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

6, 10, 5, 21, 28, 12, 45, 55, 22, 78, 91, 35, 120, 136, 51, 171, 190, 70, 231, 253, 92, 300, 325, 117, 378, 406, 145, 465, 496, 176, 561, 595, 210, 666, 703, 247, 780, 820, 287, 903, 946, 330, 1035, 1081, 376, 1176, 1225, 425, 1326, 1378, 477, 1485, 1540, 532, 1653, 1711, 590

Offset: 0

Author

Peter Bala, Feb 14 2019

Keywords

• nonn
• frac
• easy

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
• Peter Bala, A note on the sequence of numerators of a rational function .
• Wikipedia, Quasi-polynomial.
• Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Crossrefs

Cf. A000326, A060544, A081266, A306367.

Programs

GAP

List([0..100],n->NumeratorRat((n+3)*(n+4)/((n+1)*(n+2)))); # Muniru A Asiru, Feb 25 2019

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 *)

PARI

a(n) = numerator((n + 3)*(n + 4)/((n + 1)*(n + 2))); \\ Michel Marcus, Feb 26 2019

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