A056585 Eighth power of Fibonacci numbers A000045.

0, 1, 1, 256, 6561, 390625, 16777216, 815730721, 37822859361, 1785793904896, 83733937890625, 3936588805702081, 184884258895036416, 8686550888106661441, 408066367122340274881, 19170731299728100000000

Offset

0,4

Comments

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

References

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).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..107

Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.

Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.

A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.

J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.

Index to divisibility sequences

Index entries for linear recurrences with constant coefficients, signature (34,714,-4641,-12376,12376,4641,-714,-34,1).

Formula

a(n) = F(n)^8, F(n)=A000045(n).

G.f.: x*p(8, x)/q(8, x) with p(8, x) := sum_{m=0..7} A056588(7, m)*x^m = (1+x)*(1 - 34*x - 458*x^2 + 2242*x^3 - 458*x^4 - 34*x^5 + x^6) and q(8, x) := sum_{m=0..9} A055870(9, 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) (denominator factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): sum_{m=0..9} A055870(9, m)*a(n-m) = 0, n >= 9; inputs: a(n), n=0..8. a(n) = 34*a(n-1) + 714*a(n-2) - 4641*a(n-3) - 12376*a(n-4) + 12376*a(n-5) + 4641*a(n-6) - 714*a(n-7) - 34*a(n-8) + a(n-9).

a(n+1) = 8*F(n)^2*F(n+1)^2*[F(n)^4+F(n+1)^4+4*F(n)^2*F(n+1)^2+3*F(n)*F(n+1)*F(n+2)]-[F(n)^8+F(n+2)^8]+2*[2*F(n+1)^2-(-1)^n]^4 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^8, for n>=0 (This is Theorem 2.2 (vii) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

Mathematica

lst={}; Do[f=Fibonacci[n]; AppendTo[lst, f^8], {n, 0, 4!}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
Fibonacci[Range[0, 20]]^8 (* Harvey P. Dale, Jul 03 2017 *)

Prog

(Magma) [Fibonacci(n)^8: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011
(PARI) a(n)=fibonacci(n)^8 \\ Charles R Greathouse IV, Jun 30 2015

Crossrefs

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

Sequence in context: A050755 A046457 A179645 * A321818 A231307 A206129

Adjacent sequences: A056582 A056583 A056584 * A056586 A056587 A056588

Keyword

nonn,easy

Author

Wolfdieter Lang, Jul 10 2000

Status

approved