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Tenth power of Fibonacci numbers A000045.
1

%I #30 Sep 08 2022 08:45:01

%S 0,1,1,1024,59049,9765625,1073741824,137858491849,16679880978201,

%T 2064377754059776,253295162119140625,31181719929966183601,

%U 3833759992447475122176,471584161164422542970449

%N Tenth 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="/A056587/b056587.txt">Table of n, a(n) for n = 0..96</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>

%F a(n) = F(n)^10, F(n)=A000045(n).

%F G.f.: x*p(10, x)/q(10, x) with p(10, x) := sum_{m=0..9} A056588(9, m)*x^m = (1-x)*(1 - 87*x - 4047*x^2 + 42186*x^3 + 205690*x^4 + 42186*x^5 - 4047*x^6 - 87*x^7 + x^8) and q(10, x) := sum_{m=0..11} A055870(11, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)*(1 + 47*x + x^2)*(1 - 123*x + x^2) (denominator factorization deduced from Riordan result).

%F Recursion (cf. Knuth's exercise): sum_{m=0..11} A055870(11, m)*a(n-m) = 0, n >= 11; inputs: a(n), n=0..10. a(n) = 89*a(n-1) + 4895*a(n-2) - 83215*a(n-3) - 582505*a(n-4) + 1514513*a(n-5) + 1514513*a(n-6) - 582505*a(n-7) -83215*a(n-8) + 4895*a(n-9) + 89*a(n-10) - a(n-11).

%t Fibonacci[Range[0,15]]^10 (* _Harvey P. Dale_, Jul 29 2018 *)

%o (Magma) [Fibonacci(n)^10: n in [0..20]]; // _Vincenzo Librandi_, Jun 04 2011

%o (PARI) a(n) = fibonacci(n)^10; \\ _Michel Marcus_, Sep 06 2017

%Y Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056585, A056586, A056588, A055870.

%K nonn,easy

%O 0,4

%A _Wolfdieter Lang_, Jul 10 2000

%E More terms from Larry Reeves (larryr(AT)acm.org), Jul 17 2001