A048896 a(n) = 2^(A000120(n+1) - 1), n >= 0.

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4

Offset: 0

Author

Wolfdieter Lang

Keywords

• nonn
• easy

Comments

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]

Row sums of triangle A128937. - Philippe Deléham, May 02 2007

a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009

a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011

For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012

a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015

From Gus Wiseman, Oct 30 2022: (Start)

Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).

Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.

(End)

Examples

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,2;
  1,2,2,4;
  1,2,2,4,2,4,4,8;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32;
  ...,
the first half-rows converge to Gould's sequence A001316.
(End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Neil J. Calkin, Eunice Y. S. Chan, and Robert M. Corless, Some Facts and Conjectures about Mandelbrot Polynomials, Maple Transactions (2021) Vol. 1, No. 1 Article 1.
• Neil J. Calkin, Eunice Y. S. Chan, Robert M. Corless, David J. Jeffrey, and Piers W. Lawrence, A Fractal Eigenvector, arXiv:2104.01116 [math.DS], 2021.
• Emeric 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.
• OEIS Wiki, Montgomery's pair correlation conjecture
• Gus Wiseman, Statistics, classes, and transformations of standard compositions

Crossrefs

This is Guy Steele's sequence GS(3, 5) (see A135416).

Equals first right hand column of triangle A160468.

Equals A160469(n+1)/A002425(n+1).

Cf. A000079, A001316, A117972, A160476, A167364, A220466, A261363.

Standard compositions are listed by A066099.

The opposite version (counting refinements) is A080100.

The version for Heinz numbers of partitions is A317141.

Cf. A000120, A001511, A029931, A048793, A058891, A070939, A272919, A357134.

Programs

Haskell

a048896 n = a048896_list !! n
a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x,2*x])
-- Reinhard Zumkeller, Mar 07 2011

Haskell

import Data.List (transpose)
a048896 = a000079 . a000120
a048896_list = 1 : concat (transpose
   [zipWith (-) (map (* 2) a048896_list) a048896_list,
    map (* 2) a048896_list])
-- Reinhard Zumkeller, Jun 16 2013

Magma

[Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014

Maple

a := n -> 2^(add(i,i=convert(n+1,base,2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009

Mathematica

NestList[Flatten[#1 /. a_Integer -> {a, 2 a}] &, {1}, 4] // Flatten (* Robert G. Wilson v, Aug 01 2012 *)
Table[Numerator[2^n / (n + 1)!], {n, 0, 200}] (* Vincenzo Librandi, Apr 12 2014 *)
Denominator[Table[BernoulliB[2*n] / (Zeta[2*n]/Pi^(2*n)), {n, 1, 100}]] (* Terry D. Grant, May 29 2017 *)
Table[Denominator[((2 n)!/2^(2 n + 1)) (-1)^n], {n, 1, 100}]/4 (* Terry D. Grant, May 29 2017 *)
2^IntegerExponent[CatalanNumber[Range[0,100]],2] (* Harvey P. Dale, Apr 30 2018 *)

PARI

a(n)=if(n<1,1,if(n%2,a(n/2-1/2),2*a(n-1)))

PARI

a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022

Formula

a(n) = 2^A048881(n).

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

It appears that a(n) = Sum_{k=0..n} binomial(2*(n+1), k) mod 2. - Christopher Lenard (c.lenard(AT)bendigo.latrobe.edu.au), Aug 20 2001

a(0) = 1; a(2*n) = 2*a(2*n-1); a(2*n+1) = a(n).

a(n) = (1/2) * A001316(n+1). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004

It appears that a(n) = Sum_{k=0..2n} floor(binomial(2n+2, k+1)/2)(-1)^k = 2^n - Sum_{k=0..n+1} floor(binomial(n+1, k)/2). - Paul Barry, Dec 24 2004

a(n) = Sum_{k=0..n} (T(n,k) mod 2) where T = A039598, A053121, A052179, A124575, A126075, A126093. - Philippe Deléham, May 02 2007

a(n) = numerator(b(n)), where sin(x)^2/x = Sum_{n>0} b(n)*(-1)^n x^(2*n-1). - Vladimir Kruchinin, Feb 06 2013

a((2*n+1)*2^p-1) = A001316(n), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 12 2013

a(n) = numerator(2^n / (n+1)!). - Vincenzo Librandi, Apr 12 2014

a(2n) = (2n+1)!/(n!n!)/A001803(n). - Richard Turk, Aug 23 2017

a(2n-1) = (2n-1)!/(n!(n-1)!)/A001790(n). - Richard Turk, Aug 23 2017

Extensions

New definition from N. J. A. Sloane, Mar 01 2008