

A037097


Periodic vertical binary vectors of powers of 3, starting from bitcolumn 2 (halved).


5



0, 12, 120, 57120, 93321840, 10431955353116229600, 8557304989566294213168677685339060480, 102743047168201563425402150421568484707810385382513037790885688657488312400960
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OFFSET

2,2


COMMENTS

Conjecture: For n>=3, each term a(n), when considered as a GF(2)[X]polynomial, is divisible by GF(2)[X] polynomial (x + 1) ^ A000225(n1) (= A051179(n2)). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A000225(n1))) = A048720bi(A136386(n),A051179(n2)).


REFERENCES

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.


LINKS

A. Karttunen, Table of n, a(n) for n = 2..12
A. Karttunen, C program for computing this sequence
S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.


FORMULA

a(n) = Sum_{k=0..A000225(n1)} ([A000244(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function.


EXAMPLE

When powers of 3 are written in binary (see A004656), under each other as:
000000000001 (1)
000000000011 (3)
000000001001 (9)
000000011011 (27)
000001010001 (81)
000011110011 (243)
001011011001 (729)
100010001011 (2187)
it can be seen that, starting from the column 2 from the right, the bits in the nth column can be arranged in periods of 2^(n1): 4, 8, ... This sequence is formed from those bits: 0011, reversed is 11100, which is binary for 12, thus a(3) = 12, 00011110, reversed is 011110000, which is binary for 120, thus a(4) = 120.


MAPLE

a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n1))1);
bit_n := (x, n) > `mod`(floor(x/(2^n)), 2);


CROSSREFS

a(n) = floor(A037096(n)/(2^(2^(n1)))). See also A036284, A136386.
Sequence in context: A012564 A012442 A262204 * A299823 A222634 A018204
Adjacent sequences: A037094 A037095 A037096 * A037098 A037099 A037100


KEYWORD

nonn,base


AUTHOR

Antti Karttunen, Jan 29 1999. Entry revised Dec 29 2007.


STATUS

approved



