A056571 Fourth power of Fibonacci numbers A000045.

0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056

Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

• nonn
• easy

Comments

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

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/4,3/4)-fences and quarter-squares (1/4 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/8,3/8)-fences and (1/8,7/8)-fences. - Michael A. Allen, Jan 11 2022

References

• A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 31.
• Donald 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).
• Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp. 4623-4627.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..151
• 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.
• Alfred Brousseau, A sequence of power formulas, Fib. Quart., Vol. 6, No. 1 (1968), pp. 81-83.
• Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math., Vol. 199, No. 1-3 (1999), pp. 217-220. MR1675924 (99k:05016).
• Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
• Hideyuki Ohtsuka, Problem N-1220, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), p. 368; Gelin-Cesàro Identity Yields a Telescoping Product, Solution to Problem H-790 by Ramya Dutta, ibid., Vol. 56, No. 4 (2018), p. 372.
• John Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J., Vol. 29, No. 1 (1962), pp. 5-12.
• Index to divisibility sequences
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

Cf. A000045, A001622, A007598, A056570, A056588, A055870.

First differences of A005969.

Fourth row of array A103323.

Programs

Magma

[Fibonacci(n)^4: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

Maple

A056571 := proc(n)
    combinat[fibonacci](n)^4 ;
end proc:
seq(A056571(n),n=0..10) ; # R. J. Mathar, Jan 23 2022

Mathematica

Fibonacci[Range[0, 25]]^4 (* Wesley Ivan Hurt, Jan 11 2022 *)

PARI

a(n)=fibonacci(n)^4 \\ Charles R Greathouse IV, Oct 07 2016

Formula

a(n) = F(n)^4 = A007598(n)^2, F(n) = A000045(n).

G.f.: x*p(4, x)/q(4, x) with p(4, x) := sum(A056588(3, m)*x^m, m=0..3) = 1 - 4*x - 4*x^2 + x^3 = (1+x)*(1 - 5*x + x^2) and q(4, x) := Sum_{m=0..5} A055870(5, m)*x^m = 1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2) (denominator factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): 1*a(n) - 5*a(n-1) - 15*a(n-2) + 15*a(n-3) + 5*a(n-4) - 1*a(n-5) = 0, n >= 5; inputs: a(n), n=0..4.

(1/25)*((-1)^n*(2*F(2*n-2) - 6*F(2*n+1)) + 2*F(4*n-1) + F(4*n) + 6). - Ralf Stephan, May 14 2004

a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) + 1 = A244855(n)+1.

Sum_{j=0..n} binomial(n,j)*a(j) = (3^n*A005248(n) - 4*(-1)^n*A000032(n) + 6*2^n)/25. Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = -5^((n+1)/2-2)*(A001906(n) + 4*A000045(n)) if n odd. - R. J. Mathar, Oct 16 2006

a(n) = (F(n)*F(3n) - 3*F(n)^2*(-1)^n)/5. - Gary Detlefs, Dec 26 2010

Product_{n>=3} (1 - 1/a(n)) = phi^5/12, where phi is the golden ratio (A001622)(Ohtsuka, 2017). - Amiram Eldar, Dec 02 2021