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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 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.)