A217787 a(n) = (a(n-1)*a(n-3) + 1) / a(n-4) with a(0) = a(1) = a(2) = a(3) = 1.
1, 1, 1, 1, 2, 3, 4, 9, 14, 19, 43, 67, 91, 206, 321, 436, 987, 1538, 2089, 4729, 7369, 10009, 22658, 35307, 47956, 108561, 169166, 229771, 520147, 810523, 1100899, 2492174, 3883449, 5274724, 11940723, 18606722, 25272721, 57211441, 89150161, 121088881
Offset: 0
Examples
G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 9*x^7 + 14*x^8 + 19*x^9 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..4411
- 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 (better) also. See Example 6.2.18.
- Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-1).
Programs
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Magma
[n le 3 select 1 else (Self(n)*Self(n-2)+1)/Self(n-3): n in [0..40]]; // Bruno Berselli, Mar 25 2013
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Mathematica
a[ n_] := With[{m = If [n < 0, 3 - n, n]}, SeriesCoefficient[ (1 + x + x^2 - 4 x^3 - 3 x^4 - 2 x^5) / (1 - 5 x^3 + x^6), {x, 0, m}]]; (* Michael Somos, Jan 18 2015 *) LinearRecurrence[{0,0,5,0,0,-1},{1,1,1,1,2,3},40] (* Harvey P. Dale, Nov 20 2016 *) -
PARI
{a(n) = if( n<0, n = 3-n); polcoeff( (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6) + x * O(x^n), n)};
Formula
G.f.: (1 + x + x^2 - 4*x^3 - 3*x^4 - 2*x^5) / (1 - 5*x^3 + x^6).
a(n) = a(3-n) for all n in Z.
a(n+3) + a(n-3) = 5*a(n) for all n in Z.
a(n+1) + a(n-1) = a(n) * (2 + [n mod 3 == 0]) for all n in Z.
a(n+3k)+a(n-3k) = A003501(k)*a(n) for n>=3k. - Bruno Berselli, Mar 25 2013
Comments