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A036284
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Periodic vertical binary vectors of Fibonacci numbers.
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19
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6, 24, 1440, 5728448, 92568198012160, 26494530374406845814111659520, 2095920895719545919920115988669687683503034097906010941440, 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480
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OFFSET
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0,1
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COMMENTS
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The sequence can be also computed with a recurrence that does not explicitly refer to Fibonacci numbers. See the given Maple and C programs.
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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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When Fibonacci numbers are written in binary (see A004685), under each other as:
0000000 (0)
0000001 (1)
0000001 (1)
0000010 (2)
0000011 (3)
0000101 (5)
0001000 (8)
0001101 (13)
0010101 (21)
0100010 (34)
0110111 (55)
1011001 (89)
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.
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MAPLE
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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:
bit_i := (x, i) -> `mod`(floor(x/(2^i)), 2);
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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