A056571 -id:A056571 - cp's OEIS frontend

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

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}) ] ... ]].

A005969 Sum of fourth powers of Fibonacci numbers.

Original entry on oeis.org

1, 2, 18, 99, 724, 4820, 33381, 227862, 1564198, 10714823, 73457064, 503438760, 3450734281, 23651386922, 162109796922, 1111115037483, 7615701104764, 52198777931900, 357775783071021, 2452231602371646, 16807845698458702

Offset: 1

N. J. A. Sloane

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 19.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Kunle Adegoke, Sums of fourth powers of Fibonacci and Lucas numbers, arXiv:1706.00407 [math.NT], 2017.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (6,10,-30,10,6,-1)

Cf. A001654, A098531, A098532, A098533, A119285, A000071, A005968, A128697.

[(1/25)*(Fibonacci(4*n+2)-(-1)^n*4*Fibonacci(2*n+1)+6*n+3): n in [1..25]];// Vincenzo Librandi, Jun 02 2017

with(combinat): l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+fibonacci(i)^4; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000
A005969:=(z+1)*(z**2-5*z+1)/(z**2-7*z+1)/(z**2+3*z+1)/(z-1)**2; # Simon Plouffe in his 1992 dissertation, offset zero

CoefficientList[Series[(1+x)*(x^2-5*x+1)/((x^2+3*x+1)*(x^2-7*x+1)*(x- 1)^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 02 2017 *)
LinearRecurrence[{6,10,-30,10,6,-1}, {1,2,18,99,724,4820}, 30] (* G. C. Greubel, Jan 17 2018 *)

a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -1,6,10,-30,10,6]^n*[0;1;2;18;99;724])[1,1] \\ Charles R Greathouse IV, Sep 28 2015

a(n) = Sum_{i=0..n} A056571(i).
G.f.: x*(1+x)*(x^2-5*x+1)/ ( (x^2+3*x+1)*(x^2-7*x+1)*(x-1)^2 ). - Ralf Stephan, Apr 23 2004
a(n) = (1/25)*(F(4n+2)-(-1)^n*4*F(2n+1)+6n+3) where F(n)=A000045(n). - Benoit Cloitre, Sep 13 2004. [Corrected by David Lambert (dave.lambert(AT)comcast.net), Mar 28 2008]

More terms from James Sellers, May 29 2000

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

A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.

Original entry on oeis.org

1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000, 1600, 2560, 4096, 6656, 10816, 17576, 28561, 46137, 74529, 120393, 194481, 314874, 509796, 825384, 1336336, 2161720, 3496900, 5656750, 9150625, 14807375, 23961025, 38773295, 62742241, 101515536

Offset: 1

Jeff Burch

Two consecutive Fibonacci numbers are coprime. This sequence satisfies a 14th-order linear difference equation. Note that it is the fourth sequence in the sequences that begin with the Fibonacci numbers, A006498, and A006500. Subsequent sequences will have orders 22, 32, and 44. - T. D. Noe, Mar 05 2012

Also the number of subsets of the set {1,2,...,n-1} which do not contain two elements whose difference is 4. - David Nacin, Mar 07 2012

			Since F_5 = 5 and F_6 = 8 are consecutive Fibonacci numbers, 8^4 = 4096, 8^3*5 = 2560, 8^2*5^2 = 1600, 8*5^3 = 1000, and 5^4 = 625 are in the sequence.
The number 3^3*8 = 216 is not in the sequence since 3 and 8 are not consecutive.
If n = 6 then this gives the number of subsets of {1,...,5} not containing both 1 and 5. There are 2^3 subsets containing 1 and 5, giving us 2^5 - 2^3 = 24. Thus a(5) = 24. - _David Nacin_, Mar 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings, arXiv:2409.00624 [math.CO], 2024. See p. 16.
M. El-Mikkawy and T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=4.
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-2,2,2,0,2,-2,-2,0,1,-1,-1).

Cf. A000045, A006498, A006500.

A031923 := proc(n)
    local n0,i,r,s,m ;
    n0 := n-1 ;
    i := floor(n0/4) ;
    r := combinat[fibonacci](i+2) ;
    s := combinat[fibonacci](i+3) ;
    m := modp(n0,4) ;
    r^(4-m)*s^m ;
end proc:
seq(A031923(n),n=1..50) ; # R. J. Mathar, Jan 23 2022

f = Fibonacci[Range[12]]; m = Most[f]; r = Rest[f]; Union[m^4, m^3 r, m^2 r^2, m r^3] (* T. D. Noe, Mar 05 2012 *)
LinearRecurrence[{1, 1, 0, -2, 2, 2, 0, 2, -2, -2, 0, 1, -1, -1}, {1, 2, 4, 8, 16, 24, 36, 54, 81, 135, 225, 375, 625, 1000}, 40] (* T. D. Noe, Mar 05 2012 *)
Table[Fibonacci[Floor[n/4] + 3]^Mod[n, 4]*Fibonacci[Floor[n/4] + 2]^(4 - Mod[n, 4]), {n, 0, 40}] (* David Nacin, Mar 07 2012 *)
cfn[{a_,b_}]:={a^4,a^3 b,a^2 b^2,a b^3}; Flatten[cfn/@Partition[ Fibonacci[ Range[20]],2,1]]//Union (* Harvey P. Dale, Feb 03 2019 *)

