cp's OEIS Frontend

A306368 a(n) = numerator of (n + 3)*(n + 4)/((n + 1)*(n + 2)).

%I A306368 #23 Aug 11 2022 08:53:34
%S A306368 6,10,5,21,28,12,45,55,22,78,91,35,120,136,51,171,190,70,231,253,92,
%T A306368 300,325,117,378,406,145,465,496,176,561,595,210,666,703,247,780,820,
%U A306368 287,903,946,330,1035,1081,376,1176,1225,425,1326,1378,477,1485,1540,532,1653,1711,590
%N A306368 a(n) = numerator of (n + 3)*(n + 4)/((n + 1)*(n + 2)).
%C A306368 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).
%H A306368 Muniru A Asiru, <a href="/A306368/b306368.txt">Table of n, a(n) for n = 0..10000</a>
%H A306368 Peter Bala, <a href="/A306367/a306367.pdf">A note on the sequence of numerators of a rational function </a>.
%H A306368 Wikipedia, <a href="https://en.wikipedia.org/wiki/Quasi-polynomial">Quasi-polynomial</a>.
%H A306368 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-3,0,0,1).
%F A306368 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).
%F A306368 a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) for n >= 9.
%F A306368 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.
%F A306368 a(n) = (n + 3)*(n + 4)/gcd((n + 3)*(n + 4), (n + 1)*(n + 2)).
%F A306368 a(3*n)   = (3*n + 3)*(3*n + 4)/2 = A081266(n+1).
%F A306368 a(3*n+1) = (3*n + 4)*(3*n + 5)/2 = A060544(n+2).
%F A306368 a(3*n+2) = (n + 2)*(3*n + 5)/2   = A000326(n+2).
%F A306368 Sum_{n>=0} 1/a(n) = 2*log(3) - 2*Pi/(3*sqrt(3)). - _Amiram Eldar_, Aug 11 2022
%p A306368 seq((n + 3)*(n + 4)/gcd((n + 3)*(n + 4), (n + 1)*(n + 2)), n = 0..100);
%t A306368 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 *)
%o A306368 (GAP) List([0..100],n->NumeratorRat((n+3)*(n+4)/((n+1)*(n+2)))); # _Muniru A Asiru_, Feb 25 2019
%o A306368 (PARI) a(n) = numerator((n + 3)*(n + 4)/((n + 1)*(n + 2))); \\ _Michel Marcus_, Feb 26 2019
%Y A306368 Cf. A000326, A060544, A081266, A306367.
%K A306368 nonn,frac,easy
%O A306368 0,1
%A A306368 _Peter Bala_, Feb 14 2019