A006491 Generalized Lucas numbers.

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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

From Petros Hadjicostas, Jan 10 2019: (Start)

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)

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
• Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).

Crossrefs

Cf. A006490, A119458, A320341.

Programs

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

Maple

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

Mathematica

LinearRecurrence[{3, 0, -5, 0, 3, 1}, {1, 0, 4, 5, 15, 28}, 50] (* G. C. Greubel, Jan 01 2018 *)

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

Formula

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

Extensions

More terms from Emeric Deutsch, Feb 07 2006