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A056570
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Third power of Fibonacci numbers (A000045).
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20
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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; internal format)
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OFFSET
| 0,4
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COMMENTS
| Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
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REFERENCES
| 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.
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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).
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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 (qntmpkt(AT)yahoo.com), 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 (mathar(AT)strw.leidenuniv.nl), Oct 16 2006
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MAPLE
| A056570 := proc(n) combinat[fibonacci](n)^3 ; end proc:
seq(A056570(n), n=0..20) ;
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MATHEMATICA
| Table[Fibonacci[n]^3, {n, 0, 20}] (Vladimir Orlovsky, Jul 21 2008)
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PROG
| (MAGMA) [Fibonacci(n)^3: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011
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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
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KEYWORD
| nonn,easy
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AUTHOR
| Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Jul 10 2000
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