A056572 -id:A056572 - cp's OEIS frontend

A056588 Coefficient triangle of certain polynomials.

1, 1, -1, 1, -2, -1, 1, -4, -4, 1, 1, -7, -16, 7, 1, 1, -12, -53, 53, 12, -1, 1, -20, -166, 318, 166, -20, -1, 1, -33, -492, 1784, 1784, -492, -33, 1, 1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1, 1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1

Offset: 0

Wolfdieter Lang, Jul 10 2000

G.f. for column m: see column sequences: A000012, A000071, A056589-91, for m=0..4.
The row polynomials p(n,x) := sum(a(n,m)*x^m) occur as numerators of the g.f. for the (n+1)-th power of Fibonacci numbers A000045. The corresponding denominator polynomials are the row polynomials q(n+2,x) = Sum_{m=0..n+2} A055870(n+2, m)*x^m (signed Fibonomial triangle).
The row polynomials p(n,x) and the companion denominator polynomials q(n,x) can be deduced from Riordan's recursion result.
The explicit formula is found from the recursion relation for powers of Fibonacci numbers (see Knuth's exercise with solution). - Roger L. Bagula, Apr 03 2010

			Row polynomial for n=4: p(4,x) = 1 - 7*x - 16*x^2 + 7*x^3 + x^4. x*p(4,x) is the numerator of the g.f. for A056572(n), n >= 0 (fifth power of Fibonacci numbers) {0,1,1,32,243,...}. The denominator polynomial is Sum_{m=0..6} A055870(6,m)*x^m (n=6 row polynomial of signed fibonomial triangle).
From _Roger L. Bagula_, Apr 03 2010: (Start)
1;
1,  -1;
1,  -2,    -1;
1,  -4,    -4,     1;
1,  -7,   -16,     7,      1;
1, -12,   -53,    53,     12,      -1;
1, -20,  -166,   318,    166,     -20,     -1;
1, -33,  -492,  1784,   1784,    -492,    -33,    1;
1, -54, -1413,  9288,  17840,   -9288,  -1413,   54,  1;
1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1; (End)

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 84, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
S. Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.

Cf. A055870, A000012, A000071, A056589-91, A056592 (row sums), A000045, A007598, A056570-4, A056585-7.

A056588 := proc(n,k)
    if k = 0 then
        1;
    elif k >n then
        0;
    else
        combinat[fibonacci](k+1)^(n+1)+add( A055870(n+2, j)*(combinat[fibonacci](k+1-j)^(n+1)), j=1..k) ;
    end if;
end proc: # R. J. Mathar, Jun 14 2015

p[x_, n_] = Sum[(((1 + Sqrt[5])^k - (1 - Sqrt[5])^k)/(2^k*Sqrt[5]))^n*x^k, {k, 0, Infinity}];
a = Table[CoefficientList[FullSimplify[Numerator[p[ x, n]]/x], x]/2^(1 + Floor[n/2]), {n, 1, 10}];
Table[a[[n]]/a[[n]][[1]], {n, 1, 10}];
Flatten[%] (* Roger L. Bagula, Apr 03 2010 *)

S(n, k) = (-1)^floor((k+1)/2)*(prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j)));
T(n, k) = sum(j=0, k, fibonacci(k+1-j)^(n+1) * S(n+2, j));
tabl(m) = for (n=0, m, for (k=0, n, print1(T(n, k), ", ")); print);
tabl(9); \\ Tony Foster III, Aug 20 2018

a(n, m)=0 if nA000045(n) (Fibonacci) and sfibonomial(n, m) := A055870(n, m).
From Roger L. Bagula, Apr 03 2010: (Start)
p(x,n) = Sum_{k>=0} (((1 + sqrt(5))^k - (1 - sqrt(5))^k)/(2^k*sqrt(5)))^n*x^k;
t(n,m) = Numerator_coefficients(p(x,n)/x)/2^(1 + floor(n/2));
out(n,m) = t(n,m)/t(n,1). (End)
T(n, k) = Sum_{j=0..k} Fibonacci(k+1-j)^(n+1) * A055870(n+2, j). - Tony Foster III, Aug 20 2018
Sum_{j=0..n-1} a(n-1, n-1-j)*A010048(k+j, n) = Fibonacci(k)^n. - Tony Foster III, Jul 24 2018

A103323 Square array T(n,k) read by antidiagonals: powers of Fibonacci numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 9, 5, 1, 1, 16, 27, 25, 8, 1, 1, 32, 81, 125, 64, 13, 1, 1, 64, 243, 625, 512, 169, 21, 1, 1, 128, 729, 3125, 4096, 2197, 441, 34, 1, 1, 256, 2187, 15625, 32768, 28561, 9261, 1156, 55, 1, 1, 512, 6561, 78125, 262144, 371293, 194481, 39304, 3025, 89

Offset: 1

Ralf Stephan, Feb 02 2005

Number of ways to create subsets S(1), S(2),..., S(k-1) such that S(1) is in [n] and for 2<=i<=k-1, S(i) is in [n] and S(i) is disjoint from S(i-1).

