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A005187
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a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.
(Formerly M2330)
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175
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0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128
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OFFSET
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0,3
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COMMENTS
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Also the power of 2 in (2n)! and in (2n)!!.
Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003
Also numbers having a partition into distinct Mersenne numbers > 0; A079559(a(n))=1; complement of A055938. - Reinhard Zumkeller, Mar 18 2009
Equals row sums of triangle A129264, starting with "1". - Gary W. Adamson, Nov 27 2009
Wikipedia's article on the "Whitney Immersion theorem" mentions that the quantity 2n - a(n) is involved in the differential topology of manifolds, notably the Immersion Conjecture proved by Ralph Cohen in 1985." - Jonathan Vos Post, Jan 25 2010
a(n) is the sum of progressive halves and floors. E.g., for a(10) we consider 20. First we halve and floor, so 10, and continue, 5,2,1, so a(10)=18. a(15) we go from 30, so 15+7+3+1 = 26. - Jon Perry, Jul 16 2010
a(n) is the sum of binary Hamming distances between consecutive integers up to n. - Anunay Kulshrestha, Jun 16 2012
For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016
a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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N. J. A. Sloane and T. D. Noe, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
Laurent Alonso; Edward M. Reingold; René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso; Edward M. Reingold; René Schott, The average-case complexity of determining the majority, SIAM J. Comput. 26 (1997), no. 1, 1-14.
Sung-Hyuk Cha, On Integer Sequences Derived from Balanced k-ary Trees, Applied Mathematics in Electrical and Computer Engineering, 2012.
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From N. J. A. Sloane, Dec 24 2012
Ralph L. Cohen, The Immersion Conjecture for Differentiable Manifolds, The Annals of Mathematics, 1985: 237-328. [From Jonathan Vos Post, Jan 25 2010]
Hsien-Kuei Hwang, S. Janson, T. H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016.
Hsien-Kuei Hwang, S. Janson, T. H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Keith Johnson and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
A. Kulshrestha, On Hamming Distance between base-n representations of whole numbers, arXiv:1203.4547 [cs.DM], 2012.
Michael E. Saks; Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
Wikipedia, Whitney Immersion Theorem (version of March 23 2013).
Allan Wilks, Email to N. J. A. Sloane, Jul 7 1988.
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FORMULA
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For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len Smiley
a(n) = Sum_{k >= 0} floor(n/2^k) = n + A011371(n). - Henry Bottomley, Jul 03 2001
G.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002
a(n) = Sum_{k=1..n} A001511(k) - (number of 1's in binary representation of n). - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003
Sum_{n=2^k..2^(k+1)-1} a(n) = 3*4^k - (k+4)*2^(k-1) = A085354(k). - Philippe Deléham, Feb 19 2004
a(n) = A011371(n) + n, n >= 0.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
a(n) = 2n - A000120(n). - Paul Barry, Oct 26 2007
PURRS demo results: Bounds for a(n) = n + a(1/2*n) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008
If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000225(k+1). - Philippe Deléham, Oct 16 2011
a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011
a(2n+1) = a(2n) + 1. - M. F. Hasler, Jan 24 2015
a(n) = A004134(n) - n. - Cyril Damamme, Aug 04 2015
G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017
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EXAMPLE
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G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...
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MAPLE
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A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014
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MATHEMATICA
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a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
Table[IntegerExponent[(2n)!, 2], {n, 0, 65}] (* Robert G. Wilson v, Apr 19 2006 *)
Table[2n-DigitCount[2n, 2, 1], {n, 0, 70}] (* Harvey P. Dale, May 24 2014 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */
(PARI) {a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */
(PARI) {a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */
(PARI) a(n)=my(s=n); while(n>>=1, s+=n); s \\ Charles R Greathouse IV, Apr 07 2012
(PARI) a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013
(Haskell)
a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
-- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011
(Sage)
@CachedFunction
def A005187(n): return A005187(n//2) + n if n > 0 else 0
[A005187(n) for n in range(66)] # Peter Luschny, Dec 13 2012
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CROSSREFS
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Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899. A001511(n) = a(n+1) - a(n), A046161(n) = 2^a(n), a(n) = A011371(2n+1).
Partial sums of A001511.
Cf. A004134.
Sequence in context: A046541 A219897 A047344 * A184835 A184823 A242921
Adjacent sequences: A005184 A005185 A005186 * A005188 A005189 A005190
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999
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STATUS
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approved
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