A174401 Sequence showing kinds of "waves", built as follows in comments.

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, 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, 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

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.

Links

Table of n, a(n) for n=0..87.

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

Cf. A174353, A174354.

Sequence in context: A041064 A005938 A157025 * A329819 A337400 A036175

Adjacent sequences: A174398 A174399 A174400 * A174402 A174403 A174404

Keyword

easy,nonn

Author

Richard Choulet, Mar 18 2010

Status

approved