A056573 Sixth power of Fibonacci numbers A000045.

0, 1, 1, 64, 729, 15625, 262144, 4826809, 85766121, 1544804416, 27680640625, 496981290961, 8916100448256, 160005726539569, 2871098559212689, 51520374361000000, 924491486192068809, 16589354847268067929

Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

• nonn
• easy

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..124
• 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 (13,104,-260,-260,104,13,-1).

Crossrefs

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

Sixth row of array A103323.

Programs

Magma

[Fibonacci(n)^6: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

Maple

with(combinat): A056573:=n->fibonacci(n)^6: seq(A056573(n), n=0..30); # Wesley Ivan Hurt, Jun 29 2015

Mathematica

f[n_]:=Fibonacci[n]^6; lst={}; Do[AppendTo[lst,f[n]],{n,0,5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Fibonacci[Range[0,20]]^6 (* Harvey P. Dale, Sep 21 2024 *)

PARI

a(n)=fibonacci(n)^6 \\ Charles R Greathouse IV, Jun 29 2015

Formula

a(n) = F(n)^6, where F(n) = A000045(n).

G.f.: x*p(6, x)/q(6, x) with p(6, x) := sum_{m=0..5} A056588(5, m)*x^m = (1-x)*(1 - 11*x - 64*x^2 - 11*x^3 + x^4) and q(6, x) := sum_{m=0..7} A055870(7, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) (denominator factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): sum_{m=0..7} A055870(7, m)*a(n-m) = 0, n >= 7; inputs: a(n), n=0..6. a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7).

From Gary Detlefs, Jan 07 2013: (Start)

a(n) = (F(3*n)^2 - (-1)^n*6*F(n)*F(3*n) + 9*F(n)^2)/25.

a(n) = (10*F(n)^3*F(3*n) - F(3*n)^2 + 9*F(n)^2)/25. (End)

a(n+1) = 2*[2*F(n+1)^2-(-1)^n]^3+3*F(n)^2*F(n+1)^2*F(n+2)^2-[F(n)^6+F(n+2)^6] = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^6, for n (this is Theorem 2.2 (vi) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015