A056572 -id:A056572 - cp's OEIS frontend

A215045 a(n) = F(2*n+1)^5 with n >= 0, F=A000045 (Fibonacci numbers).

1, 32, 3125, 371293, 45435424, 5584059449, 686719856393, 84459630100000, 10387823949447757, 1277617458486664901, 157136551895768914976, 19326518128014212635057, 2377004590722802744140625, 292352238096435536675521568

Offset: 0

Wolfdieter Lang, Aug 31 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (144,-2640,6930,-2640,144,-1).

Cf. A000045, A056572, A215044 (even part).

[Fibonacci(2*n+1)^5: n in [0..15]]; // Vincenzo Librandi, Sep 02 2012

Table[Fibonacci[2*n+1]^5, {n,0,201}] (* Vincenzo Librandi, Sep 02 2012 *)
Fibonacci[Range[1,35,2]]^5 (* Harvey P. Dale, Apr 05 2019 *)

O.g.f.: (1-x)*(1-111*x+1046*x^2-111*x^3+x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)). From the odd part of the bisection of A056572.

A358934 a(n) = Fibonacci(n+1)^5 - Fibonacci(n-1)^5.

Original entry on oeis.org

0, 1, 31, 242, 3093, 32525, 368168, 4051333, 45064131, 499200274, 5538624025, 61414079849, 681135796944, 7553728681433, 83772910243607, 929052526388050, 10303364319347757, 114266002348885717, 1267229634537217144, 14053790947047408701, 155858934437282250075

Offset: 0

Feryal Alayont, Dec 06 2022

a(n) is the number of edge covers of a spider graph with five branches where each branch has n vertices besides the center vertex. The idea is each branch is treated as a path P_(n+2). Each branch acts independently then and has F_(n+1) covers (P_n has F(n-1) covers), hence F_(n+1)^5 total. Except we remove the cases where each branch is missing the connecting edge to the center, which is when that edge cover comes from P_n , hence the minus F_(n-1)^5.

An edge cover of a graph is a subset of edges for which each vertex is incident to at least one edge in the subset.

			Case n=1 is a star graph with five branches and one edge cover (all edges).
   *   *
    \ /
  *__C__*
     |
     *
For n=2, there are 31 edge covers of the graph obtained by gluing five P_3 paths at one single vertex. Each of the pendant edges of the P_3's have to be in the edge cover for the pendants to be incident with an edge. The middle vertices are then automatically incident with at least one edge. There remains the center vertex. We then need at least one of the remaining five edges to be in the subset, giving us 2^5-1 choices.
   *__ *   *__*
       \ /
  *__*__C__*__*
        |
        *__*

Michael De Vlieger, Table of n, a(n) for n = 0..957
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Index entries for linear recurrences with constant coefficients, signature (8,40,-60,-40,8,1).

Cf. A000045, A056572.

LinearRecurrence[{8, 40, -60, -40, 8, 1}, {0, 1, 31, 242, 3093, 32525}, 20] (* Amiram Eldar, Dec 07 2022 *)
Join[{0},#[[3]]-#[[1]]&/@Partition[Fibonacci[Range[0,30]]^5,3,1]] (* Harvey P. Dale, Aug 05 2024 *)

from sympy import fibonacci
def a(n):
    return fibonacci(n+1)**5-fibonacci(n-1)**5

from gmpy2 import fib2
def A358934(n): return sum(f:=fib2(n))**5-f[1]**5 # Chai Wah Wu, Jan 04 2023

From Stefano Spezia, Dec 07 2022: (Start)
G.f.: x*(1 + 23*x - 46*x^2 - 23*x^3 + x^4)/((1 + 4*x - x^2)*(1 - x - x^2)*(1 - 11*x - x^2)).
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) for n > 5. (End)
5*a(n) = 2*A000045(n)+11*A049666(n)+8*(-1)^n*A001076(n). - R. J. Mathar, May 07 2024

A056587 Tenth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 1024, 59049, 9765625, 1073741824, 137858491849, 16679880978201, 2064377754059776, 253295162119140625, 31181719929966183601, 3833759992447475122176, 471584161164422542970449

Offset: 0

Wolfdieter Lang, Jul 10 2000

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

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..96
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 (89,4895,-83215,-582505,1514513,1514513,-582505,-83215,4895,89,-1).

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056585, A056586, A056588, A055870.

[Fibonacci(n)^10: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

Fibonacci[Range[0,15]]^10 (* Harvey P. Dale, Jul 29 2018 *)

a(n) = fibonacci(n)^10; \\ Michel Marcus, Sep 06 2017

a(n) = F(n)^10, F(n)=A000045(n).
G.f.: x*p(10, x)/q(10, x) with p(10, x) := sum_{m=0..9} A056588(9, m)*x^m = (1-x)*(1 - 87*x - 4047*x^2 + 42186*x^3 + 205690*x^4 + 42186*x^5 - 4047*x^6 - 87*x^7 + x^8) and q(10, x) := sum_{m=0..11} A055870(11, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)*(1 + 47*x + x^2)*(1 - 123*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..11} A055870(11, m)*a(n-m) = 0, n >= 11; inputs: a(n), n=0..10.  a(n) = 89*a(n-1) + 4895*a(n-2) - 83215*a(n-3) - 582505*a(n-4) + 1514513*a(n-5) + 1514513*a(n-6) - 582505*a(n-7) -83215*a(n-8) + 4895*a(n-9) + 89*a(n-10) - a(n-11).

