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 A026625 #21 Jun 17 2024 07:02:58 %S A026625 1,1,2,4,7,13,21,36,58,96,155,253,409,664,1074,1740,2815,4557,7373, %T A026625 11932,19306,31240,50547,81789,132337,214128,346466,560596,907063, %U A026625 1467661,2374725,3842388,6217114,10059504,16276619,26336125,42612745,68948872,111561618 %N A026625 a(n) = Sum_{k=0..floor(n/2)} A026615(n-k,k). %H A026625 Colin Barker, <a href="/A026625/b026625.txt">Table of n, a(n) for n = 0..1000</a> %H A026625 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-1,-1). %F A026625 For n>1, a(n) = Fibonacci(n-1) + Lucas(n) - (3 + (-1)^n)/2. - _Ralf Stephan_, May 13 2004 %F A026625 From _Colin Barker_, Jul 12 2017: (Start) %F A026625 G.f.: (1 - x^2 + x^3 + x^4 + x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)). %F A026625 a(n) = 2^(-1-n)*(-5*((-2)^n + 3*2^n) - (-15+sqrt(5))*(1+sqrt(5))^n + (1-sqrt(5))^n*(15+sqrt(5))) / 5 for n>1. %F A026625 a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>5. %F A026625 (End) %t A026625 Join[{1,1},Table[Fibonacci[n-1]+LucasL[n]-(3+(-1)^n)/2,{n,2,40}]] (* or *) Join[{1,1},LinearRecurrence[{1,2,-1,-1},{2,4,7,13},40]] (* _Harvey P. Dale_, Sep 27 2011 *) %o A026625 (PARI) Vec((1 - x^2 + x^3 + x^4 + x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)) + O(x^50)) \\ _Colin Barker_, Jul 12 2017 %o A026625 (Magma) [n eq 1 select 1 else 3*Fibonacci(n+1) - 2*Fibonacci(n) - (3+(-1)^n)/2: n in [0..40]]; // _G. C. Greubel_, Jun 16 2024 %o A026625 (SageMath) [3*fibonacci(n+1) -2*fibonacci(n) -(3+(-1)^n)//2 + int(n==1) for n in range(41)] # _G. C. Greubel_, Jun 16 2024 %Y A026625 Cf. A000032, A000045, A026615, A026616, A026617, A026618, A026619. %Y A026625 Cf. A026620, A026621, A026622, A026623, A026624, A026956, A026957. %Y A026625 Cf. A026958, A026959, A026960. %K A026625 nonn,easy %O A026625 0,3 %A A026625 _Clark Kimberling_