| Displaying 1-5 of 5 results found.
page
1
Sequence A136382 shown in octal base.
+20
5
6, 50, 5660, 57266500, 5745056664272600, 57725105055132666451322727325000, 5774565132122505051225325134566664426132512553272725535132426000, 57772261341352513254262525526505050275531226553251312456427326666445105642653132512245531220572727245225254562513252136134550000
Periodic vertical binary vectors of Fibonacci numbers.
+10
19
6, 24, 1440, 5728448, 92568198012160, 26494530374406845814111659520, 2095920895719545919920115988669687683503034097906010941440, 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480
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(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.
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 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.
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^(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);
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}] (* Jean-François Alcover, Mar 04 2016 *)
Quotient obtained when A036284(n) is considered as a GF(2)[X]-polynomial and it is divided by (x^3 + 1) ^ A000225(n-1).
+10
6
24, 160, 11968, 49657088, 837028380268032, 237269922100748727235760269312, 18811253173629696438994877569412700111469395859003555753984, 118178826602781220665226658680265194908312590801831513776333330179329649495708436476846379030238467286212637486694400
Quotient obtained when A136380(n)/2 is considered as a GF(2)[X]-polynomial and it is divided by (x + 1).
+10
6
4, 48, 3360, 14043520, 233515838757120, 65982595605873500894008888320, 5233741023536997251047595348728205456443682897303843358720, 32837130684987081672210288030183520098814938795984162933658101468543499651419210151303128996446334767341864627691520
COMMENTS
Note that each term a(n) fits into A007283(n-1) bits.
Quotient obtained when A037097(n) is considered as a GF(2)[X]-polynomial and it is divided by (x + 1) ^ A000225(n-1) (= A051179(n-2)).
+10
6
4, 8, 352, 3728, 7269662752, 761166466256046848, 390022035611646394530728097023856870592, 91600670557117582933643002658167825054614175029432880501373395030525438396928, 13417853484388319477475698658536993288839029124735549539652836318808118017743106800015257954250357092148394821846783842030516713870361254572407216621548672
Search completed in 0.008 seconds
| 