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 A166876 #15 May 27 2016 18:07:40
%S A166876 1983,1984,1986,1989,1994,2002,2015,2036,2070,2125,2214,2358,2591,
%T A166876 2968,3578,4565,6162,8746,12927,19692,30638,48349,77006,123374,198399,
%U A166876 319792,516210,834021,1348250,2180290,3526559,5704868,9229446,14932333,24159798
%N A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.
%C A166876 Starting at some a(1)=s and creating further terms with the recurrence a(n)=a(n-1)+A000045(n) defines a family of sequences with recurrences a(n)= 2*a(n-1) -a(n-3).
%C A166876 The generating functions are x*( s+(1-s)*x+(1-s)*x^2 )/((1-x) * (1-x-x^2)).
%C A166876 The terms are a(n) = A000045(n+2)+s-2 = s + A001911(n-1) = (2*s+1+k)/2 where k=A166863(n-1), n>=1.
%C A166876 Examples: Up to offsets, s=1 yields A000071, s=2 yields A000045 shifted left thrice, s=3 yields A001611 shifted left thrice, s=4 yields A018910.
%C A166876 I appreciate the editing by R. J. Mathar. However I would like further analysis of the following formula. The sequence which I call GAP can have any integer as its first term, not just 1983. Thus a(1) can be 0, 1, 2, 3,... Then a(2) is always a(1)+ 1, while a(3) is a(1) + k(n)/2; where k(n) = k(n-2)+ k(n-1)+4 (This is a separate sequence submitted for consideration). [_Geoff Ahiakwo_, Nov 19 2009]
%H A166876 G. C. Greubel, <a href="/A166876/b166876.txt">Table of n, a(n) for n = 1..1000</a>
%H A166876 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1)
%F A166876 a(n) = 2*a(n-1) - a(n-3).
%F A166876 G.f.: x*(-1983 + 1982*x + 1982*x^2)/((1-x)*(x^2+x-1)).
%F A166876 Let a(n)= a(1)+ k(n)/2, then G.f.: k(n)= k(n-2)+ k(n-1) + 4. - _Geoff Ahiakwo_, Nov 19 2009
%e A166876 For s=1983, n=3, we have k= A166863(2)= 5, a(3) = (2s+1+k)/2 = (2*1983+1+5)/2 = 1986.
%e A166876 For n=3, a(3)= a(1)+ k(3)/2 = a(1)+ [K(3-2)+ k(3-1)]/2 + 2 = a(1)+ 1 + 2 thus if a(1)is 0, a(3)= 3; if a(1)= 5, a(3)= 8; if a(1)=1983, a(3)= 1986, etc. [_Geoff Ahiakwo_, Nov 19 2009]
%t A166876 LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* _G. C. Greubel_, May 27 2016 *)
%Y A166876 Cf. A001911, A000071
%K A166876 nonn,easy,less
%O A166876 1,1
%A A166876 _Geoff Ahiakwo_, Oct 22 2009
%E A166876 Definition and comments edited by _R. J. Mathar_, Oct 26 2009