			Square array T(n,k) begins:
  1, 1,  2,   3,     5,      8, ...
  1, 1,  4,   9,    25,     64, ...
  1, 1,  8,  27,   125,    512, ...
  1, 1, 16,  81,   625,   4096, ...
  1, 1, 32, 243,  3125,  32768, ...
  1, 1, 64, 729, 15625, 262144, ...
  ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 138.

Alois P. Heinz, Antidiagonals n = 1..100, flattened

Rows include A000045, A007598, A056570, A056571, A056572, A056573, A056574.
Main diagonal gives A100399.
Cf. A244003.
Cf. A105317, A254719.

A:= (n, k)-> (<<1|1>, <1|0>>^n)[1, 2]^k:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Jun 17 2014

T[n_, k_] := Fibonacci[k]^n; Table[T[n-k+1, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 16 2015 *)

PARI
```
T(n,k)=fibonacci(k)^n
```

T(n, k) = A000045(k)^n, n, k > 0.

T(n, k) = Sum[i_1>=0, Sum[i_2>=0, ... Sum[i_{k-1}>=0, C(n, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{k-2}, i_{k-1}) ] ... ]].

A098531 Sum of fifth powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 34, 277, 3402, 36170, 407463, 4491564, 49926988, 553211363, 6137270812, 68054635036, 754774491429, 8370420537086, 92830050637086, 1029498223070793, 11417322172518550, 126619992693837974, 1404237451180502875, 15573231068749231000

Offset: 0

Benoit Cloitre, Sep 12 2004

Prime p divides a((p-1)/2) for p = {29,89,101,181,229,...} = A047650[n]. Primes for which golden mean tau is a quadratic residue or Primes of the form x^2 + 20y^2. - Alexander Adamchuk, Aug 07 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (9,32,-100,20,48,-7,-1).

Cf. A000071, A001654, A005968, A005969, A047650, A056572, A098532, A098533, A119286, A128697.

[(&+[Fibonacci(k)^5:k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

Accumulate[Fibonacci[Range[0,20]]^5]  (* Harvey P. Dale, Jan 14 2011 *)
CoefficientList[Series[x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 13 2012 *)

PARI
```
a(n)=sum(i=0,n,fibonacci(i)^5)
```

a(n) = -7/22 + 2*F(n+2)/5 + (F(5*(n+1)) + F(5*n))/(5*55) - (-1)^n*(F(3*(n+1)) - F(3*n))/(2*10), where F=A000045. One may use F(5*(n+1)) + F(5*n) = F(5*n+1) + 4*F(5*n+2) (due to the Binet-de Moivre formula).

G.f.: x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)). - Bruno Berselli, Oct 12 2012

Formula corrected, with the author's consent, by Wolfdieter Lang, Oct 12 2012

A056573 Sixth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 64, 729, 15625, 262144, 4826809, 85766121, 1544804416, 27680640625, 496981290961, 8916100448256, 160005726539569, 2871098559212689, 51520374361000000, 924491486192068809, 16589354847268067929

Offset: 0

Wolfdieter Lang, Jul 10 2000

A 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..124
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 (13,104,-260,-260,104,13,-1).

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

Sixth row of array A103323.

[Fibonacci(n)^6: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

with(combinat): A056573:=n->fibonacci(n)^6: seq(A056573(n), n=0..30); # Wesley Ivan Hurt, Jun 29 2015

f[n_]:=Fibonacci[n]^6; lst={}; Do[AppendTo[lst,f[n]],{n,0,5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Fibonacci[Range[0,20]]^6 (* Harvey P. Dale, Sep 21 2024 *)

a(n)=fibonacci(n)^6 \\ Charles R Greathouse IV, Jun 29 2015

a(n) = F(n)^6, where F(n) = A000045(n).
G.f.: x*p(6, x)/q(6, x) with p(6, x) := sum_{m=0..5} A056588(5, m)*x^m = (1-x)*(1 - 11*x - 64*x^2 - 11*x^3 + x^4) and q(6, x) := sum_{m=0..7} A055870(7, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..7} A055870(7, m)*a(n-m) = 0, n >= 7; inputs: a(n), n=0..6. a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7).
From Gary Detlefs, Jan 07 2013: (Start)
a(n) = (F(3*n)^2 - (-1)^n*6*F(n)*F(3*n) + 9*F(n)^2)/25.
a(n) = (10*F(n)^3*F(3*n) - F(3*n)^2 + 9*F(n)^2)/25. (End)
a(n+1) = 2*[2*F(n+1)^2-(-1)^n]^3+3*F(n)^2*F(n+1)^2*F(n+2)^2-[F(n)^6+F(n+2)^6] = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^6, for n (this is Theorem 2.2 (vi) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A056574 Seventh power of Fibonacci numbers A000045.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Jul 10 2000

A 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..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).

