OFFSET
0,1
COMMENTS
If P(x) and Q(x) are coprime integral polynomials such that Q(n) > 0 for n >= 0 then the sequence of numerators of the rational numbers P(n)/Q(n) for n >= 0 and the sequence of denominators of P(n)/Q(n) for n >= 0 are both quasi-polynomial in n. In fact, there exists a purely periodic sequence b(n) such that numerator(P(n)/Q(n)) = P(n)/b(n) and denominator(P(n)/Q(n)) = Q(n)/b(n). Here we take P(n) = (n + 3)*(n + 4) and Q(n) = (n + 1)*(n + 2).
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..10000
Wikipedia, Quasi-polynomial.
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).
FORMULA
O.g.f.: (x^8 + x^7 - 3*x^5 - 2*x^4 + 3*x^3 + 5*x^2 + 10*x + 6)/((1 - x)^3*(x^2 + x + 1)^3).
a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) for n >= 9.
a(n) = (n + 3)*(n + 4)/b(n), where (b(n))n>=0 is the purely periodic sequence [2, 2, 6, 2, 2, 6, ...] with period 3.
a(n) = (n + 3)*(n + 4)/gcd((n + 3)*(n + 4), (n + 1)*(n + 2)).
a(3*n) = (3*n + 3)*(3*n + 4)/2 = A081266(n+1).
a(3*n+1) = (3*n + 4)*(3*n + 5)/2 = A060544(n+2).
a(3*n+2) = (n + 2)*(3*n + 5)/2 = A000326(n+2).
Sum_{n>=0} 1/a(n) = 2*log(3) - 2*Pi/(3*sqrt(3)). - Amiram Eldar, Aug 11 2022
MAPLE
seq((n + 3)*(n + 4)/gcd((n + 3)*(n + 4), (n + 1)*(n + 2)), n = 0..100);
MATHEMATICA
Table[((n+3)(n+4))/((n+1)(n+2)), {n, 0, 60}]//Numerator (* or *) LinearRecurrence[{0, 0, 3, 0, 0, -3, 0, 0, 1}, {6, 10, 5, 21, 28, 12, 45, 55, 22}, 60] (* Harvey P. Dale, Mar 28 2020 *)
PROG
(GAP) List([0..100], n->NumeratorRat((n+3)*(n+4)/((n+1)*(n+2)))); # Muniru A Asiru, Feb 25 2019
(PARI) a(n) = numerator((n + 3)*(n + 4)/((n + 1)*(n + 2))); \\ Michel Marcus, Feb 26 2019
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Peter Bala, Feb 14 2019
STATUS
approved