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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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230
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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 exponent of the largest power of 2 dividing (2n)! and (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
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
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
a(n) is the position of the n'th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
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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)
J.-P. Allouche, J. Betrema, and J. Shallit, Sur des points fixes de morphismes d'un monoïde libre, RAIRO-Theor. Inf. Appl. 23 (1989), 235-249.
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso, Edward M. Reingold, and 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, and 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, and 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 and Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Whitney Immersion Theorem.
Allan Wilks, Email to N. J. A. Sloane, Jul 7 1988.
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FORMULA
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a(n) = A011371(2n+1) = A011371(n) + n, n >= 0.
A046161(n) = 2^a(n).
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), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019]
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
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(n/2) 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
(MAGMA) [n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019
(Python)
def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021
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CROSSREFS
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Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899, A046161.
Cf. A001511 (first differences), A122247 (partial sums).
Cf. A004134.
Sequence in context: A324470 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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