%I A011371
%S A011371 0,0,1,1,3,3,4,4,7,7,8,8,10,10,11,11,15,15,16,16,18,18,19,19,22,22,23,
%T A011371 23,25,25,26,26,31,31,32,32,34,34,35,35,38,38,39,39,41,41,42,42,46,46,
%U A011371 47,47,49,49,50,50,53,53,54,54,56,56,57,57,63,63,64,64,66,66,67,67,70
%N A011371 n minus (number of 1's in binary expansion of n). Also highest power
of 2 dividing n!.
%C A011371 a(n) = A007814(A000142(n)). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com),
Apr 09 2004
%C A011371 Entries of A005187 appearing twice. - Lekraj Beedassy (blekraj(AT)yahoo.com),
Jul 06 2004
%C A011371 This sequence shows why in binary 0 and 1 are the only two numbers n
such that n equals the sum of its digits raised to the consecutive
powers (equivalent to the base 10 sequence A032799). 1 raised to
any consecutive power is still 1 and thus any the sum of digits raised
to consecutive powers for any n > 1 falls short of equaling the value
of n by the n-th number of this sequence. - Alonso Delarte (alonso.delarte(AT)gmail.com),
Jul 27 2004
%C A011371 Also the number of trailing zeros in the base-2 representation of n!.
- Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 18 2007
%D A011371 Laurent Alonso, Edward M. Reingold and Ren\`e Schott, Determining the
majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
%D A011371 Laurent Alonso, Edward M. Reingold and Ren\`e Schott, The average-case
complexity of determining the majority, SIAM J. Comput. 26 (1997),
no. 1, 1-14.
%D A011371 K. Atanassov, On Some of Smarandache's Problems, section 7, on the 61-st
problem, page 42, American Research Press, 1999, 16-21.
%D A011371 G. Bachman, Introduction to p-Adic Numbers and Valuation Theory, Academic
Press, 1964; see Lemma 3.1.
%D A011371 L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 305.
%D A011371 Michael E. Saks and Michael Werman, On computing majority by comparisons.
Combinatorica 11 (1991), no. 4, 383-387.
%H A011371 T. D. Noe, <a href="b011371.txt">Table of n, a(n) for n=0..1000</a>
%H A011371 K. Atanassov, <a href="http://www.gallup.unm.edu/~smarandache/Atanassov-SomeProblems.pdf">
On Some of Smarandache's Problems</a>
%H A011371 K. Matthews, <a href="http://www.numbertheory.org/php/factorial.html">
Computing the prime-power factorization of n!</a>
%H A011371 R. Stephan, <a href="somedcgf.html">Some divide-and-conquer sequences
...</a>
%H A011371 R. Stephan, <a href="a079944.ps">Table of generating functions</a>
%F A011371 a(n) =a([n/2])+[n/2] =[n/2]+[n/4]+[n/8]+[n/16]+... - Henry Bottomley
(se16(AT)btinternet.com), Apr 24 2001
%F A011371 G.f.: A(x) = 1/(1-x)*Sum(k=1, infinity, x^(2^k)/(1-x^(2^k))). - Ralf
Stephan (ralf(AT)ark.in-berlin.de), Apr 11 2002
%F A011371 a(n) = n - A000120(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Sep
01 2003
%F A011371 a(n)=A005187(n)-n, n>=0.
%F A011371 a(n)=sum{2<=k<=n, sum{j|k,j>=2, floor(log_2(j))-floor(log_2(j-1))}}.
- Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 18 2007
%F A011371 The g.f. can be expressed in terms of a Lambert series, in that g(x)=L[b(k)](x)/
(1-x) where L[b(k)](x)=sum{k>=0, b(k)*x^k/(1-x^k)} is a Lambert series
with b(k)=1, if k is a power of 2, else b(k)=0. - Hieronymus Fischer
(Hieronymus.Fischer(AT)gmx.de), Jun 25 2007
%F A011371 G.f.: g(x)=sum{k>0, c(k)*x^k}/(1-x), where c(k)=sum{j>1,j|k, floor(log_2(j))-floor(log_2(j-1))}.
- Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 25 2007
%F A011371 Recurrence: a(n)=floor(n/2)+a(floor(n/2)); a(2*n)=n+a(n); a(n*2^m)=n*(2^m-1)+a(n).
- Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
%F A011371 a(2^m)=2^m-1, m>=0. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%F A011371 Asymptotic behavior: a(n)=n+O(log(n)), a(n+1)-a(n)=O(log(n)), which follows
from the inequalities below. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%F A011371 a(n)<=n-1; equality holds for powers of 2. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%F A011371 a(n)>=n-1-floor(log_2(n)); equality holds for n=2^m-1, m>0. - Hieronymus
Fischer (Hieronymus.Fischer(AT)gmx.de), Aug 13 2007
%F A011371 lim inf (n-a(n))=1, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%F A011371 lim sup (n-log_2(n)-a(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%F A011371 lim sup (a(n+1)-a(n)-log_2(n))=0, for n-->oo. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
Aug 13 2007
%p A011371 a(n) = RETURN(((2^(l))-1)+sum('(j*floor((n-(2^l)+2^j)/(2^(j+1))))','j'=1..l));
# after K. Atanassov. Here l is [ log2(n) ].
%p A011371 a := n -> n - add(i,i=convert(n,base,2)) [From Peter Luschny (peter(AT)luschny.de),
May 02 2009]
%t A011371 -1+Length[ Last[ Split[ IntegerDigits[ 2(n!), 2 ] ] ] ], FoldList[ Plus,
0, Fold[ Flatten[ {#1, #2, #1} ]&, 0, Range[ 6 ] ] ]
%t A011371 Table[IntegerExponent[n!, 2], {n, 0, 127}]; Table[n-DigitCount[n, 2,
1], {n, 0, 127}]
%t A011371 Table[t = 0; p = 2; While[s = Floor[n/p]; t = t + s; s > 0, p *= 2];
t, {n, 0, 100} ]
%o A011371 (PARI) a(n)=if(n<0,0,valuation(n!,2))
%o A011371 (PARI) a(n)=if(n<0,0,sum(k=1,n,n\2^k))
%o A011371 (PARI) {a(n) = if( n < 0, 0, n - subst( Pol( binary( n ) ), x, 1) ) }
/* Michael Somos Aug 28 2007 */
%Y A011371 Cf. A000120, A005187, A054861.
%Y A011371 a(n)=sum(A007814(k), k=1..n), n >= 1, a(0)=0.
%Y A011371 Cf. A032799.
%Y A011371 Cf. A067080, A098844, A132027.
%Y A011371 Sequence in context: A025561 A145814 A007768 this_sequence A097355 A003860
A108216
%Y A011371 Adjacent sequences: A011368 A011369 A011370 this_sequence A011372 A011373
A011374
%K A011371 nonn,nice,easy
%O A011371 0,5
%A A011371 N. J. A. Sloane (njas(AT)research.att.com).
%E A011371 More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Oct 02 2001
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