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Number of leading 1's in binary expansion of n.
15

%I #52 Mar 06 2023 21:24:25

%S 0,1,1,2,1,1,2,3,1,1,1,1,2,2,3,4,1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,4,5,6,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2

%N Number of leading 1's in binary expansion of n.

%C Mirror of triangle A065120. See example. - _Omar E. Pol_, Oct 17 2013

%C a(n) is also the least part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - _Emeric Deutsch_, Jul 24 2017

%H Alois P. Heinz, <a href="/A090996/b090996.txt">Table of n, a(n) for n = 0..65535</a> (first 10001 terms from Vincenzo Librandi)

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(2^k-1)=k; a(A004754(k))=1; a(A004758(k))=2.

%F a(2^k-1)=k; for any other n, a(n) = a(floor(n/2)).

%F a(n) = f(n, 0) with f(n, x) = if n < 2 then n + x else f([n/2], (x+1)*(n mod 2)). - _Reinhard Zumkeller_, Feb 02 2007

%F Conjecture: a(n) = w(n+1)*(w(n+1)-w(n)+1) - w(2^(w(n+1)+1)-n-1) for n>0, where w(n) = floor(log_2(n)), that is, A000523(n). - _Velin Yanev_, Dec 21 2016

%F a(n) = A360189(n-1,floor(log_2(n))). - _Alois P. Heinz_, Mar 06 2023

%e In binary : 14=1110 and there are 3 leading 1's, so a(14)=3.

%e From _Omar E. Pol_, Oct 17 2013: (Start)

%e Written as an irregular triangle with row lengths A011782 the sequence begins:

%e 0;

%e 1;

%e 1,2;

%e 1,1,2,3;

%e 1,1,1,1,2,2,3,4;

%e 1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5;

%e 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,4,4,5,6;

%e Right border gives A001477. Row sums give A000225.

%e (End)

%p a := proc(n) if type(log[2](n+1), integer) then log[2](n+1) else a(floor((1/2)*n)) end if end proc: seq(a(n), n = 0 .. 200); # _Emeric Deutsch_, Jul 24 2017

%p # second Maple program:

%p b:= proc(n, t) `if`(n=0, t,

%p b(iquo(n, 2, 'm'), m*(t+1)))

%p end:

%p a:= n-> b(n, 0):

%p seq(a(n), n=0..127); # _Alois P. Heinz_, Mar 06 2023

%t Join[{0},Table[Length@First@Split@IntegerDigits[n,2],{n,30}]] (* _Birkas Gyorgy_, Mar 09 2011 *) (* adapted by _Vincenzo Librandi_, Dec 23 2016 *)

%o (PARI) a(n) = if(n==0, 0); b=binary(n+1); if(hammingweight(b) == 1, #b-1, a(n\2)) \\ _David A. Corneth_, Jul 24 2017

%o (PARI) a(n) = if(n==0, 0); my(b = binary(n), r = #b); for(i=2, #b, if(!b[i], return(i-1))); r \\ _David A. Corneth_, Jul 24 2017

%Y a(n) = A007814(1+A030101(n)).

%Y Cf. A279209, A279210, A360189.

%K nonn,base

%O 0,4

%A _Benoit Cloitre_, Feb 29 2004

%E Edited and corrected by _Franklin T. Adams-Watters_, Apr 08 2006

%E Sequence had accidentally been shifted left by one step, which was corrected and term a(0)=0 added by _Antti Karttunen_, Jan 01 2007