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A010914 Pisot sequence E(5,17), a(n)=[ a(n-1)^2/a(n-2)+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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

REFERENCES

D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory (Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.

LINKS

Table of n, a(n) for n=0..25.

D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295-305.

Index to sequences with linear recurrences with constant coefficients, signature (4,-2).

FORMULA

For positive n, a(n) equals [1,1;1,3]^n.[1,2].[1.2], which means calculate the n-th power of the specific 2x2 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]. [From John M. Campbell, Jul 09 2011]

a(n) is the 3rd sub-diagonal of array A228405.  - Richard R. Forberg, Sep 02 2013

From Max Alekseyev, Sep 03 2013: (Start)

It is not hard to show that A010914 indeed satisfies the following simple linear recurrence relation: a(n) = 4*a(n-1) - 2*a(n-2).

Let a(n) = A010914(n) be the sequence defined by the recurrence

a(n)=floor( a(n-1)^2/a(n-2)+1/2 ) and initial terms a(0)=5 and a(1)=17.

Define a sequence b(n) by the recurrence b(n) = 4*b(n-1) - 2*b(n-2) 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() satisfied 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(n-1)^2 - b(n)*b(n-2) ) < b(n-2)/2.

To do this, we will use the explicit formula for b(n):

b(n) = (10+7*sqrt(2))/4 * (2+sqrt(2))^n +  (10-7*sqrt(2))/4 * (2-sqrt(2))^n

Explicitly computing abs( b(n-1)^2 - b(n)*b(n-2) ), 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(n-1)^2 - b(n)*b(n-2) ) < b(n-2)/2. (End)

a(n) = A079496(n+3) * A016116(n). - Michael Somos, Sep 03 2013

a(n) - a(n-1)^2 / a(n-2) = 2^floor(n/2) / A143609(n+1) < 1/2 if n>1. - Michael Somos, Sep 03 2013

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 */

CROSSREFS

Cf. A079496, A143609.

Sequence in context: A054113 A146697 A009229 * A180502 A171838 A105392

Adjacent sequences:  A010911 A010912 A010913 * A010915 A010916 A010917

KEYWORD

nonn,easy

AUTHOR

Simon Plouffe

STATUS

approved

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Last modified August 30 00:29 EDT 2014. Contains 246212 sequences.