

A216801


a(n) = 13*a(n1)  65*a(n2) + 156*a(n3)  182*a(n4) + 91*a(n5)  13*a(n6).


5



2, 22, 117, 468, 1755, 6513, 24336, 91988, 351689, 1357408, 5277363, 20625774, 80909257, 318173258, 1253243498, 4941450657, 19495914360, 76945654032, 303737001009, 1199041027587, 4733273752831, 18683644465447, 73743457866962
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OFFSET

1,1


COMMENTS

a(n) is equal to the rational part of the number sqrt(2*(13 + 3*sqrt(13))/13)*X(2*n1), where X(n) = sqrt((13 3*sqrt(13))/2)*X(n1) + sqrt(13)*X(n2)  sqrt((13 + 3*sqrt(13))/2)*X(n3), with X(0) = 3, X(1) = sqrt((13  3*sqrt(13))/2), and X(2) = (13 + sqrt(13))/2.
Let us observe that all numbers of the form a(n)*13^(floor((n+3)/6)) are integers.
We note that the sequence X(n) is defined "similarly" to sequence Y(n) in the comments to A216540. The only difference between them is in initial condition: X(2) = Y(2).


REFERENCES

R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).


LINKS



FORMULA

G.f.: x*(52*x^5520*x^4+689*x^3299*x^2+48*x2) / (13*x^691*x^5+182*x^4156*x^3+65*x^213*x+1).  Colin Barker, Jun 01 2013


EXAMPLE

We have 4*a(3)=a(4), 4*a(4)=a(5)+a(3). The 3valuation of a(n) for n=1,...,10 is contained in A167366. Moreover it can be obtained X(7)  22*X(3) = 4*sqrt(2*(133*sqrt(13))), 4*X(5)  X(7) = 2*sqrt(26(133*sqrt(13))), and 15*X(5)  X(9) = 20*sqrt(26(133*sqrt(13))), which implies (15*X(5)  X(9))/(4*X(5)  X(7)) = 10.


CROSSREFS



KEYWORD

sign,easy


AUTHOR



STATUS

approved



