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%I #48 Sep 08 2022 08:45:48
%S 6,8,11,15,20,26,33,41,50,60,71,83,96,110,125,141,158,176,195,215,236,
%T 258,281,305,330,356,383,411,440,470,501,533,566,600,635,671,708,746,
%U 785,825,866,908,951,995,1040,1086,1133,1181,1230,1280,1331,1383,1436,1490
%N a(n) = n*(n+3)/2 + 6.
%C Numbers m > 5 such that 8*m - 39 is a square. - _Bruce J. Nicholson_, Jul 25 2017
%H Vincenzo Librandi, <a href="/A167499/b167499.txt">Table of n, a(n) for n = 0..1000</a>
%H Ângela Mestre and José Agapito, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Mestre/mestre2.html">Square Matrices Generated by Sequences of Riordan Arrays</a>, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
%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) = 8.
%F G.f.: (6 - 10*x + 5*x^2)/(1-x)^3. - _Vincenzo Librandi_, Sep 16 2013
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Sep 16 2013
%F Sum_{n>=0} 1/a(n) = 2*Pi*tanh(Pi*sqrt(39)/2)/sqrt(39) - 1/5. - _Amiram Eldar_, Jan 17 2021
%p A167499:=n->n*(n+3)/2+6: seq(A167499(n), n=0..100); # _Wesley Ivan Hurt_, Jul 25 2017
%t Table[n*(n + 3)/2 + 6, {n, 0, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 03 2011 *)
%t CoefficientList[Series[(6 - 10 x + 5 x^2) / (1 - x)^3, {x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 16 2013 *)
%t LinearRecurrence[{3,-3,1},{6,8,11},60] (* _Harvey P. Dale_, Jun 21 2022 *)
%o (Magma) [n*(n+3)/2+6: n in [0..60]]; // _Vincenzo Librandi_, Sep 16 2013
%o (PARI) a(n)=n*(n+3)/2+6 \\ _Charles R Greathouse IV_, Jun 17 2017
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Nov 07 2009
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