(i)) \\ _Gheorghe Coserea_, Sep 17 2015
%o (Python)
%o from itertools import groupby
%o def a(n): return len(set([len(list(g)) for k, g in groupby(bin(n)[2:])]))
%o print([a(n) for n in range(1, 106)]) # _Michael S. Branicky_, Jan 04 2021
%Y Cf. A140690 (locations of 1's), A165933 (locations of new highs).
%Y Cf. A005811, A165414.
%Y Cf. A030308, A044813.
%K base,nonn
%O 1,4
%A _Leroy Quet_, Sep 17 2009
%E More terms from _Robert G. Wilson v_, Sep 30 2009
|
