A048881 a(n) = A000120(n+1) - 1 = wt(n+1) - 1.

0, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3

Offset: 0

Author

Wolfdieter Lang

Keywords

• nonn
• easy

Comments

Highest power of 2 dividing n-th Catalan number (A000108).

a(n) = 0 iff n = 2^k - 1, k=0,1,...

Appears to be number of binary left-rotations (iterations of A006257) to reach fixed point of form 2^k-1. Right-rotation analog is A063250. This would imply that for n >= 0, a(n)=f(n), recursively defined to be 0 for n=0, otherwise as f( ( (1-n)(1-p)(1-s) - (1-n-p-s) ) / 2) + p (s+1) / 2, where p = n mod 2 and s = - signum(n) (f(n<0) is A000120(-n)). - Marc LeBrun, Jul 11 2001

Let f(0) = 01, f(1) = 12, f(2) = 23, f(3) = 34, f(4) = 45, etc. Sequence gives concatenation of 0, f(0), f(f(0)), f(f(f(0))), ... Also f(f(...f(0)...)) converges to A000120. - Philippe Deléham, Aug 14 2003

C(n, k) is the number of occurrence of k in the n-th group of terms in this sequence read by rows: {0}; {0, 1}; {0, 1, 1, 2}; {0, 1, 1, 2, 1, 2, 2, 3}; {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 }; ... - Philippe Deléham, Jan 01 2004

Highest power of 2 dividing binomial(n,floor(n/2)). - Benoit Cloitre, Oct 20 2003

2^a(n) are numerators in the Maclaurin series for (sin x)^2. - Jacob A. Siehler, Nov 11 2009

Conjecture: a(n) is the sum of digits of the n-th word in A076478, for n >= 1; has been confirmed for n up to 20000. - Clark Kimberling, Jul 14 2021

Examples

			From _Omar E. Pol_, Mar 08 2011: (Start)
Sequence can be written in the following form (irregular triangle):
  0,
  0,1,
  0,1,1,2,
  0,1,1,2,1,2,2,3,
  0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,
  0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  ...
Row sums are A001787.
(End)
		

Links

• R. Alter and K. K. Kubota, Prime and Prime Power Divisibility of Catalan Numbers, J. Comb. Th. A 15 (1973) pp. 243-256.
• E. Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215.

Crossrefs

Cf. A000120, A006257, A007318, A063250.

First differences of A078903.

Programs

Haskell

a048881 n = a048881_list !! n
a048881_list = c [0] where c (x:xs) = x : c (xs ++ [x,x+1])
-- Reinhard Zumkeller, Mar 07 2011
(Python 3.10+)
def A048881(n): return (n+1).bit_count()-1 # Chai Wah Wu, Nov 15 2022

Maple

A048881 := proc(n)
    A000120(n+1)-1 ;
end proc:
seq(A048881(n),n=0..200) ; # R. J. Mathar, Mar 12 2018

Mathematica

a[n_] := IntegerExponent[ CatalanNumber[n], 2]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jun 21 2013 *)

PARI

{ a(n) = if( n<0, 0, n++; n /= 2^valuation(n,2); subst( Pol( binary( n ) ), x, 1) - 1 ) } /* Michael Somos, Aug 23 2007 */

PARI

{a(n) = if( n<0, 0, valuation( (2*n)! / n! / (n+1)!, 2 ) ) } /* Michael Somos, Aug 23 2007 */

PARI

a(n) = hammingweight(n+1) - 1; \\ Michel Marcus, Nov 15 2022

Formula

Writing n as 2^m+k with -1 <= k < 2^m-1, then a(n) = A000120(k+1). - Henry Bottomley, Mar 28 2000

a(n) = k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n).

a(2*n) = a(n-1)+1, a(2*n+1) = a(n). - Vladeta Jovovic, Oct 10 2002

G.f.: (1/(x-x^2)) * (x^2/(1-x) - Sum_{k>=1} x^(2^k)/(1-x^(2^k))). - Ralf Stephan, Apr 13 2002

a(n) = A000120(A129760(n+1)). - Reinhard Zumkeller, Jun 30 2010

a(n+k) = A240857(n,k), 0 <= k <= n; in particular: a(n) = A240857(n,0). - Reinhard Zumkeller, Apr 14 2014

a(n) = (n+1)*2 - A101925(n+1). - Gleb Ivanov, Jan 12 2022

Extensions

Entry revised by N. J. A. Sloane, Jun 07 2009