

A174401


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


0



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
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OFFSET

0,2


COMMENTS

With the particular values: a(2^k)=(74*4^(k1))+4)/3 and also for example: a(3*2^(k1))=(8*4^(k2)+4)/3, a(7*2^(k2))=a(5*2^(k2))=(8*4^(k3)+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^(k1)+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 * A036175 A239178 A154869
Adjacent sequences: A174398 A174399 A174400 * A174402 A174403 A174404


KEYWORD

easy,nonn


AUTHOR

Richard Choulet, Mar 18 2010


STATUS

approved



