login
a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=4.
3

%I #38 Nov 06 2020 08:47:27

%S 1,1,1,1,5,13,33,217,617,1633,10813,30805,81601,540401,1539601,

%T 4078401,27009205,76949213,203838433,1349919817,3845921017,

%U 10187843233,67468981613,192219101605,509188323201,3372099160801,9607109159201,25449228316801,168537489058405

%N a(n) = (1 + Sum_{j=1..K-1} a(n-j) + a(n-1)*a(n-K+1))/a(n-K) with a(1),...,a(K)=1, where K=4.

%H Seiichi Manyama, <a href="/A283329/b283329.txt">Table of n, a(n) for n = 1..1770</a>

%H Matthew Christopher Russell, <a href="http://www.math.rutgers.edu/~zeilberg/Theses/MatthewRussellThesis.pdf">Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences</a>, PhD Dissertation, Mathematics Department, Rutgers University, May 2016; see <a href="https://pdfs.semanticscholar.org/fdeb/e20954dacb7ec7a24afe2cf491b951c5a28d.pdf">also</a>.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,51,0,0,-51,0,0,1).

%F From _Seiichi Manyama_, Mar 18 2017: (Start)

%F a(3*n) = 3*a(3*n-1) - a(3*n-2) - 1,

%F a(3*n+1) = 3*a(3*n) - a(3*n-1) - 1,

%F a(3*n+2) = 7*a(3*n+1) - a(3*n) - 1. (End)

%F From _Colin Barker_, Nov 03 2020: (Start)

%F G.f.: x*(1 + x + x^2 - 50*x^3 - 46*x^4 - 38*x^5 + 33*x^6 + 13*x^7 + 5*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 50*x^3 + x^6)).

%F a(n) = 51*a(n-3) - 51*a(n-6) + a(n-9).

%F (End)

%o (Ruby)

%o def A(k, n)

%o a = Array.new(k, 1)

%o ary = [1]

%o while ary.size < n

%o j = (1..k - 1).inject(1){|s, i| s + a[-i]} + a[1] * a[-1]

%o break if j % a[0] > 0

%o a = *a[1..-1], j / a[0]

%o ary << a[0]

%o end

%o ary

%o end

%o def A283329(n)

%o A(4, n)

%o end # _Seiichi Manyama_, Mar 18 2017

%Y Cf. A276123, A283330.

%K nonn

%O 1,5

%A _N. J. A. Sloane_, Mar 17 2017

%E More terms from _Seiichi Manyama_, Mar 17 2017