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.

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

Author

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

Keywords

• nonn
• easy
• nice

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 a(n)-dimensional sphere arises in 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

Numbers that can be expressed as the sum of distinct numbers of the form 2^k - 1 (lenient Mersenne numbers, A000225). This follows from the 2N - Hamming weight definition. A corollary is that these are the numbers with no 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 25 2025

Examples

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

References

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

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
• Jean-Paul Allouche, Jean Bétréma, and Jeffrey 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.
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi 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, Svante Janson, and Tsung-Hsi 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.
• Anunay 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.

Crossrefs

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

Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).

Cf. A004134, A010050, A000165.

Cf. A000120, A079559, A085354, A030308, A000225.

Programs

Haskell

a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
-- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011

Magma

[n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019

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 *)

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

Python

def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021

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

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