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A005246
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a(n)=(1+a(n-1)a(n-2))/a(n-3).
(Formerly M0829)
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10
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1, 1, 1, 2, 3, 7, 11, 26, 41, 97, 153, 362, 571, 1351, 2131, 5042, 7953, 18817, 29681, 70226, 110771, 262087, 413403, 978122, 1542841, 3650401, 5757961, 13623482, 21489003, 50843527, 80198051, 189750626, 299303201, 708158977, 1117014753
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| For n >= 4 we have the linear recurrence a(n) = 4*a(n-2) - a(n-4). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 04 2001
Integer solutions to the equation floor(sqrt(3)*x^2)=x*floor(sqrt(3)*x) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 18 2004
For n>2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). i.e. a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n))<frac(sqrt(3)*a(n-1)). - Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 20 2003
The lower principal and intermediate convergents to 3^(1/2), beginning with 1/1, 3/2, 5/3, 12/7, 19/11, form a strictly increasing sequence; essentially, numerators=A143643 and denominators=A005246. - Clark Kimberling (ck6(AT)evansville.edu), Aug 27 2008
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)) [From Richard Choulet (richardchoulet(AT)yahoo.fr), Feb 24 2010]
In the closed form formula (sqrt(2+sqrt(3))^n)=((sqrt(6)+sqrt(2))/2)^n;(-sqrt(2+sqrt(3))^n)=((-sqrt(6)-sqrt(2))/2)^n;(sqrt(2-sqrt(3))^n)=((sqrt(6)-sqrt(2))/2)^n ;(-sqrt(2-sqrt(3))^n)=((sqrt(2)-sqrt(6))/2)^n. - Tim Monahan, Jul 07 2011
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REFERENCES
| T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
Clark Kimberling, "Best lower and upper approximates to irrational numbers," Elemente der Mathematik, 52 (1997) 122-126.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..500
Enrica Duchi, Andrea Frosini, Renzo Pinzani and Simone Rinaldi, A Note on Rational Succession Rules, J. Integer Seqs., Vol. 6, 2003.
S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
Index to sequences with linear recurrences with constant coefficients, signature (0,4,0,-1).
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FORMULA
| G.f.: (1+x-3*x^2-2*x^3)/(1-4*x^2+x^4).
lim n ->infinity a(2n+1)/a(2n) = (3+sqrt(3))/3 =1.5773502... lim n ->infinity a(2n)/a(2n-1) = (3+sqrt(3))/2 = 2.3660254.... - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 07 2002
A101265(n) = a(n)*a(n+1). - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Apr 24 2006
a(2-n)=a(n). - Michael Somos Nov 15 2006
For n>2: a(n) = a(n-1) + SUM(a(2*k): 1 <= k < n/2). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 16 2007
a(2*m)=sum((-1)^p*binomial(m-p,p)*4^(m-2*p),p=0..floor(m/2))-3*sum((-1)^p*binomial(m-1-p,p)*4^(m-1-2*p),p=0..floor((m-1)/2)). a(2*m+1)=sum((-1)^p*binomial(m-p,p)*4^(m-2*p),p=0..floor(m/2))-2*sum((-1)^p*binomial(m-1-p,p)*4^(m-1-2*p),p=0..floor((m-1)/2)) [From Richard Choulet (richardchoulet(AT)yahoo.fr), Feb 24 2010]
Contribution by Tim Monahan, Jul 01 2011: (Start)
Closed form without extra leading 1 ((sqrt(6)+3)*(sqrt(2+sqrt(3))^n+(sqrt(2-sqrt(3))^n))+(3-sqrt(6))*(-sqrt(2+sqrt(3))^n+(-sqrt(2-sqrt(3))^n)))/12.
Closed form with extra leading 1 ((6+3*sqrt(6)-2*sqrt(3)-3*sqrt(2))*(sqrt(2+sqrt(3))^n)+(6+3*sqrt(6)+2*sqrt(3)+3*sqrt(2))*(sqrt(2-sqrt(3))^n)+(6-3*sqrt(6)-2*sqrt(3)+3*sqrt(2))*(-sqrt(2+sqrt(3))^n)+(6-3*sqrt(6)+2*sqrt(3)-3*sqrt(2))*(-sqrt(2-sqrt(3))^n))/24. (End)
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EXAMPLE
| a(4)=4^2-4^0-3*4^1=3. a(7)=4^3-4*binomial(2,1)-2*(4^2-1)=26. [From Richard Choulet (richardchoulet(AT)yahoo.fr), Feb 24 2010]
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MAPLE
| A005246:=-(-1-z+2*z**2+z**3)/(1-4*z**2+z**4); [Conjectured by S. Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.]
for q from 1 to 10 do :a:=1:b:=1:Q:=(a*b^2+q*b+a+q)/(a*b): for m from 0 to 15 do U(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)):od: for m from 0 to 15 do V(m):=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)):od:for m from 0 to 15 do W(2*m):=U(m):od:for m from 0 to 14 do W(2*m+1):=V(m):od:seq(W(m), m=0..30):od; [From Richard Choulet (richardchoulet(AT)yahoo.fr), Feb 24 2010]
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PROG
| (PARI) {a(n)=if(n<0, n=2-n); polcoeff((1+x-3*x^2-2*x^3)/(1-4*x^2+x^4)+x*O(x^n), n)} /* Michael Somos Nov 15 2006 */
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CROSSREFS
| Bisections are A001835 and A001075.
Cf. A101265.
Sequence in context: A007481 A121268 A101173 * A116406 A112843 A036651
Adjacent sequences: A005243 A005244 A005245 * A005247 A005248 A005249
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KEYWORD
| easy,nonn,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| More terms from Michael Somos, Aug 01 2001
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