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a(n) = n*(n + 3)/2 + 8.
5

%I #47 Oct 31 2024 18:39:34

%S 8,10,13,17,22,28,35,43,52,62,73,85,98,112,127,143,160,178,197,217,

%T 238,260,283,307,332,358,385,413,442,472,503,535,568,602,637,673,710,

%U 748,787,827,868,910,953,997,1042,1088,1135,1183,1232,1282,1333,1385,1438,1492

%N a(n) = n*(n + 3)/2 + 8.

%C 2*a(i) + 3 is prime for i = 0..14. - _Vincenzo Librandi_, Jun 01 2014

%C Numbers m >= 8 such that 8*m - 55 is a square. - _Bruce J. Nicholson_, Jul 26 2017

%H Vincenzo Librandi, <a href="/A167487/b167487.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 a(n) = n + a(n-1) + 1 with n > 1, a(1)=10.

%F G.f.: (8 - 14*x + 7*x^2)/(1 - x)^3. - _Vincenzo Librandi_, Sep 16 2013

%F a(n) = Sum_{i=n-5..n+7} i*(i+1)/26. - _Bruno Berselli_, Oct 20 2016

%F Sum_{n>=0} 1/a(n) = -1/7 + 2*Pi*tanh(sqrt(55)*Pi/2)/sqrt(55). - _Amiram Eldar_, Dec 13 2022

%F From _Elmo R. Oliveira_, Oct 31 2024: (Start)

%F E.g.f.: exp(x)*(8 + 2*x + x^2/2).

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

%t Table[n (n + 3)/2 + 8, {n, 0, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 03 2011 *)

%t CoefficientList[Series[(8 - 14 x + 7 x^2) / (1 - x)^3, {x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 16 2013 *)

%t LinearRecurrence[{3,-3,1},{8,10,13},60] (* _Harvey P. Dale_, Jul 05 2020 *)

%o (Magma) [n*(n+3)/2+8: n in [0..60]]; // _Vincenzo Librandi_, Sep 16 2013

%o (PARI) a(n)=n*(n+3)/2+8 \\ _Charles R Greathouse IV_, Jun 16 2017

%Y Cf. A167499.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 07 2009