OFFSET
0,1
COMMENTS
If we denote "A" the finite sequence between a(2^(n-1)-2) and a(2^n-2), the subsequence of a between
a(2^(n)-2) and a(2^(n+1)-2) is given by: " A - a(3*2^(n-1)-2) - A" for every n>=2.
FORMULA
a(2n+1)=2. a(2^(n+1)-2)=9*5^n. a(3*2^n-2)=(9*5^n-1)/4.
EXAMPLE
a(4)=a(2*3-2)=(9*4-1)/4=11. a(14)=a(2^4-2)==9*5^3=125*9=1125.
Between a(2) and a(6) the subsequence is "2, 11, 2"; then between a(6) and a(14) the subsequence of a is:
"2, 11, 2, a(10)=56, 2, 11, 2".
It seems that this new sequence gives the number of 2 in the sets of 2 of the sequence A174835.
CROSSREFS
KEYWORD
easy,nonn,uned
AUTHOR
Richard Choulet, Mar 30 2010
STATUS
approved