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 A027005 #17 Jul 25 2025 19:00:23 %S A027005 1,19,101,370,1148,3278,8967,23993,63483,167040,438346,1148844, %T A027005 3009181,7879855,20631713,54016798,141420392,370246298,969320643, %U A027005 2537718005,6643835991,17393792844,45537545686,119218847640,312119000953,817138159243,2139295481117 %N A027005 a(n) = T(2*n+1,n+2), T given by A026998. %C A027005 Bisection of A027963. %H A027005 Colin Barker, <a href="/A027005/b027005.txt">Table of n, a(n) for n = 1..1000</a> %H A027005 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (6,-13,13,-6,1). %F A027005 From _Colin Barker_, Feb 19 2016: (Start) %F A027005 a(n) = 2^(-1-n)*((25+11*sqrt(5))*(3+sqrt(5))^n - (25-11*sqrt(5))*(3-sqrt(5))^n )/sqrt(5) + 7*(1+n) - 6*(n+1)*(n+2) + 7*(n+1) - 6. %F A027005 a(n) = 6*a(n-1) - 13*a(n-2) + 13*a(n-3) - 6*a(n-4) + a(n-5) for n>5. %F A027005 G.f.: x*(1+13*x-2*x^3) / ((1-x)^3*(1-3*x+x^2)). (End) %F A027005 From _G. C. Greubel_, Jul 21 2025: (Start) %F A027005 a(n) = Lucas(2*n+5) - (6*n^2 + 11*n + 11). %F A027005 E.g.f.: exp(3*x/2)*(11*cosh(p*x) + 10*p*sinh(p*x)) - (11 + 17*x + 6*x^2)*exp(x), where 2*p = sqrt(5). (End) %t A027005 LinearRecurrence[{6,-13,13,-6,1},{1,19,101,370,1148},30] (* _Harvey P. Dale_, Aug 19 2020 *) %o A027005 (PARI) Vec(x*(1+13*x-2*x^3)/((1-x)^3*(1-3*x+x^2)) + O(x^40)) \\ _Colin Barker_, Feb 19 2016 %o A027005 (Magma) %o A027005 A027005:= func< n | Lucas(2*n+5) -(6*n^2+11*n+11) >; %o A027005 [A027005(n): n in [1..40]]; // _G. C. Greubel_, Jul 21 2025 %o A027005 (SageMath) %o A027005 def A027005(n): return lucas_number2(2*n+5,1,-1) -(6*n**2 +11*n +11) %o A027005 print([A027005(n) for n in range(1,41)]) # _G. C. Greubel_, Jul 21 2025 %Y A027005 Cf. A000032, A026998, A027963. %K A027005 nonn,easy %O A027005 1,2 %A A027005 _Clark Kimberling_