

A036284


Periodic vertical binary vectors of Fibonacci numbers.


18



6, 24, 1440, 5728448, 92568198012160, 26494530374406845814111659520, 2095920895719545919920115988669687683503034097906010941440, 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480
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OFFSET

0,1


COMMENTS

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(n1). If this holds, then for n>=1, a(n) = A048720bi(A136380(n),A048723bi(9,A000225(n1))). Conjecture 2: there is also one extra (x^1 + 1) factor present, see A136384.


LINKS

A. Karttunen, Table of n, a(n) for n = 0..10
A. Karttunen, C program for computing this sequence


FORMULA

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.


EXAMPLE

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 nth 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.


MAPLE

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^(n1)); l := 3*(2^n); j := 0; for i from 0 to l do z := bit_i(A036284(n1), (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);


MATHEMATICA

a[n_] := Sum[Mod[Fibonacci[k]/2^n // Floor, 2]* 2^k, {k, 0, 3*2^n  1}]; Table[a[n], {n, 0, 7}] (* JeanFrançois Alcover, Mar 04 2016 *)


CROSSREFS

Same sequence in octal base: A036285. Bits reversed: A036286. See also A136378, A136379, A136380, A136382, A136384, A037096, A037093, A000045.
Sequence in context: A052524 A267032 A234635 * A139235 A184388 A136606
Adjacent sequences: A036281 A036282 A036283 * A036285 A036286 A036287


KEYWORD

nonn,base


AUTHOR

Antti Karttunen, Nov 01 1998. Entry revised Dec 29 2007.


STATUS

approved



