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Fourth power of Fibonacci numbers A000045.
13

%I #76 Sep 08 2022 08:45:01

%S 0,1,1,16,81,625,4096,28561,194481,1336336,9150625,62742241,429981696,

%T 2947295521,20200652641,138458410000,949005240561,6504586067281,

%U 44583076827136,305577005139121,2094455819300625,14355614096087056

%N Fourth power of Fibonacci numbers A000045.

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

%C 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

%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 31.

%D 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).

%D 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.

%H Vincenzo Librandi, <a href="/A056571/b056571.txt">Table of n, a(n) for n = 0..151</a>

%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/41-44-2012/azarianIJCMS41-44-2012.pdf">Fibonacci Identities as Binomial Sums II</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.

%H Alfred Brousseau, <a href="http://www.fq.math.ca/Scanned/6-1/brousseau3.pdf">A sequence of power formulas</a>, Fib. Quart., Vol. 6, No. 1 (1968), pp. 81-83.

%H Andrej Dujella, <a href="http://dx.doi.org/10.1016/S0012-365X(98)00341-0">A bijective proof of Riordan's theorem on powers of Fibonacci numbers</a>, Discrete Math., Vol. 199, No. 1-3 (1999), pp. 217-220. MR1675924 (99k:05016).

%H Kenneth Edwards and Michael A. Allen, <a href="https://www.fq.math.ca/Papers1/58-5/edwards.pdf">A new combinatorial interpretation of the Fibonacci numbers cubed</a>, Fib. Q. 58:5 (2020) 128-134.

%H Hideyuki Ohtsuka, <a href="https://www.fq.math.ca/Problems/ElemProbSolnNov2017.pdf">Problem N-1220</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), p. 368; <a href="https://www.fq.math.ca/Problems/ElemProbSolnNov2018.pdf">Gelin-Cesàro Identity Yields a Telescoping Product</a>, Solution to Problem H-790 by Ramya Dutta, ibid., Vol. 56, No. 4 (2018), p. 372.

%H John 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., Vol. 29, No. 1 (1962), pp. 5-12.

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).

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

%F 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).

%F 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.

%F (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

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

%F 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

%F a(n) = (F(n)*F(3n) - 3*F(n)^2*(-1)^n)/5. - _Gary Detlefs_, Dec 26 2010

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

%p A056571 := proc(n)

%p combinat[fibonacci](n)^4 ;

%p end proc:

%p seq(A056571(n),n=0..10) ; # _R. J. Mathar_, Jan 23 2022

%t Fibonacci[Range[0, 25]]^4 (* _Wesley Ivan Hurt_, Jan 11 2022 *)

%o (Magma) [Fibonacci(n)^4: n in [0..30]]; // _Vincenzo Librandi_, Jun 04 2011

%o (PARI) a(n)=fibonacci(n)^4 \\ _Charles R Greathouse IV_, Oct 07 2016

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

%Y First differences of A005969.

%Y Fourth row of array A103323.

%K nonn,easy

%O 0,4

%A _Wolfdieter Lang_, Jul 10 2000