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A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3. 3
1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000, 1600, 2560, 4096, 6656, 10816, 17576, 28561, 46137, 74529, 120393, 194481, 314874, 509796, 825384, 1336336, 2161720, 3496900, 5656750, 9150625, 14807375, 23961025, 38773295, 62742241, 101515536 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Two consecutive Fibonacci numbers are coprime. This sequence satisfies a 14th-order linear difference equation. Note that it is the fourth sequence in the sequences that begin with the Fibonacci numbers, A006498, and A006500. Subsequent sequences will have orders 22, 32, and 44. - T. D. Noe, Mar 05 2012

Also the number of subsets of the set {1,2,...,n-1} which do not contain two elements whose difference is 4. - David Nacin, Mar 07 2012

REFERENCES

M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=4.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.

Index entries for linear recurrences with constant coefficients, signature (1,1,0,-2,2,2,0,2,-2,-2,0,1,-1,-1).

FORMULA

a(n) = F(floor((n-1)/4) + 3)^(n-1 mod 4)*F(floor((n-1)/4) + 2)^(4 - (n-1 mod 4)) where F(n) is the n-th Fibonacci number. - David Nacin, Mar 07 2012

a(n) = a(n-1) + a(n-2) - 2*a(n-4) + 2*a(n-5) + 2*a(n-6) + 2*a(n-8) - 2*a(n-9) - 2*a(n-10) + a(n-12) - a(n-13) - a(n-14). - David Nacin, Mar 07 2012

G.f.: x*(2 + 2*x + 2*x^2 + 4*x^3 + 4*x^4 - 2*x^6 - 1*x^7 - 4*x^8 - 3*x^9 - x^10 - x^11 - 2*x^12 - x^13)/((1 - x)*(1 + x)*(1 + x^2)*(1 - x - x^2)*(1 + 3*x^4 + x^8)). - David Nacin, Mar 08 2012

EXAMPLE

Since F_5=5 and F_6=8 are consecutive Fibonacci numbers, 8^4=4096, 8^3*5=2560, 8^2*5^2=1600, 8*5^3=1000, and 5^4=625 are in the sequence.

The number 3^3*8=216 is not in the sequence since 3 and 8 are not consecutive.

If n=6 then this gives the number of subsets of {1,...,5} not containing both 1 and 5.  There are 2^3 subsets containing 1 and 5, giving us 2^5 - 2^3 = 24.  Thus a(5) = 24. - David Nacin, Mar 07 2012

MATHEMATICA

f = Fibonacci[Range[12]]; m = Most[f]; r = Rest[f]; Union[m^4, m^3 r, m^2 r^2, m r^3] (* T. D. Noe, Mar 05 2012 *)

LinearRecurrence[{1, 1, 0, -2, 2, 2, 0, 2, -2, -2, 0, 1, -1, -1}, {1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000}, 40] (* T. D. Noe, Mar 05 2012 *)

Table[Fibonacci[Floor[n/4] + 3]^Mod[n, 4]*Fibonacci[Floor[n/4] + 2]^(4 - Mod[n, 4]), {n, 0, 40}] (* David Nacin, Mar 07 2012 *)

PROG

(PARI) for(m=2, 10, r=fibonacci(m); s=fibonacci(m+1); print(r^4, " ", r^3*s, " ", r^2*s^2, " ", r*s^3)) \\ Michael B. Porter, 04 Mar 2012

(Python)

def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:4, 6:15, 7:37, 8:87, 9:200}):

.if n in adict:

..return adict[n]

.adict[n]=3*a(n-1)-2*a(n-2)+2*a(n-3)-4*a(n-4)+2*a(n-5)-2*a(n-6)-4*a(n-7)-a(n-8)+a(n-9)+2*a(n-10)

.return adict[n] # David Nacin, Mar 07 2012

CROSSREFS

Cf. A000045, A006498, A006500.

Sequence in context: A005943 A008233 A224815 * A304076 A225549 A138278

Adjacent sequences:  A031920 A031921 A031922 * A031924 A031925 A031926

KEYWORD

nonn,easy

AUTHOR

Jeff Burch

EXTENSIONS

a(19) changed from 10416 to 10816 by David Nacin, Mar 04 2012

STATUS

approved

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Last modified August 19 07:51 EDT 2018. Contains 313857 sequences. (Running on oeis4.)