A072105 Let c(1)=x, c(n+1) = c(n)/2 + n if c(n) is even, c(n+1)= 2c(n) - n otherwise; then a(n)=c(n) for c(1)=3.

3, 5, 8, 7, 10, 10, 11, 15, 22, 20, 20, 21, 30, 28, 28, 29, 42, 38, 37, 55, 90, 66, 55, 87, 150, 100, 76, 65, 102, 80, 70, 66, 65, 102, 80, 70, 66, 65, 97, 160, 115, 194, 134, 105, 171, 302, 192, 138, 112, 100, 95, 144, 119, 190, 144, 122, 112, 108, 107, 160, 135

Offset

1,1

Comments

It seems that for any n, 2n <= a(n) <16n. If x=0,1,2,4 or 6 we have c(k+1)-c(k)=2 for k large enough and then lim k -> infinity c(k)/k=2. For x=3,5 and for any x >6 there is a conjectured constant 4 < C < 5 such that lim N -> infinity (1/N)*sum(k=1,N,c(k)/k) = C. Hence lim N -> infinity (1/N)*sum(k=1,N,a(k)/k) should be C=4.6...

Links

Table of n, a(n) for n=1..61.

Example

a(1)=3 is odd hence a(2)=2*a(1)-1= 2*3-1 =5

Crossrefs

Sequence in context: A304026 A110641 A121729 * A350693 A364847 A320451

Adjacent sequences: A072102 A072103 A072104 * A072106 A072107 A072108

Keyword

easy,nonn

Author

Benoit Cloitre, Jul 30 2002

Status

approved