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%I #10 Mar 04 2016 03:17:34
%S 6,24,1440,5728448,92568198012160,26494530374406845814111659520,
%T 2095920895719545919920115988669687683503034097906010941440,
%U 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480
%N Periodic vertical binary vectors of Fibonacci numbers.
%C The sequence can be also computed with a recurrence that does not explicitly refer to Fibonacci numbers. See the given Maple and C programs.
%C Conjecture: For n>=1, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x^3 + 1) ^ A000225(n-1). If this holds, then for n>=1, a(n) = A048720bi(A136380(n),A048723bi(9,A000225(n-1))). Conjecture 2: there is also one extra (x^1 + 1) factor present, see A136384.
%H A. Karttunen, <a href="/A036284/b036284.txt">Table of n, a(n) for n = 0..10</a>
%H A. Karttunen, <a href="/A036284/a036284.c.txt">C program for computing this sequence</a>
%F a(n) = Sum_{k=0..A007283(n)-1} ([A000045(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function, i.e. Sum (bit n of Fibonacci(k))*(2^k), k = 0 ... (3*(2^n))-1.
%e When Fibonacci numbers are written in binary (see A004685), under each other as:
%e 0000000 (0)
%e 0000001 (1)
%e 0000001 (1)
%e 0000010 (2)
%e 0000011 (3)
%e 0000101 (5)
%e 0001000 (8)
%e 0001101 (13)
%e 0010101 (21)
%e 0100010 (34)
%e 0110111 (55)
%e 1011001 (89)
%e it can be seen that the bits in the n-th column from right repeat after a period of A007283(n): 3, 6, 12, 24, ... (See also A001175). This sequence is formed from those bits: 011, reversed is 110, is binary for 6, thus a(0) = 6. 000110, reversed is 11000, is binary for 24, thus a(1) = 24, 000001011010, reversed is 10110100000, is binary for 1440, thus a(2) = 1440.
%p A036284:=proc(n) option remember; local a, b, c, i, j, k, l, s, x, y, z; if (0 = n) then (6) else a := 0; b := 0; s := 0; x := 0; y := 0; k := 3*(2^(n-1)); l := 3*(2^n); j := 0; for i from 0 to l do z := bit_i(A036284(n-1),(j)); c := (a + b + (`if`((x = y),x,(z+1))) mod 2); if(c <> 0) then s := s + (2^i); fi; a := b; b := c; x := y; y := z; j := j + 1; if(j = k) then j := 0; fi; od; RETURN(s); fi; end:
%p bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);
%t a[n_] := Sum[Mod[Fibonacci[k]/2^n // Floor, 2]* 2^k, {k, 0, 3*2^n - 1}]; Table[a[n], {n, 0, 7}] (* _Jean-François Alcover_, Mar 04 2016 *)
%Y Same sequence in octal base: A036285. Bits reversed: A036286. See also A136378, A136379, A136380, A136382, A136384, A037096, A037093, A000045.
%K nonn,base
%O 0,1
%A _Antti Karttunen_, Nov 01 1998. Entry revised Dec 29 2007.
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