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

Seventh row of array A103323.

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

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

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

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

A056585 Eighth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 256, 6561, 390625, 16777216, 815730721, 37822859361, 1785793904896, 83733937890625, 3936588805702081, 184884258895036416, 8686550888106661441, 408066367122340274881, 19170731299728100000000

Offset: 0

Wolfdieter Lang, Jul 10 2000

A 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..107
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 (34,714,-4641,-12376,12376,4641,-714,-34,1).

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

[Fibonacci(n)^8: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

lst={};Do[f=Fibonacci[n];AppendTo[lst, f^8], {n, 0, 4!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
Fibonacci[Range[0,20]]^8 (* Harvey P. Dale, Jul 03 2017 *)

a(n)=fibonacci(n)^8 \\ Charles R Greathouse IV, Jun 30 2015

a(n) = F(n)^8, F(n)=A000045(n).
G.f.: x*p(8, x)/q(8, x) with p(8, x) := sum_{m=0..7} A056588(7, m)*x^m = (1+x)*(1 - 34*x - 458*x^2 + 2242*x^3 - 458*x^4 - 34*x^5 + x^6) and q(8, x) := sum_{m=0..9} A055870(9, 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) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..9} A055870(9, m)*a(n-m) = 0, n >= 9; inputs: a(n), n=0..8. a(n) = 34*a(n-1) + 714*a(n-2) - 4641*a(n-3) - 12376*a(n-4) + 12376*a(n-5) + 4641*a(n-6) - 714*a(n-7) - 34*a(n-8) + a(n-9).
a(n+1) = 8*F(n)^2*F(n+1)^2*[F(n)^4+F(n+1)^4+4*F(n)^2*F(n+1)^2+3*F(n)*F(n+1)*F(n+2)]-[F(n)^8+F(n+2)^8]+2*[2*F(n+1)^2-(-1)^n]^4 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^8, for n>=0 (This is Theorem 2.2 (vii) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A215044 a(n) = F(2*n)^5 with F=A000045 (Fibonacci numbers).

Original entry on oeis.org

0, 1, 243, 32768, 4084101, 503284375, 61917364224, 7615646045657, 936668172433707, 115202670521319424, 14168993617568728125, 1742671044798615789551, 214334370099947863277568, 26361384861716322814590193

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, A215045 (odd part).

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

Table[Fibonacci[2*n]^5, {n,0,15}] (* Vincenzo Librandi, Sep 02 2012 *)

a(n)=fibonacci(2*n)^5 \\ Charles R Greathouse IV, Oct 16 2015

O.g.f.: x*(1 + 99*x + 416*x^2 + 99*x^3 + x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)), (from the even part of the bisection of A056572).
a(n) = (5*F(4*n) - 4*F(8*n) + F(12*n))/(5^2*L(2*n)), with L=A000032 (Lucas). See the third row in the signed triangle A039598, called in a general comment S.
a(n) = (10*F(2*n) - 5*F(6*n) + F(10*n))/5^2, from the partial fraction decomposition of the o.g.f. - Wolfdieter Lang, Oct 11 2012

A055519 a(n) = 9a(n-1) + 33a(n-2) - 76a(n-3) - 33a(n-4) + 9*a(n-5) + a(n-6), a(0)=a(1)=1, a(2)=2, a(3)=35, a(4)=312, a(5)=3779.

Original entry on oeis.org

1, 1, 2, 35, 312, 3779, 41590, 474169, 5342808, 60450145, 682988978, 7720432691, 87256315920, 986227664411, 11146765278382, 125986353493225, 1423957841588232, 16094263592763889, 181905138292910570, 2055979904686591259, 23237679087969620328, 262643489044489470155

Offset: 0

Barry Cipra, Jul 04 2000

Index entries for linear recurrences with constant coefficients, signature (9,33,-76,-33,9,1).

Cf. A000045, A054894, A055518, A056572, A215928.

