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A166876
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a(n) = a(n-1) + Fibonacci(n), a(1)=1983.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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)).
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]
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LINKS
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FORMULA
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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
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EXAMPLE
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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]
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MATHEMATICA
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LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* G. C. Greubel, May 27 2016 *)
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CROSSREFS
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KEYWORD
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nonn,easy,less
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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