

A010914


Pisot sequence E(5,17), a(n) = floor(a(n1)^2 / a(n2) + 1/2).


3



5, 17, 58, 198, 676, 2308, 7880, 26904, 91856, 313616, 1070752, 3655776, 12481600, 42614848, 145496192, 496755072, 1696027904, 5790601472, 19770350080, 67500197376, 230460089344, 786839962624, 2686439671808, 9172078761984, 31315435704320, 106917585293312
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OFFSET

0,1


REFERENCES

Shalosh B. Ekhad, N. J. A. Sloane and Doron Zeilberger, Automated Proof (or Disproof) of Linear Recurrences Satisfied by Pisot Sequences, Preprint, 2016.


LINKS

Colin Barker, Table of n, a(n) for n = 0..1000
D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295305.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Index entries for linear recurrences with constant coefficients, signature (4,2).


FORMULA

From Max Alekseyev, Sep 03 2013: (Start)
It is not hard to show that this sequence satisfies the following simple linear recurrence relation: a(n) = 4*a(n1)  2*a(n2).
Proof: Let a(n) = A010914(n) be the sequence defined by the recurrence a(n) = floor( a(n1)^2/a(n2) + 1/2 ) and initial terms a(0)=5 and a(1)=17. Define a sequence b(n) by the recurrence b(n) = 4*b(n1)  2*b(n2) and the same initial terms b(0)=a(0)=5 and b(1)=a(1)=17. We want to prove that a(n)=b(n) for all n.
The main trick is that instead of proving that the sequence a() satisfies the recurrence for the sequence b(), we will do it the other way around: prove that b() satisfies the recurrence for a(). This will imply that these two sequences coincide.
To prove that b() satisfies the recurrence for a(), it is enough to show that abs( b(n1)^2  b(n)*b(n2) ) < b(n2)/2. To do this, we will use the explicit formula for b(n) that
b(n) = (10+7*sqrt(2))/4 * (2+sqrt(2))^n + (107*sqrt(2))/4 * (2sqrt(2))^n.
Computing abs( b(n1)^2  b(n)*b(n2) ), one finds that it equals 2^n times a constant. At the same time, b(n) grows as (2+sqrt(2))^n (up to a constant) which allows one to easily prove (e.g., by induction on n) that abs( b(n1)^2  b(n)*b(n2) ) < b(n2)/2. (End)
For positive n, a(n) equals [1,1;1,3]^n.[1,2].[1.2], which means calculate the nth power of the specific 2 X 2 matrix, multiply from the right by the column vector [1,2], and finally take the dot product of this column vector with the vector [1,2].  John M. Campbell, Jul 09 2011
a(n) is the 3rd subdiagonal of array A228405.  Richard R. Forberg, Sep 02 2013
a(n) = A079496(n+3) * A016116(n).  Michael Somos, Sep 03 2013
a(n)  a(n1)^2 / a(n2) = 2^floor(n/2) / A143609(n+1) < 1/2 if n>1.  Michael Somos, Sep 03 2013
G.f.: (53*x) / (14*x+2*x^2).  Colin Barker, Dec 06 2015


PROG

(PARI) {a(n) = if( n<0, 0, real( (10 + 7 * quadgen(8)) / 2 * (2 + quadgen(8))^n ))} /* Michael Somos, Sep 03 2013 */
(PARI) {a(n) = if( n<0, 0, n += 3 ; 2^ceil((n  4)/2) * polcoeff( (1 + x  3*x^2  x^3) / (1  6*x^2 + x^4) + x * O(x^n), n))} /* Michael Somos, Sep 03 2013 */
(PARI) Vec((53*x)/(14*x+2*x^2) + O(x^30)) \\ Colin Barker, Dec 06 2015


CROSSREFS

Cf. A079496, A143609.
Sequence in context: A054113 A146697 A009229 * A180502 A261515 A171838
Adjacent sequences: A010911 A010912 A010913 * A010915 A010916 A010917


KEYWORD

nonn,easy


AUTHOR

Simon Plouffe


STATUS

approved



