A006500 Restricted combinations.
1, 2, 4, 8, 12, 18, 27, 45, 75, 125, 200, 320, 512, 832, 1352, 2197, 3549, 5733, 9261, 14994, 24276, 39304, 63580, 102850, 166375, 269225, 435655, 704969, 1140624, 1845504, 2985984, 4831488, 7817616, 12649337, 20466953, 33116057, 53582633
Offset: 0
Examples
For example, a_4=12 and 12 subsets are: emptyset, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {2,3}, {2,4}, {3,4}, {1,2,3}, {2,3,4}. Corresponding compositions of 7=4+3 are: 1+1+1+1+1+1+1+1, 4+1+1+1, 1+4+1+1, 1+1+4+1, 1+1+1+4, 5+1+1, 4+2+1, 1+5+1, 1+4+2, 1+1+5, 6+1 and 1+6.
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=3.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Katharine A. Ahrens, Combinatorial Applications of the k-Fibonacci Numbers: A Cryptographically Motivated Analysis, Ph. D. thesis, North Carolina State University (2020).
- 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.
- G. E. Bergum and V. E. Hoggatt, Jr., A combinatorial problem involving recursive sequences and tridiagonal matrices, Fib. Quart., 16 (1978), 113-118.
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- 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,-1,1,1,1,-1,-1).
Programs
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Maple
A006500:=-(2*z**6+z**7-z**4+z**5-3*z**3-z**2-z-1)/(z**6-z**3-1)/(z**2+z-1); # Conjectured by Simon Plouffe in his 1992 dissertation.
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Mathematica
Table[Fibonacci[Floor[n/3] + 3]^Mod[n, 3] * Fibonacci[Floor[n/3] + 2]^(3 - Mod[n, 3]), {n, 0, 40}] (* David Nacin, Feb 29 2012 *) Table[Product[Fibonacci[Floor[(n + i)/3] + 2], {i, 0, 2}], {n, 0, 30}] (* David Nacin, Mar 07 2012 *) LinearRecurrence[{1, 1, -1, 1, 1, 1, -1, -1}, {1, 2, 4, 8, 12, 18, 27, 45}, 40] (* David Nacin, Mar 07 2012 *)
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Python
def a(n, adict={0:1, 1:2, 2:4, 3:8, 4:12, 5:18, 6:27, 7:45}): if n in adict: return adict[n] adict[n]=a(n-1)+a(n-2)-a(n-3)+a(n-4)+a(n-5)+a(n-6)-a(n-7)-a(n-8) return adict[n] # David Nacin, Mar 07 2012
Formula
Recurrence: a(n) = a(n-1)+a(n-2)-a(n-3)+a(n-4)+a(n-5)+a(n-6)-a(n-7)-a(n-8) G.f.: -(x^7+2*x^6+x^5-x^4-3*x^3-x^2-x-1)/(x^8+x^7-x^6-x^5-x^4+x^3-x^2-x+1). - Vladimir Baltic, Feb 17 2003
a(n) = F(floor(n/3) + 3)^(n mod 3)*F(floor(n/3) + 2)^(3 - (n mod 3)) where F(n) is the n-th Fibonacci number. - David Nacin, Feb 29 2012
Comments