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A079935
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a(n) = 4a(n-1) - a(n-2).
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15
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1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| See A001835 for another version.
Greedy frac multiples of sqrt(3): a(1)=1, sum(n>0,frac(a(n)*x)) < 1 at x=sqrt(3).
The n-th greedy frac multiple of x is the smallest integer that does not cause sum(k=1..n,frac(a(k)*x)) to exceed unity; an infinite number of terms appear as the denominators of the convergents to the continued fraction of x.
In general, sum{k=0..n, binomial(2n-k,k)j^(n-k)}=(-1)^n*U(2n,I*sqrt(j)/2), I=sqrt(-1). - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005
The Hankel transform of this sequence is [1,2,0,0,0,0,0,0,0,0,0,...]. - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 21 2007
Contribution from Richard Choulet (richardchoulet(AT)yahoo.fr), May 09 2010: (Start)
This sequence is a particular case of the following situation:
a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n)
where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.
The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4);
so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).
The general form of a(n) is given by:
a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and
a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2))
(End)
x-values in the solution to 3*x^2-2=y^2. - Sture Sjöstedt, Nov 25 2011
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REFERENCES
| Jaime Rangel-Mondragon, Polyominoes and Related Families, The Mathematica Journal, 9:3 (2005), 609-640.
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LINKS
| Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
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FORMULA
| For n>0, a(n)= ceil( (2+sqrt(3))^n/(3+sqrt(3)) ).
G.f.: (1-x)/(1-4x+x^2); E.g.f.: exp(2x)(sinh(sqrt(3)x)/sqrt(3)+cosh(sqrt(3)x)); a(n)=(1/2+sqrt(3)/6)(2+sqrt(3))^n+(1/2-sqrt(3)/6)(2-sqrt(3))^n (offset 0). Binomial transform of A002605. - Paul Barry (pbarry(AT)wit.ie), Sep 17 2003
a(n)=sum{k=0..n, binomial(2n-k, k)2^(n-k)}; - Paul Barry (pbarry(AT)wit.ie), Jan 22 2005
a(n)=(-1)^n*U(2n, I*sqrt(2)/2), U(n, x) Chebyshev polynomial of second kind, I=sqrt(-1); - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005
a(n)=Jacobi_P(n,-1/2,1/2,2)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry (pbarry(AT)wit.ie), Feb 03 2006
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EXAMPLE
| a(4) = 41 since frac(1*x) + frac(3*x) + frac(11*x) + frac(41*x) < 1, while frac(1*x) + frac(3*x) + frac(11*x) + frac(k*x) > 1 for all k>11 and k<41.
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MATHEMATICA
| a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[1, 1]]; Table[ a[n], {n, 0, 23}]] (from Robert G. Wilson v Jan 13 2005)
LinearRecurrence[{4, -1}, {1, 3}, 30] (* or *) CoefficientList[Series[ (1-x)/(1-4x+x^2), {x, 0, 30}], x] (* From Harvey P. Dale, Apr 26 2011 *)
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PROG
| (Sage) [lucas_number1(n, 4, 1)-lucas_number1(n-1, 4, 1) for n in xrange(1, 25)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 29 2009]
(Haskell)
a079935 n = a079935_list !! (n-1)
a079935_list =
1 : 3 : zipWith (-) (map (4 *) $ tail a079935_list) a079935_list
-- Reinhard Zumkeller, Aug 14 2011
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CROSSREFS
| Cf. A002530 (denominators of convergents to sqrt(3)), A079934, A079936, A001353.
Cf. A001835 (same except for the first term).
Row 4 of array A094954.
Sequence in context: A077831 A032952 A001835 * A113437 A076540 A196472
Adjacent sequences: A079932 A079933 A079934 * A079936 A079937 A079938
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr) and Paul D. Hanna (pauldhanna(AT)juno.com), Jan 20 2003
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