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a(n) = (n^2 + 3*n + 8)/2.
4

%I #49 Feb 10 2022 11:46:28

%S 6,9,13,18,24,31,39,48,58,69,81,94,108,123,139,156,174,193,213,234,

%T 256,279,303,328,354,381,409,438,468,499,531,564,598,633,669,706,744,

%U 783,823,864,906,949,993,1038,1084,1131,1179,1228,1278,1329,1381,1434,1488

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

%H Vincenzo Librandi, <a href="/A167614/b167614.txt">Table of n, a(n) for n = 1..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)=6.

%F G.f:: x*(6-9*x+4*x^2)/(1-x)^3. - _Vincenzo Librandi_, Sep 16 2013

%F A228446(a(n)) = 7. - _Reinhard Zumkeller_, Mar 12 2014

%F a(n) = A152950(n+2) = A152949(n+3) = A016028(n+5). - _Mathew Englander_, Feb 03 2022

%t Table[(n(n+3))/2+4,{n,80}] (* _Harvey P. Dale_, Mar 24 2011 *)

%t CoefficientList[Series[(6 - 9 x + 4 x^2)/(1 - x)^3,{x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 16 2013 *)

%o (PARI) a(n)=n*(n+3)/2+4 \\ _Charles R Greathouse IV_, Jan 11 2012

%o (Magma) [(n^2+3*n+8)/2: n in [1..60]]; // _Vincenzo Librandi_, Sep 16 2013

%o (Python)

%o print([n*(n+3)//2+4 for n in range(1, 60)]) # _Gennady Eremin_, Feb 03 2022

%Y Cf. A016028, A152949, A152950, A228446.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Nov 07 2009

%E Corrected (changed one term from 1036 to 1038) by _Harvey P. Dale_, Mar 24 2011

%E New name from _Charles R Greathouse IV_, Jan 11 2012