A174401 - OEIS

%I #5 Oct 12 2013 01:52:54

%S 1,6,26,2,100,2,4,2,396,2,4,2,12,2,4,2,1580,2,4,2,12,2,4,2,44,2,4,2,

%T 12,2,4,2,6316,2,4,2,12,2,4,2,44,2,4,2,12,2,4,2,172,2,4,2,12,4,2,44,2,

%U 4,2,12,2,4,2,25260,2,4,2,12,2,4,2,44,2,4,2,12,2,4,2,172,2,4,2,12,2,4,2,44

%N Sequence showing kinds of "waves", built as follows in comments.

%C 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.

%e 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.

%Y Cf. A174353, A174354.

%K easy,nonn

%O 0,2

%A _Richard Choulet_, Mar 18 2010