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A006491
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Generalized Lucas numbers.
(Formerly M3258)
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4
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1, 0, 4, 5, 15, 28, 60, 117, 230, 440, 834, 1560, 2891, 5310, 9680, 17527, 31545, 56468, 100590, 178395, 315106, 554530, 972564, 1700400, 2964325, 5153868, 8938300, 15465497, 26700915, 46004620, 79112304, 135801105, 232715006, 398151740
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OFFSET
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1,3
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COMMENTS
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For n>2 note that (n+1)|a(n) unless n is prime, in which case (n+1)|2*a(n). This sequence is not the better-known generalized Lucas numbers V(n,a,b) defined for fixed integers a and b such that D = a^2 + 4*b is nonnegative, V(0) = 2, V(1) = a and for n>1 the recurrence V(n) = V(n-1) + V(n-2). The a = b = 1 case gives the Lucas Numbers. - Jonathan Vos Post, Mar 16 2005
Number of circular binary words of length n+1 having exactly two occurrences of 00. Example: a(4) = 5 because we have 00011, 10001, 11000, 00110 and 01100. Column 2 of A119458. - Emeric Deutsch, May 20 2006
In view of the comment by Emeric Deutsch above, we clarify the previous comment by Jonathan Vos Post. We have that 25 + 1 = 26 does not divide a(25) = 2964325 even though n = 25 is not a prime. What he probably meant is that, for n >= 1, we have (n+1) | a(n) unless n is odd, in which case (n+1)|2*a(n). (Of course, for some odd numbers n, we do have (n+1)|a(n), but not for all of them.)
From Emeric Deutsch's comment above, we have a(n) = A119458(n+1, k=2), while from the theory of marked and unmarked circular words, we have A320341(n+1, k=2) = (1/(n+1))*Sum_{d|gcd(n+1, 2)} phi(d)*A119458((n+1)/d, 2/d).
If n is even, then n+1 is odd, and hence A320341(n+1, k=2) = (1/(n+1)) * A119458(n+1, 2) = a(n)/(n+1), i.e., (n+1)|a(n).
If n=1, then (1+1)|2*a(1). Let n be odd >= 3, in which case n+1 is even and 2*A320341(n+1, k=2) = (1/(n+1))*(2*A119458(n+1, k=2) + 2*A119458((n+1)/2, k=1)). Thus, 2*A320341(n+1, k=2) = (1/(n+1))*(2*a(n) + 2*A006490((n+1)/2)) = (1/(n+1))*(2*a(n) + 2*((n+1)/2)*Fibonacci((n-3)/2)). It follows that 2*A320341(n+1, k=2) = 2*a(n)/(n+1) + Fibonacci((n-3)/2). Thus, 2*a(n)/(n+1) = 2*A320341(n+1, k=2) - Fibonacci((n-3)/2), and thus, (n+1)|2*a(n). (End)
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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G.f.: x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3. - Ralf Stephan, Apr 23 2004, corrected Feb 08 2006
a(n) = a(n-1) + a(n-2) + n*Fibonacci(n-2) - (n-1)*Fibonacci(n-3) for n >= 3; a(1)=1, a(2)=0. - Emeric Deutsch, May 20 2006
a(n) = 3*a(n-1) - 5*a(n-3) + 3*a(n-5) + a(n-6). - G. C. Greubel, Jan 01 2018
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MAPLE
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G:=x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3: Gser:=series(G, x=0, 45): seq(coeff(Gser, x^n), n=1..40); # Emeric Deutsch, Feb 07 2006
with(combinat): a[1]:=1: a[2]:=0: for n from 3 to 40 do a[n]:=a[n-1]+a[n-2]+n*fibonacci(n-2)-(n-1)*fibonacci(n-3) od: seq(a[n], n=1..40); # Emeric Deutsch, May 20 2006
A006491:=(z-1)*(1-2*z+2*z**2)/(z**2+z-1)**3; # conjectured by Simon Plouffe in his 1992 dissertation
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MATHEMATICA
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LinearRecurrence[{3, 0, -5, 0, 3, 1}, {1, 0, 4, 5, 15, 28}, 50] (* G. C. Greubel, Jan 01 2018 *)
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PROG
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(PARI) x='x+O('x^30); Vec(x*(1-x)*(1-2*x+2*x^2)/(1-x-x^2)^3) \\ G. C. Greubel, Jan 01 2018
(Magma) I:=[1, 0, 4, 5, 15, 28]; [n le 6 select I[n] else 3*Self(n-1) -5*Self(n-3) +3*Self(n-5)+Self(n-6): n in [1..30]]; // G. C. Greubel, Jan 01 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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