OFFSET
0,2
COMMENTS
With the particular values: a(2^k)=(74*4^(k-1))+4)/3 and also for example: a(3*2^(k-1))=(8*4^(k-2)+4)/3, a(7*2^(k-2))=a(5*2^(k-2))=(8*4^(k-3)+4)/3, the recurrence rule is: if we denote U the finited sequence of numbers between a(2^k) and a(2^(k+1)), the finited sequence of numbers between a(2^(k+1)) and a(2^(k+2)) is given by: U - ((8*4^(k-1)+4)/3) - U. It seems that this sequence gives the numbers of "1" in the sets of "1" in the sequence A174353.
EXAMPLE
a(8)=a(2^3)=(74*4^2+4)/3=396. Between a(8)=396 and a(16)=1580, the numbers are: 2, 4, 2, 12, 2, 4, 2. Then between a(16) and a(32)= 6316, the numbers of the sequence a are: 2, 4, 2, 12, 2, 4, 2 , 44=(8*4^2+4)/3, 2, 4, 2, 12, 2, 4, 2. So have we obtained in the next step: 1580, 2, 4, 2, 12, 2, 4, 2 , 44, 2, 4, 2, 12, 2, 4, 2, 6316.
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Richard Choulet, Mar 18 2010
STATUS
approved