A005187 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)

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

Offset

0,3

Comments

Also exponent of the largest power of 2 dividing (2n)! (A010050) and (2n)!! (A000165).

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

References

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

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.

Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.

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.

Formula

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

Example

G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...

Maple

A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014

Mathematica

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 *)

Prog

(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

Crossrefs

Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899, A046161.

Cf. A001511 (first differences), A122247 (partial sums).

Cf. A004134, A010050, A000165.

Sequence in context: A324470 A219897 A047344 * A184835 A184823 A242921

Adjacent sequences: A005184 A005185 A005186 * A005188 A005189 A005190

Keyword

nonn,easy,nice

Author

N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999

Status

approved