A056574 Seventh power of Fibonacci numbers A000045.

0, 1, 1, 128, 2187, 78125, 2097152, 62748517, 1801088541, 52523350144, 1522435234375, 44231334895529, 1283918464548864, 37281334283719577, 1082404156823183753, 31427428360210000000, 912473096871571914483

Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

• nonn
• easy

Comments

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

References

• 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).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..115
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
• A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
• 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 (21,273,-1092,-1820,1092,273,-21,-1).

Crossrefs

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056588, A055870.

Seventh row of array A103323.

Programs

Magma

[Fibonacci(n)^7: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

Mathematica

f[n_]:=Fibonacci[n]^7; lst={}; Do[AppendTo[lst,f[n]],{n,0,5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)

PARI

a(n)=fibonacci(n)^7 \\ Charles R Greathouse IV, Jan 30 2012

Formula

a(n) = F(n)^7, where F(n) = A000045(n).

G.f.: x*p(7, x)/q(7, x) with p(7, x) := sum_{m=0..6} A056588(6, m)*x^m = 1 - 20*x - 166*x^2 + 318*x^3 + 166*x^4 - 20*x^5 - x^6 and q(7, x) := sum_{m=0..8} A055870(8, m)*x^m = (1 + x - x^2)*(1 - 4*x - x^2)*(1 + 11*x - x^2)*(1 - 29*x - x^2) (factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): sum_{m=0..8} A055870(8, m)*a(n-m) = 0, n >= 8; inputs: a(n), n=0..7. a(n) = 21*a(n-1) + 273*a(n-2) - 1092*a(n-3) - 1820*a(n-4) + 1092*a(n-5) + 273*a(n-6) - 21*a(n-7) - a(n-8).

a(n+1) = F(n)^7+F(n+1)^7+7*F(n)*F(n+1)*F(n+2)*[2*F(n+1)^2-(-1)^n]^2 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^7, for n>=0 (This is Theorem 2.3 (iv) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015