0,4

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

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).

J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-1.

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

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(A055870(5, m)*x^m, m=0..5)= 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[2F(2n-2)-6F(2n+1)] + 2F(4n-1) + F(4n) + 6}. - R. Stephan, May 14 2004

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 (mathar(AT)strw.leidenuniv.nl), Oct 16 2006

with (combinat):seq(mul(fibonacci(n), k=1..4), n=0..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Sep 21 2007

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

First differences of A005969.

Fourth row of array A103323.

Sequence in context: A113849 A046453 A030514 this_sequence A053909 A030693 A118675

Adjacent sequences: A056568 A056569 A056570 this_sequence A056572 A056573 A056574

nonn,easy

Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Jul 10 2000