A283330 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=5.
1, 1, 1, 1, 1, 6, 16, 41, 106, 806, 2311, 6126, 16066, 122401, 351136, 931006, 2441881, 18604041, 53370241, 141506681, 371149801, 2827691726, 8111925376, 21508084401, 56412327826, 429790538206, 1232959286791, 3269087322166, 8574302679706, 65325334115481
Offset: 1
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..1838
- Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016; see also.
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,153,0,0,0,-153,0,0,0,1).
Programs
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Mathematica
a[n_] := a[n] = If[n <= 5, 1, With[{m = If[Mod[n, 4] == 2, 8, 3]}, m a[n-1] - a[n-2] - 1]]; Array[a, 30] (* Jean-François Alcover, Nov 03 2020 *) -
Ruby
def A(k, n) a = Array.new(k, 1) ary = [1] while ary.size < n j = (1..k - 1).inject(1){|s, i| s + a[-i]} + a[1] * a[-1] break if j % a[0] > 0 a = *a[1..-1], j / a[0] ary << a[0] end ary end def A283330(n) A(5, n) end # Seiichi Manyama, Mar 18 2017
Formula
From Seiichi Manyama, Mar 18 2017: (Start)
a(4*n-1) = 3*a(4*n-2) - a(4*n-3) - 1,
a(4*n) = 3*a(4*n-1) - a(4*n-2) - 1,
a(4*n+1) = 3*a(4*n) - a(4*n-1) - 1,
a(4*n+2) = 8*a(4*n+1) - a(4*n) - 1. (End)
From Colin Barker, Nov 03 2020: (Start)
G.f.: x*(1 + x + x^2 + x^3 - 152*x^4 - 147*x^5 - 137*x^6 - 112*x^7 + 106*x^8 + 41*x^9 + 16*x^10 + 6*x^11) / ((1 - x)*(1 + x)*(1 + x^2)*(1 - 152*x^4 + x^8)).
a(n) = 153*a(n-4) - 153*a(n-8) + a(n-12) for n>12.
(End)
Extensions
More terms from Seiichi Manyama, Mar 17 2017