A243138 - OEIS

%I #53 Oct 18 2024 20:30:02

%S 13,29,47,67,89,113,139,167,197,229,263,299,337,377,419,463,509,557,

%T 607,659,713,769,827,887,949,1013,1079,1147,1217,1289,1363,1439,1517,

%U 1597,1679,1763,1849,1937,2027,2119,2213,2309,2407,2507,2609,2713,2819,2927,3037,3149

%N a(n) = n^2 + 15*n + 13.

%C From _Klaus Purath_, Dec 13 2022: (Start)

%C Numbers m such that 4*m + 173 is a square.

%C The product of two consecutive terms belongs to the sequence, a(n)*a(n+1) = a(a(n)+n).

%C 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)

%H Vincenzo Librandi, <a href="/A243138/b243138.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: (13 - 10*x - x^2)/(1 - x)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.

%F From _Klaus Purath_, Dec 13 2022: (Start)

%F a(n) = A119412(n+2) - 13.

%F a(n) = A132759(n+1) - 1.

%F a(n) = A098847(n+1) + n. (End)

%F Sum_{n>=0} 1/a(n) = tan(sqrt(173)*Pi/2)*Pi/sqrt(173) + 742077303/604626139. - _Amiram Eldar_, Feb 14 2023

%F E.g.f.: (13 + 16*x + x^2)*exp(x). - _Elmo R. Oliveira_, Oct 18 2024

%t Table[n^2 + 15 n + 13, {n, 0, 50}] (* or *) CoefficientList[Series[(13 - 10 x - x^2)/(1 - x)^3, {x, 0, 50}], x]

%t LinearRecurrence[{3,-3,1},{13,29,47},50] (* _Harvey P. Dale_, Sep 06 2020 *)

%o (Magma) [n^2+15*n+13: n in [0..50]];

%o (PARI) a(n)=n^2+15*n+13 \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. A098847, A119412, A126719, A132759, A153422.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Jun 02 2014