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%I #10 Mar 31 2012 14:02:29
%S 0,12,120,57120,93321840,10431955353116229600,
%T 8557304989566294213168677685339060480,
%U 102743047168201563425402150421568484707810385382513037790885688657488312400960
%N Periodic vertical binary vectors of powers of 3, starting from bit-column 2 (halved).
%C 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(n-1) (= A051179(n-2)). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A000225(n-1))) = A048720bi(A136386(n),A051179(n-2)).
%D S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.
%H A. Karttunen, <a href="/A037097/b037097.txt">Table of n, a(n) for n = 2..12</a>
%H A. Karttunen, <a href="/A036284/a036284.c.txt">C program for computing this sequence</a>
%H S. Wolfram, <a href="http://www.wolframscience.com/nksonline/page-119">A New Kind of Science, Wolfram Media Inc., (2002), p. 119.</a>
%F a(n) = Sum_{k=0..A000225(n-1)} ([A000244(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function.
%e When powers of 3 are written in binary (see A004656), under each other as:
%e 000000000001 (1)
%e 000000000011 (3)
%e 000000001001 (9)
%e 000000011011 (27)
%e 000001010001 (81)
%e 000011110011 (243)
%e 001011011001 (729)
%e 100010001011 (2187)
%e it can be seen that, starting from the column 2 from the right, the bits in the n-th column can be arranged in periods of 2^(n-1): 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.
%p a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n-1))-1);
%p bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);
%Y a(n) = floor(A037096(n)/(2^(2^(n-1)))). See also A036284, A136386.
%K nonn,base
%O 2,2
%A _Antti Karttunen_, Jan 29 1999. Entry revised Dec 29 2007.
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