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A167614 a(n) = (n^2 + 3*n + 8)/2.

%I A167614 #53 Nov 15 2024 11:13:03
%S A167614 6,9,13,18,24,31,39,48,58,69,81,94,108,123,139,156,174,193,213,234,
%T A167614 256,279,303,328,354,381,409,438,468,499,531,564,598,633,669,706,744,
%U A167614 783,823,864,906,949,993,1038,1084,1131,1179,1228,1278,1329,1381,1434,1488,1543
%N A167614 a(n) = (n^2 + 3*n + 8)/2.
%H A167614 Vincenzo Librandi, <a href="/A167614/b167614.txt">Table of n, a(n) for n = 1..1000</a>
%H A167614 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A167614 a(n) = n + a(n-1) + 1, with n > 1, a(1)=6.
%F A167614 G.f.: x*(6 - 9*x + 4*x^2)/(1-x)^3. - _Vincenzo Librandi_, Sep 16 2013
%F A167614 A228446(a(n)) = 7. - _Reinhard Zumkeller_, Mar 12 2014
%F A167614 a(n) = A152950(n+2) = A152949(n+3) = A016028(n+5). - _Mathew Englander_, Feb 03 2022
%F A167614 From _Elmo R. Oliveira_, Nov 15 2024: (Start)
%F A167614 E.g.f.: exp(x)*(4 + 2*x + x^2/2) - 4.
%F A167614 a(n) = A027691(n+1)/2.
%F A167614 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 4. (End)
%t A167614 Table[(n(n+3))/2+4,{n,80}]  (* _Harvey P. Dale_, Mar 24 2011 *)
%t A167614 CoefficientList[Series[(6 - 9 x + 4 x^2)/(1 - x)^3,{x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 16 2013 *)
%o A167614 (PARI) a(n)=n*(n+3)/2+4 \\ _Charles R Greathouse IV_, Jan 11 2012
%o A167614 (Magma) [(n^2+3*n+8)/2: n in [1..60]]; // _Vincenzo Librandi_, Sep 16 2013
%o A167614 (Python)
%o A167614 print([n*(n+3)//2+4 for n in range(1, 60)]) # _Gennady Eremin_, Feb 03 2022
%Y A167614 Cf. A016028, A027691, A152949, A152950, A228446.
%K A167614 nonn,easy
%O A167614 1,1
%A A167614 _Vincenzo Librandi_, Nov 07 2009
%E A167614 Corrected (changed one term from 1036 to 1038) by _Harvey P. Dale_, Mar 24 2011
%E A167614 New name from _Charles R Greathouse IV_, Jan 11 2012