cp's OEIS Frontend

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.

A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

Original entry on oeis.org

1983, 1984, 1986, 1989, 1994, 2002, 2015, 2036, 2070, 2125, 2214, 2358, 2591, 2968, 3578, 4565, 6162, 8746, 12927, 19692, 30638, 48349, 77006, 123374, 198399, 319792, 516210, 834021, 1348250, 2180290, 3526559, 5704868, 9229446, 14932333, 24159798
Offset: 1

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Author

Geoff Ahiakwo, Oct 22 2009

Keywords

Comments

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).
The generating functions are x*( s+(1-s)*x+(1-s)*x^2 )/((1-x) * (1-x-x^2)).
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.
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.
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]

Examples

			For s=1983, n=3, we have k= A166863(2)= 5, a(3) = (2s+1+k)/2 = (2*1983+1+5)/2 = 1986.
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]
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* G. C. Greubel, May 27 2016 *)

Formula

a(n) = 2*a(n-1) - a(n-3).
G.f.: x*(-1983 + 1982*x + 1982*x^2)/((1-x)*(x^2+x-1)).
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

Extensions

Definition and comments edited by R. J. Mathar, Oct 26 2009