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A000100
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a(n) is the number of compositions of n in which the maximal part is 3.
(Formerly M1394 N0543)
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5
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0, 0, 0, 1, 2, 5, 11, 23, 47, 94, 185, 360, 694, 1328, 2526, 4781, 9012, 16929, 31709, 59247, 110469, 205606, 382087, 709108, 1314512, 2434364, 4504352, 8328253, 15388362, 28417385, 52451811, 96771787, 178473023, 329042890, 606466009, 1117506500
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OFFSET
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0,5
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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G.f.: x^3/((1-x-x^2)*(1-x-x^2-x^3)).
a(n+3) = Sum_{k=0..n} F(k)*T(n-k), F(i)=A000045(i+1), T(i)=A000073(i+2).
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EXAMPLE
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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 + ...
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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)
(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 */
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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