More terms from Larry Reeves (larryr(AT)acm.org), Jul 17 2001

A213809 Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5.

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 3, 3, 3, 5, 5, 3, 5, 5, 5, 5, 5, 5, 5, 7, 7, 5, 7, 7, 7, 7, 7, 7, 7, 9, 9, 7, 9, 9, 9, 9, 9, 9, 9, 11, 11, 9, 11, 11, 11, 11, 11, 11, 11, 13, 13, 11, 13, 13, 13, 13, 13, 13, 13, 15, 15, 13, 15, 15, 15, 15, 15, 15, 15, 17, 17, 15, 17, 17, 17, 17, 17, 17, 17, 19, 19, 17, 19, 19, 19, 19, 19, 19, 19, 21, 21, 19, 21, 21, 21, 21, 21, 21, 21, 23, 23, 21, 23, 23, 23, 23, 23, 23, 23

Offset: 1

Art DuPre, Jun 20 2012

nonn

The maximum elements themselves are in A113506.

The fractions F(n+1)^5/F(n)^5 are 1, 32, 243/32, 3125/243,... (see A056572). The continued fractions are [1], [32], [7,1,1,2,6], [12,1,6,6,1,4], [10,2,17,17,1,4]..., and for the first 4 of these the maximum element is the first, for the 5th of these the maximum element is the third.

			The continued fraction of the fraction corresponding to [1,1,1,1,1,1,1,1,1,1,1,1,1]^5 is
[11,11,7,1,39282,2,5,11,11,1,11,11] and the maximum occurs at place 5, which according to the formula, should be 3+2k, and since 13=10k+3, k=1 and 3+2k=3+2=5.

A213809 := proc(n)
        local c,a,i;
        (combinat[fibonacci](n+1)/combinat[fibonacci](n))^5 ;
        c := numtheory[cfrac](%,quotients) ;
        a := 1 ;
        for i from 2 to nops(c) do
                if op(i,c) > op(a,c) then
                        a := i ;
                end if;
        end do:
        a ;
end proc: # R. J. Mathar, Jul 06 2012

a(10k+m)=3+2k if m=0,1,3,4,5,6,7,8,9, k>0.

a(10k+2)=1+2k, k>0.

A216755 Digital root of the fifth power of Fibonacci(n).

Original entry on oeis.org

1, 1, 5, 9, 2, 8, 7, 9, 4, 1, 8, 9, 8, 8, 4, 9, 7, 1, 2, 9, 5, 8, 1, 9, 1, 1, 5, 9, 2, 8, 7, 9, 4, 1, 8, 9, 8, 8, 4, 9, 7, 1, 2, 9, 5, 8, 1, 9, 1, 1, 5, 9, 2, 8, 7, 9, 4, 1, 8, 9, 8, 8, 4, 9, 7, 1, 2, 9, 5, 8, 1, 9, 1, 1, 5, 9, 2, 8, 7, 9, 4, 1, 8, 9, 8, 8, 4, 9, 7, 1, 2, 9, 5, 8, 1, 9, 1, 1, 5, 9

Offset: 1

Ravi Bhandari, Sep 15 2012

This sequence is periodic with period 24, i.e. gcd(period of digital roots of squares of Fibonacci, period of digital roots of cubes of Fibonacci)

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 1).

Cf. A030132, A216676, A216699

(* First run program for A211821 to define digitalRoot *) Table[digitalRoot[Fibonacci[n]^5], {n, 90}] (* Alonso del Arte, Sep 15 2012 *)
LinearRecurrence[{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 1},{1, 1, 5, 9, 2, 8, 7, 9, 4, 1, 8, 9, 8, 8, 4, 9},100] (* Ray Chandler, Aug 27 2015 *)

a(n) = A010888(A056572(n)).
a(n) = a(n-4) - a(n-12) + a(n-16). - R. J. Mathar, Sep 15 2012
G.f. x*( -1-x-5*x^2-9*x^3-x^4-7*x^5-2*x^6-2*x^8+7*x^9-x^10-5*x^12-8*x^13-x^14-9*x^15 ) / ( (x-1) *(1+x) *(x^2+1) *(x^4+1) *(x^8-x^4+1) ). - R. J. Mathar, Sep 15 2012

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A215045 a(n) = F(2*n+1)^5 with n >= 0, F=A000045 (Fibonacci numbers).

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A358934 a(n) = Fibonacci(n+1)^5 - Fibonacci(n-1)^5.

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A056587 Tenth power of Fibonacci numbers A000045.

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A213809 Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5.

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A216755 Digital root of the fifth power of Fibonacci(n).

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