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A056571 Fourth power of Fibonacci numbers A000045. 10
0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

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

REFERENCES

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

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

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

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).

Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, 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

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

Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).

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 (5,15,-15,-5,1).

FORMULA

a(n)= F(n)^4, 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.

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

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

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

First differences of A005969.

Fourth row of array A103323.

Sequence in context: A113849 A046453 A030514 * A053909 A151502 A030693

Adjacent sequences:  A056568 A056569 A056570 * A056572 A056573 A056574

KEYWORD

nonn,easy

AUTHOR

Wolfdieter Lang, Jul 10 2000

STATUS

approved

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Last modified March 20 21:49 EDT 2019. Contains 321352 sequences. (Running on oeis4.)