This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A110325 #27 Nov 02 2024 14:58:15 %S A110325 1,0,-5,-14,-27,-44,-65,-90,-119,-152,-189,-230,-275,-324,-377,-434, %T A110325 -495,-560,-629,-702,-779,-860,-945,-1034,-1127,-1224,-1325,-1430, %U A110325 -1539,-1652,-1769,-1890,-2015,-2144,-2277,-2414,-2555,-2700,-2849,-3002,-3159,-3320,-3485,-3654,-3827,-4004,-4185,-4370 %N A110325 Row sums of number triangle related to the Jacobsthal numbers. %C A110325 Essentially the same sequence as A014106. %C A110325 Rows sums of A110324. Results from a general construction: the row sums of the inverse of the number triangle whose columns have e.g.f. (x^k/k!)/(1 - a*x - b*x^2), g.f. (1 - (a+2)*x - (2*b-a-1)*x^2)/(1-x)^3, and general term 1 + (b-a)*n - b*n^2. This is the binomial transform of (1, -a, -2b, 0, 0, 0, ...). %H A110325 Vincenzo Librandi, <a href="/A110325/b110325.txt">Table of n, a(n) for n = 0..1000</a> %H A110325 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1). %F A110325 a(n) = 1 + n - 2*n^2. %F A110325 G.f.: (1 - 3*x - 2*x^2)/(1-x)^3. %F A110325 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Jul 08 2012 %F A110325 From _Elmo R. Oliveira_, Nov 02 2024: (Start) %F A110325 E.g.f.: exp(x)*(1 - x - 2*x^2). %F A110325 a(n) = -A005408(n)*A110325(n). (End) %t A110325 CoefficientList[Series[(1-3x-2x^2)/(1-x)^3,{x,0,50}],x] (* _Vincenzo Librandi_, Jul 08 2012 *) %t A110325 LinearRecurrence[{3,-3,1},{1,0,-5},50] (* _Harvey P. Dale_, Oct 20 2024 *) %o A110325 (Magma) [1+n-2*n^2: n in [0..50]]; // _Vincenzo Librandi_, Jul 08 2012 %o A110325 (PARI) a(n)=1+n-2*n^2 \\ _Charles R Greathouse IV_, Jun 17 2017 %Y A110325 Cf. A014106 (essentially the same sequence), A110324. %Y A110325 Cf. A005408, A110325. %K A110325 easy,sign %O A110325 0,3 %A A110325 _Paul Barry_, Jul 20 2005