A123087 Sequence of numbers such that a(2*n) + a(n) = n and a(n) is the smallest number such that a(n) >= a(n-1).

0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 24, 24, 25, 25, 25, 25, 26

Offset: 0

Author

Philippe Deléham, Sep 27 2006

Keywords

• nonn

Comments

If the value a(n) = m >= 1 is appearing for the first time, then n is of the form n = 2^k*s, where k,s are odd numbers. Therefore every m occurs 2 or 4 times consecutively. More exactly, if n+2 has the same form as n (i.e., 2^k*s with odd k,s), then a(n) = m occurs 2 times, otherwise, m occurs 4 times. - Vladimir Shevelev, Aug 25 2010

a(n) is the number of those numbers not exceeding n for which 2 is an infinitary divisor (for definition see comment at A037445). - Vladimir Shevelev, Feb 21 2011

Examples

			  a(2*0) + a(0) = 0 -----> a(0)  = 0
  a(1)  >= a(0) ---------> a(1)  = 0
  a(2*1) + a(1) = 1 -----> a(2)  = 1
  a(3)  >= a(2) ---------> a(3)  = 1
  a(2*2) + a(2) = 2 -----> a(4)  = 1
  a(5)  >= a(4) ---------> a(5)  = 1
  a(2*3) + a(3) = 3 -----> a(6)  = 2
  a(7)  >= a(6) ---------> a(7)  = 2
  a(2*4) + a(4) = 4 -----> a(8)  = 3
  a(9)  >= a(8) ---------> a(9)  = 3
  a(2*5) + a(5) = 5 -----> a(10) = 4
  a(11) >= a(10) --------> a(11) = 4
  a(2*6) + a(6) = 6 -----> a(12) = 4
  a(13) >= a(12) --------> a(13) = 4
  a(2*7) + a(7) = 7 -----> a(14) = 5
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Partial sums of A096268 and of A328979.

Cf. A050292, A065359.

Programs

Haskell

a123087 n = a123087_list !! n
a123087_list = scanl (+) 0 a096268_list
-- Reinhard Zumkeller, Jul 29 2014

PARI

a(n)=if(n<1,0,floor(n/2)-a(floor(n/2))) \\ Benoit Cloitre, Sep 30 2006

Formula

a(0)=0, a(n) = floor(n/2) - a(floor(n/2)); partial sums of A096268; a(2n) = A050292(n); a(n) is asymptotic to n/3. - Benoit Cloitre, Sep 30 2006

a(2*n+1) = a(2*n); a(n) = n/3 + O(log(n)), moreover, the equation a(3m) = m has infinitely many solutions, e.g., a(3*2^k) = 2^k; on the other hand, a((4^k-1)/3) = (4^k-1)/9 - k/3, i.e., limsup|a(n) - n/3| = infinity. - Vladimir Shevelev, Aug 25 2010

a(n) = (n - A065359(n))/3. - Velin Yanev, Jul 13 2021

a(n) = n - A050292(n). - Max Alekseyev, Mar 05 2023