A056572 Fifth power of Fibonacci numbers A000045.

0, 1, 1, 32, 243, 3125, 32768, 371293, 4084101, 45435424, 503284375, 5584059449, 61917364224, 686719856393, 7615646045657, 84459630100000, 936668172433707, 10387823949447757, 115202670521319424, 1277617458486664901

Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

• nonn
• easy

Comments

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..135
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
• A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
• 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.
• 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 (8,40,-60,-40,8,1).

Crossrefs

Cf. A000045, A007598, A056570-1, A056588, A055870.

Fifth row of array A103323.

Programs

Magma

[Fibonacci(n)^5: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

Mathematica

Table[f=Fibonacci[n];f^5,{n,0,12}] (* Vladimir Joseph Stephan Orlovsky, Jul 22 2008 *)
Fibonacci[Range[0,20]]^5 (* Harvey P. Dale, Aug 07 2022 *)

PARI

a(n)=fibonacci(n)^5 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = F(n)^5, F(n)=A000045(n).

G.f.: x*p(5, x)/q(5, x) with p(5, x) := sum(A056588(4, m)*x^m, m=0..4)= 1-7*x-16*x^2+7*x^3+x^4 and q(5, x) := sum(A055870(6, m)*x^m, m=0..6)= 1-8*x-40*x^2+60*x^3+40*x^4-8*x^5-x^6 = (1-x-x^2)*(1+4*x-x^2)*(1-11*x-x^2) (factorization deduced from Riordan result).

Recursion (cf. Knuth's exercise): sum(A055870(6, m)*a(n-m), m=0..6) = 0, n >= 6; inputs: a(n), n=0..5. a(n) = +8*a(n-1) +40*a(n-2) -60*a(n-3) -40*a(n-4) +8*a(n-5) +a(n-6).

a(n) = (10*F(n) + 5*(-1)^(n+1)*F(3*n) + F(5*n))/25, n >= 0. See the general comment on A111418 regarding the Ozeki reference; here the row 10, 5, 1 of that triangle applies. - Wolfdieter Lang, Aug 25 2012

a(n) = (F(n)^2*(F(3*n)-(-1)^n*3*F(n)))/5. - Gary Detlefs, Jan 07 2013

a(n) = F(n-2)*F(n-1)*F(n)*F(n+1)*F(n+2) + F(n). - Tony Foster III, Apr 11 2018