OFFSET
0,1
COMMENTS
From Klaus Purath, Dec 13 2022: (Start)
Numbers m such that 4*m + 173 is a square.
The product of two consecutive terms belongs to the sequence, a(n)*a(n+1) = a(a(n)+n).
The prime terms in this sequence are listed in A153422. Each prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -15 (mod p). (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: (13 - 10*x - x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
From Klaus Purath, Dec 13 2022: (Start)
a(n) = A119412(n+2) - 13.
a(n) = A132759(n+1) - 1.
a(n) = A098847(n+1) + n. (End)
Sum_{n>=0} 1/a(n) = tan(sqrt(173)*Pi/2)*Pi/sqrt(173) + 742077303/604626139. - Amiram Eldar, Feb 14 2023
E.g.f.: (13 + 16*x + x^2)*exp(x). - Elmo R. Oliveira, Oct 18 2024
MATHEMATICA
Table[n^2 + 15 n + 13, {n, 0, 50}] (* or *) CoefficientList[Series[(13 - 10 x - x^2)/(1 - x)^3, {x, 0, 50}], x]
LinearRecurrence[{3, -3, 1}, {13, 29, 47}, 50] (* Harvey P. Dale, Sep 06 2020 *)
PROG
(Magma) [n^2+15*n+13: n in [0..50]];
(PARI) a(n)=n^2+15*n+13 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jun 02 2014
STATUS
approved