A090895 a(1)=1 then a(n)=a(n-1)/2 if a(n-1) is even, a(n)=a(n-1)+n otherwise.

1, 3, 6, 3, 8, 4, 2, 1, 10, 5, 16, 8, 4, 2, 1, 17, 34, 17, 36, 18, 9, 31, 54, 27, 52, 26, 13, 41, 70, 35, 66, 33, 66, 33, 68, 34, 17, 55, 94, 47, 88, 44, 22, 11, 56, 28, 14, 7, 56, 28, 14, 7, 60, 30, 15, 71, 128, 64, 32, 16, 8, 4, 2, 1, 66, 33, 100, 50, 25, 95, 166, 83, 156, 78, 39

Offset

1,2

Comments

Does a(n)=1 for infinitely many values of n ?

It seems that the answer is yes (see A185038). The number a(n) is always in the range on 1 to 3*a(n), and there is an average of 2 addition steps for every 5 steps. In order to reach '1', the sequence must reach a power of two after an addition step, which is likely to happen on an exponential basis. [Sergio Pimentel, Mar 01 2012]

a(A208852(n)) = n and a(m) != n for m < A208852(n); A185038(a(n)) = 1. [Reinhard Zumkeller, Mar 02 2012]

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Formula

sum(k=1, n, a(k)) seems to be asymptotic to c*n^2 where c=0.57....

Mathematica

nxt[{n_, a_}]:={n+1, If[EvenQ[a], a/2, a+n+1]}; Transpose[NestList[nxt, {1, 1}, 80]][[2]] (* Harvey P. Dale, Aug 25 2015 *)

Prog

(PARI) a(n)=if(n<2, 1, if(a(n-1)%2, a(n-1)+n, a(n-1)/2))
(Haskell)
a090895 n = a090895_list !! (n-1)
a090895_list = 1 : f 2 1 where
   f x y = z : f (x + 1) z where
        z = if m == 0 then y' else x + y; (y', m) = divMod y 2
-- Reinhard Zumkeller, Mar 02 2012

Crossrefs

Cf. A135287, A185038.

Sequence in context: A326935 A135003 A350877 * A337903 A358787 A264767

Adjacent sequences: A090892 A090893 A090894 * A090896 A090897 A090898

Keyword

nonn

Author

Benoit Cloitre, Feb 25 2004

Status

approved