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A217052
a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=a(1)=1, and a(2)=19.
5
1, 1, 19, 82, 703, 4096, 29242, 186733, 1266103, 8309143, 55500634, 367187437, 2441886670, 16193659132, 107553444913, 713750040577, 4738726458775, 31453733795086, 208804386436435, 1386041496850144, 9200883498819958, 61076450807299765, 405436597890428431
OFFSET
0,3
COMMENTS
The Ramanujan type sequence number 10 for the argument 2Pi/9 defined by the relation a(n) = ((1/3 - c(1))^n + (1/3 - c(2))^n + (1/3 - c(4))^n)*3^(n-1), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are sequences A217053 and A217069.
For more informations about connections a(n) with these two sequences - see comments in A217053.
The 3-valuation of the sequence a(n) is equal to (1).
REFERENCES
R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.
FORMULA
G.f.: (1-2*x-8*x^2)/(1-3*x-24*x^2-x^3).
EXAMPLE
We have a(4)=37*a(2) and a(5) = 2^(12), which implies ((1/3 - c(1))^4 + (1/3 - c(2))^4 + (1/3 - c(4))^4 = (37/9)*((1/3 - c(1))^2 + (1/3 - c(2))^2 + (1/3 - c(4))^2) = (37/27)*19 = 703/27, (1/3 - c(1))^5 + (1/3 - c(2))^5 + (1/3 - c(4))^5 = (8/3)^4. Moreover we have a(10) = 676837*a(3)
MATHEMATICA
LinearRecurrence[{3, 24, 1}, {1, 1, 19}, 30]
PROG
(PARI) Vec((1-2*x-8*x^2)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Sep 25 2012
STATUS
approved