cp's OEIS Frontend

A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.

%I A031923 #78 Sep 11 2024 00:39:58
%S A031923 1,2,4,8,16,24,36,54,81,135,225,375,625,1000,1600,2560,4096,6656,
%T A031923 10816,17576,28561,46137,74529,120393,194481,314874,509796,825384,
%U A031923 1336336,2161720,3496900,5656750,9150625,14807375,23961025,38773295,62742241,101515536
%N A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.
%C A031923 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
%C A031923 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
%H A031923 Vincenzo Librandi, <a href="/A031923/b031923.txt">Table of n, a(n) for n = 1..1000</a>
%H A031923 Michael A. Allen, <a href="https://arxiv.org/abs/2209.01377">On a Two-Parameter Family of Generalizations of Pascal's Triangle</a>, arXiv:2209.01377 [math.CO], 2022.
%H A031923 Michael A. Allen, <a href="https://arxiv.org/abs/2409.00624">Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings</a>, arXiv:2409.00624 [math.CO], 2024. See p. 16.
%H A031923 M. El-Mikkawy and T. Sogabe, <a href="https://doi.org/10.1016/j.amc.2009.12.069">A new family of k-Fibonacci numbers</a>, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=4.
%H A031923 M. Tetiva, <a href="http://www.jstor.org/stable/10.4169/math.mag.84.4.296">Subsets that make no difference d</a>, Mathematics Magazine 84 (2011), no. 4, 300-301.
%H A031923 <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,-2,2,2,0,2,-2,-2,0,1,-1,-1).
%F A031923 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
%F A031923 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
%F A031923 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
%F A031923 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
%F A031923 a(4n+1)= A056571(n+2). a(4n+3)=A197424(n). - _R. J. Mathar_, Jan 23 2022
%e A031923 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.
%e A031923 The number 3^3*8 = 216 is not in the sequence since 3 and 8 are not consecutive.
%e A031923 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
%p A031923 A031923 := proc(n)
%p A031923     local n0,i,r,s,m ;
%p A031923     n0 := n-1 ;
%p A031923     i := floor(n0/4) ;
%p A031923     r := combinat[fibonacci](i+2) ;
%p A031923     s := combinat[fibonacci](i+3) ;
%p A031923     m := modp(n0,4) ;
%p A031923     r^(4-m)*s^m ;
%p A031923 end proc:
%p A031923 seq(A031923(n),n=1..50) ; # _R. J. Mathar_, Jan 23 2022
%t A031923 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 *)
%t A031923 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 *)
%t A031923 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 *)
%t A031923 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 *)
%o A031923 (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
%o A031923 (Python)
%o A031923 def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:4, 6:15, 7:37, 8:87, 9:200}):
%o A031923     if n in adict:
%o A031923         return adict[n]
%o A031923     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)
%o A031923     return adict[n] # _David Nacin_, Mar 07 2012
%Y A031923 Cf. A000045, A006498, A006500.
%K A031923 nonn,easy
%O A031923 1,2
%A A031923 _Jeff Burch_
%E A031923 a(19) changed from 10416 to 10816 by _David Nacin_, Mar 04 2012