LinearRecurrence[{9,33,-76,-33,9,1},{1,1,2,35,312,3779},20] (* Harvey P. Dale, Oct 20 2021 *)

a(n) = Sum_{k=1..n} Fibonacci(k)^5*a(n-k), a(0)=1. - Vladeta Jovovic, Apr 23 2003

G.f.: (x^2+x-1)*(x^2+11*x-1)*(x^2-4*x-1)/(x^6+9*x^5-33*x^4-76*x^3+33*x^2+9*x-1). - Alois P. Heinz, Oct 24 2021

a(0)=1 prepended and edited by Alois P. Heinz, Oct 24 2021

A217471 Partial sum of fifth power of the even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 244, 33012, 4117113, 507401488, 62424765712, 7678070811369, 944346243245076, 116147016764564500, 14285140634333292625, 1756956185432949082176, 216091326285380812359744, 26577476188001703626949937

Offset: 0

Wolfdieter Lang, Oct 11 2012

For the o.g.f. for general powers of Fibonacci numbers F=A000045 see A056588 (row polynomials as numerators) and A055870 (row polynomials as denominator). The even part of the bisection leads to the o.g.f. for powers of F(2*n), and the partial sums of these powers are then given by dividing this o.g.f. by (1-x). For the o.g.f.s for F(n)^5 and F(2*n)^5 see A056572 and A215044, respectively.

The tables of the coefficient of the polynomials which appear in Ozeki's formula and in Melham's conjecture are found in A217472 and A217475, respectively (see References).

			a(2) = 244 = 2*(8-3)/5 - 610/20 + (832040-6765)/55^2 - 7/22.
a(2) = 244 = (1/11)*5^5 - (15/44)*5^3 + (25/44)*5 - 7/22.
a(2) = 244 = (5-1)^2*(4*5^3 + 8*5^2 - 3*5 - 14)/44
           = (4*5^3 + 8*5^2 - 3*5 - 14)*(4/11).

G. C. Greubel, Table of n, a(n) for n = 0..475
R. S. Melham, Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 (2008/2009), no. 3, 207-215.

Cf. A163198 (third powers).

Table[Sum[Fibonacci[2*k]^5, {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Apr 12 2017 *)
Accumulate[Fibonacci[Range[0,30,2]]^5] (* Harvey P. Dale, Jun 30 2025 *)

a(n) = sum(k=1, n, fibonacci(2*k)^5); \\ Michel Marcus, Feb 29 2016

a(n) = Sum_{k=0..n} F(2*k)^5, n>=0.
O.g.f.: x*(1+99*x+416*x^2+99*x^3+x^4)/((1-3*x+x^2)*(1-18*x+x^2)*(1-123*x+x^2)*(1-x)).
a(n) = 2*(F(2*(n+1)) - F(2*n))/5 - F(3*(2*n+1))/20 +
  (F(10*(n+1)) - F(10*n))/F(10)^2 - 7/22 (from the partial fraction decomposition of the o.g.f.).
a(n) = (1/11)*F(2*n+1)^5 - (15/44)*F(2*n+1)^3 + (25/44)*F(2*n+1) - 7/22 (from Ozeki reference, Theorem 2, p. 109 --- with a misprint -- and from Prodinger reference, p. 207).
a(n) =(F(2*n+1)-1)^2*(4*F(2*n+1)^3 + 8*F(2*n+1)^2 - 3*F(2*n+1) - 14)/44 (an example for Melham's conjecture, see the reference, eq. (2.7) for m=2).

A056586 Ninth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 512, 19683, 1953125, 134217728, 10604499373, 794280046581, 60716992766464, 4605366583984375, 350356403707485209, 26623333280885243904, 2023966356928852115753, 153841020405122283630137

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..101
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 (55,1870,-19635,-85085,136136,85085,-19635,-1870,55,1).

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

[Fibonacci(n)^9: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

a(n) = fibonacci(n)^9; \\ Michel Marcus, Sep 06 2017

a(n) = F(n)^9, F(n)=A000045(n).
G.f.: x*p(9, x)/q(9, x) with p(9, x) := sum_{m=0..8} A056588(8, m)*x^m = 1 - 54*x - 1413*x^2 + 9288*x^3 + 17840*x^4 - 9288*x^5 - 1413*x^6 + 54*x^7 + x^8 and q(9, x) := sum_{m=0..10} A055870(10, m)*x^m = (1 - x - x^2)*(1 + 4*x - x^2)*(1 - 11*x - x^2)*(1 + 29*x - x^2)*(1 - 76*x - x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..10} A055870(10, m)*a(n-m) = 0, n >= 10; inputs: a(n), n=0..9. a(n) = 55*a(n-1) + 1870*a(n-2) - 19635*a(n-3) - 85085*a(n-4) + 136136*a(n-5) + 85085*a(n-6) - 19635*a(n-7) - 1870*a(n-8) + 55*a(n-9) + a(n-10).

cp's OEIS Frontend

A056588 Coefficient triangle of certain polynomials.

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A056573 Sixth power of Fibonacci numbers A000045.

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A056574 Seventh power of Fibonacci numbers A000045.

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A056585 Eighth power of Fibonacci numbers A000045.

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A055519 a(n) = 9a(n-1) + 33a(n-2) - 76a(n-3) - 33a(n-4) + 9*a(n-5) + a(n-6), a(0)=a(1)=1, a(2)=2, a(3)=35, a(4)=312, a(5)=3779.