OFFSET
0,5
COMMENTS
For n > 5, a(n) - (a(n-3)+a(n-2)+a(n-1)) = F(n-2) where F(i) is the i-th Fibonacci number; e.g., 11 - (1+2+5) = 3, 23 - (2+5+11) = 8; also lim_{n->oo} a(n)/(a(n-1)+a(n-2)+a(n-3)) = 1 and lim_{n->oo} a(n)*a(n-2)/a(n-1)^2 = 1. - Gerald McGarvey, Jun 26 2004
a(n) is also the number of binary sequences of length n-1 in which the longest run of 0's is exactly two. - Geoffrey Critzer, Nov 06 2008
a(n) is also the difference between the n-th tribonacci number and the n-th Fibonacci number; i.e., a(n) = A000073(n) - A000045(n). - Gregory L. Simay, Jan 31 2018
Let F_0(n) be the n-th Fibonacci number, A000045(n). Let F_1(n) = Sum_{j=1..n} A000045(n+1-j)*A000045(j). Let F_r(n) = Sum_{j=1..n} F_(r-1)(n+1-j)*A000045(j). Then the number of compositions of n having exactly r 3's as the highest part is F_r(n), and a(n+1) = F_1(n-3) + F_1(n-6) + ... - Gregory L. Simay, Apr 17 2018
The Apr 17 2018 comment can be generalized. Let F(n,k) be the n-th k-step Fibonacci number, with the convention that F(0,k)=0 and F(1,k)=1. Let F(n,k,0)= F(n,k) Let F(n, k, 1) = Sum_{j=1..n} F(n+1-j,k)*F(j,k). Let F(n,k,r) = Sum_{j=1..n} F(n+1-j, k, r-1) * A000045(j, k). Let G(n,k,r) be the number of compositions of n having k as the largest part exactly r times. Then G(n,k,r) = F(n+1 - kr, k-1, r). - Gregory L. Simay, May 17 2018
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29.
LINKS
T. D. Noe, Table of n, a(n) for n = 0..200
Nick Hobson, Python program for this sequence.
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29. (Annotated scanned copy)
Index entries for linear recurrences with constant coefficients, signature (2,1,-1,-2,-1).
FORMULA
G.f.: x^3/((1-x-x^2)*(1-x-x^2-x^3)).
a(n) = 2*a(n-1)+a(n-2)-a(n-3)-2*a(n-4)-a(n-5). Convolution of Fibonacci and tribonacci numbers (A000045 and A000073). - Franklin T. Adams-Watters, Jan 13 2006
EXAMPLE
For example, a(5)=5 counts 1+1+3, 2+3, 3+2, 3+1+1, 1+3+1. - David Callan, Dec 09 2004
a(5)=5 because there are 5 binary sequences of length 4 in which the longest run of consecutive 0's is exactly two: 0010, 0011, 0100, 1001, 1100. - Geoffrey Critzer, Nov 06 2008
G.f.: x^3 + 2*x^4 + 5*x^5 + 11*x^6 + 23*x^7 + 47*x^8 + 94*x^9 + 185*x^10 + 360*x^11 + ...
MAPLE
a:= n -> (Matrix(5, (i, j)-> if (i=j-1) then 1 elif j=1 then [2, 1, -1, -2, -1][i] else 0 fi)^(n))[1, 4]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 04 2008
MATHEMATICA
a[n_] := a[n] = a[n-1] + a[n-2] + a[n-3] + Fibonacci[n-2]; a[n_ /; n < 3] = 0; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Aug 03 2012, after Gerald McGarvey *)
a[ n_] := SeriesCoefficient[ If[ n > 0, x^3 / ((1 - x - x^2) (1 - x - x^2 - x^3)), -x^2 / ((1 + x - x^2) (1 + x + x^2 - x^3))], {x, 0, Abs@n}]; (* Michael Somos, Jun 01 2013 *)
LinearRecurrence[{2, 1, -1, -2, -1}, {0, 0, 0, 1, 2}, 40] (* Harvey P. Dale, Jul 22 2013 *)
PROG
(Haskell)
a000100 n = a000100_list !! (n-1)
a000100_list = f (tail a000045_list) [head a000045_list] where
f (x:xs) ys = (sum $ zipWith (*) ys a000073_list) : f xs (x:ys)
-- Reinhard Zumkeller, Jul 31 2012
(PARI) {a(n) = polcoeff( if( n>0, x^3 / ((1 - x - x^2) * (1 - x - x^2 - x^3)), -x^2 / ((1 + x - x^2) * (1 + x + x^2 - x^3))) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Jun 01 2013 */
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
More terms from Henry Bottomley, Dec 15 2000
Better definition from David Callan and Franklin T. Adams-Watters
STATUS
approved