A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^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
Offset: 1
Examples
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
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
- Michael A. Allen, Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings, arXiv:2409.00624 [math.CO], 2024. See p. 16.
- 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).
Programs
-
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 *)
-
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
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
Extensions
a(19) changed from 10416 to 10816 by David Nacin, Mar 04 2012
Comments