login
This site is supported by donations to The OEIS Foundation.
Logo

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A056570 Third power of Fibonacci numbers (A000045). 27
0, 1, 1, 8, 27, 125, 512, 2197, 9261, 39304, 166375, 704969, 2985984, 12649337, 53582633, 226981000, 961504803, 4073003173, 17253512704, 73087061741, 309601747125, 1311494070536, 5555577996431, 23533806109393 (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. 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) 5-12.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..173

D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux

Index to divisibility sequences

Index to sequences with linear recurrences with constant coefficients, signature (3,6,-3,-1).

FORMULA

a(n) = A000045(n)^3.

G.f.: x*p(3, x)/q(3, x) with p(3, x):=sum(A056588(2, m)*x^m, m=0..2)=1-2*x-x^2 and q(3, x):=sum(A055870(4, m)*x^m, m=0..4)=1-3*x-6*x^2+3*x^3+x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): 1*a(n)-3*a(n-1)-6*a(n-2)+3*a(n-3)+1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5-th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4

a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - R. Stephan, May 14 2004

a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g. Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679...an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004

Sum_(j=0..n) binomial(n,j) a(j)= [2^n A001906(n)+3 A000045(n)]/5. sum_(j=0..n) (-1)^j binomial(n,j) a(j)=[(-2)^n A000045(n)-3 A001906(n)]/5. - R. J. Mathar, Oct 16 2006

G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). [Colin Barker, Feb 28 2012]

EXAMPLE

a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.

a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.

MAPLE

A056570 := proc(n) combinat[fibonacci](n)^3 ; end proc:

seq(A056570(n), n=0..20) ;

MATHEMATICA

Table[Fibonacci[n]^3, {n, 0, 20}] (* Vladimir Orlovsky, Jul 21 2008 *)

PROG

(MAGMA) [Fibonacci(n)^3: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

CROSSREFS

Cf. A000045, A007598, A056588, A055870.

Cf. A066259, A066258.

First differences of A005968.

Third row of array A103323.

Sequence in context: A051751 A133042 A181361 * A165048 A066963 A067813

Adjacent sequences:  A056567 A056568 A056569 * A056571 A056572 A056573

KEYWORD

nonn,easy

AUTHOR

Wolfdieter Lang Jul 10 2000

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
Recent Additions | More pages | Superseeker | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified May 19 14:00 EDT 2013. Contains 225430 sequences.