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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. 7
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
LINKS
Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
M. El-Mikkawy and 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.
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
a(4*k-3) = F(k+1)^4, a(4*k-2) = F(k+1)^3*F(k+2), a(4*k-1) = F(k+1)^2*F(k+2)^2, a(4*k) = F(k+1)*F(k+2)^3, k >= 1, where F = A000045. - Jianing Song, Feb 06 2019
a(4n+1)= A056571(n+2). a(4n+3)=A197424(n). - R. J. Mathar, Jan 23 2022
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
MAPLE
A031923 := proc(n)
local n0, i, r, s, m ;
n0 := n-1 ;
i := floor(n0/4) ;
r := combinat[fibonacci](i+2) ;
s := combinat[fibonacci](i+3) ;
m := modp(n0, 4) ;
r^(4-m)*s^m ;
end proc:
seq(A031923(n), n=1..50) ; # R. J. Mathar, Jan 23 2022
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 *)
cfn[{a_, b_}]:={a^4, a^3 b, a^2 b^2, a b^3}; Flatten[cfn/@Partition[ Fibonacci[ Range[20]], 2, 1]]//Union (* Harvey P. Dale, Feb 03 2019 *)
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, Mar 04 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
Sequence in context: A330131 A008233 A224815 * A304076 A225549 A138278
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
a(19) changed from 10416 to 10816 by David Nacin, Mar 04 2012
STATUS
approved

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Last modified March 19 06:32 EDT 2024. Contains 370953 sequences. (Running on oeis4.)