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A056571
Fourth power of Fibonacci numbers A000045.
13
0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056
OFFSET
0,4
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
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.
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
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 *)
PROG
(Magma) [Fibonacci(n)^4: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011
(PARI) a(n)=fibonacci(n)^4 \\ Charles R Greathouse IV, Oct 07 2016
CROSSREFS
First differences of A005969.
Fourth row of array A103323.
Sequence in context: A113849 A046453 A030514 * A053909 A151502 A030693
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jul 10 2000
STATUS
approved