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Ninth power of Fibonacci numbers A000045.
2

%I #28 Sep 08 2022 08:45:01

%S 0,1,1,512,19683,1953125,134217728,10604499373,794280046581,

%T 60716992766464,4605366583984375,350356403707485209,

%U 26623333280885243904,2023966356928852115753,153841020405122283630137

%N Ninth power of Fibonacci numbers A000045.

%C Divisibility sequence; that is, if n divides m, then a(n) divides a(m).

%D D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

%H Vincenzo Librandi, <a href="/A056586/b056586.txt">Table of n, a(n) for n = 0..101</a>

%H A. Brousseau, <a href="http://www.fq.math.ca/Scanned/6-1/brousseau3.pdf">A sequence of power formulas</a>, Fib. Quart., 6 (1968), 81-83.

%H J. Riordan, <a href="http://dx.doi.org/10.1215/S0012-7094-62-02902-2">Generating functions for powers of Fibonacci numbers</a>, Duke. Math. J. 29 (1962) 5-12.

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (55,1870,-19635,-85085,136136,85085,-19635,-1870,55,1).

%F a(n) = F(n)^9, F(n)=A000045(n).

%F G.f.: x*p(9, x)/q(9, x) with p(9, x) := sum_{m=0..8} A056588(8, m)*x^m = 1 - 54*x - 1413*x^2 + 9288*x^3 + 17840*x^4 - 9288*x^5 - 1413*x^6 + 54*x^7 + x^8 and q(9, x) := sum_{m=0..10} A055870(10, m)*x^m = (1 - x - x^2)*(1 + 4*x - x^2)*(1 - 11*x - x^2)*(1 + 29*x - x^2)*(1 - 76*x - x^2) (factorization deduced from Riordan result).

%F Recursion (cf. Knuth's exercise): sum_{m=0..10} A055870(10, m)*a(n-m) = 0, n >= 10; inputs: a(n), n=0..9. a(n) = 55*a(n-1) + 1870*a(n-2) - 19635*a(n-3) - 85085*a(n-4) + 136136*a(n-5) + 85085*a(n-6) - 19635*a(n-7) - 1870*a(n-8) + 55*a(n-9) + a(n-10).

%o (Magma) [Fibonacci(n)^9: n in [0..20]]; // _Vincenzo Librandi_, Jun 04 2011

%o (PARI) a(n) = fibonacci(n)^9; \\ _Michel Marcus_, Sep 06 2017

%Y Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056585, A056588, A055870.

%K nonn,easy

%O 0,4

%A _Wolfdieter Lang_, Jul 10 2000