for(m=2,10,r=fibonacci(m);s=fibonacci(m+1);print(r^4," ",r^3*s," ",r^2*s^2," ",r*s^3)) \\ Michael B. Porter, Mar 04 2012

def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:4, 6:15, 7:37, 8:87, 9:200}):
    if n in adict:
        return adict[n]
    adict[n]=3*a(n-1)-2*a(n-2)+2*a(n-3)-4*a(n-4)+2*a(n-5)-2*a(n-6)-4*a(n-7)-a(n-8)+a(n-9)+2*a(n-10)
    return adict[n] # David Nacin, Mar 07 2012

a(n) = F(floor((n-1)/4) + 3)^(n-1 mod 4)*F(floor((n-1)/4) + 2)^(4 - (n-1 mod 4)) where F(n) is the n-th Fibonacci number. - David Nacin, Mar 07 2012
a(n) = a(n-1) + a(n-2) - 2*a(n-4) + 2*a(n-5) + 2*a(n-6) + 2*a(n-8) - 2*a(n-9) - 2*a(n-10) + a(n-12) - a(n-13) - a(n-14). - David Nacin, Mar 07 2012
G.f.: x*(2 + 2*x + 2*x^2 + 4*x^3 + 4*x^4 - 2*x^6 - 1*x^7 - 4*x^8 - 3*x^9 - x^10 - x^11 - 2*x^12 - x^13)/((1 - x)*(1 + x)*(1 + x^2)*(1 - x - x^2)*(1 + 3*x^4 + x^8)). - David Nacin, Mar 08 2012
a(4*k-3) = F(k+1)^4, a(4*k-2) = F(k+1)^3*F(k+2), a(4*k-1) = F(k+1)^2*F(k+2)^2, a(4*k) = F(k+1)*F(k+2)^3, k >= 1, where F = A000045. - Jianing Song, Feb 06 2019
a(4n+1)= A056571(n+2). a(4n+3)=A197424(n). - R. J. Mathar, Jan 23 2022

a(19) changed from 10416 to 10816 by David Nacin, Mar 04 2012

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

A248822 Number of integers k^4 that divide 1!2!3!...n!.

Original entry on oeis.org

1, 1, 1, 2, 3, 8, 10, 36, 64, 200, 432, 630, 1088, 4800, 7590, 32448, 47040, 114240, 164160, 835920, 1302840, 4804800, 7091712, 25243920, 39168000, 171555840, 320973840, 667447200, 1113944832, 3338108928, 5181926400, 19372953600, 31804416000, 132562944000

Offset: 1

Clark Kimberling, Oct 15 2014

			a(6) counts these integers k^4 that divide 24883200:  1^4, 2^4, 4^4, 8^4, 6^4, 12^4, 24^4.

Alois P. Heinz, Table of n, a(n) for n = 1..1000 (first 400 terms from Clark Kimberling)

Cf. A000178, A056571, A248784, A248821, A248823.

b:= proc(n) option remember; add(i[2]*x^numtheory[pi](i[1]),
      i=ifactors(n)[2])+`if`(n=1, 0, b(n-1))
    end:
c:= proc(n) option remember; b(n)+`if`(n=1, 0, c(n-1)) end:
a:= n->(p->mul(iquo(coeff(p, x, i), 4)+1, i=1..degree(p)))(c(n)):
seq(a(n), n=1..30);  # Alois P. Heinz, Oct 16 2014

z = 40; p[n_] := Product[k!, {k, 1, n}];
f[n_] := f[n] = FactorInteger[p[n]];
r[m_, x_] := r[m, x] = m*Floor[x/m]
u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
t[m_, n_] := Apply[Times, 1 + r[m, v[n]]/m]
m = 4; Table[t[m, n], {n, 1, z}] (* A248822 *)

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

A244855 a(n) = Fibonacci(n)^4-1.

Original entry on oeis.org

0, 0, 15, 80, 624, 4095, 28560, 194480, 1336335, 9150624, 62742240, 429981695, 2947295520, 20200652640, 138458409999, 949005240560, 6504586067280, 44583076827135, 305577005139120, 2094455819300624, 14355614096087055, 98394841894789440, 674408281676875200

Offset: 1

Michel Lagneau, Jul 25 2014

For n > 1, a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) with the property F(n-2)+F(n-1)+ F(n+1)+ F(n+2) = F(n) + F(n+3) = 2*F(n+2).
F(n)^2 - 1 = F(n-1)*F(n+1) if n odd, and F(n)^2 - 1 = F(n-2)*F(n+2)if n even ;
F(n)^2 + 1 = F(n-2)*F(n+2) if n odd, and F(n)^2 + 1 = F(n-1)*F(n+1) if n even, hence the product (F(n)^2 - 1)*(F(n)^2 + 1)= F(n-2)*F(n-1)*F(n+1)*F(n+2).

			a(5) = Fibonacci(5)^4-1 = 624 = 3*8*2*13 because Fibonacci(5)^2-1=3*8 and Fibonacci(5)^2+1 = 2*13.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Hideyuki Ohtsuka, Problem B-1174, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 3 (2015), p. 273; A Summation of Recipricals, Solution to Problem B-1174 by Steve Edwards, ibid., Vol. 54, No. 3 (2016), pp. 277-278.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001622, A056571, .

[Fibonacci(n)^4-1: n in [1..30]]; // Vincenzo Librandi, Jul 26 2014

with(numtheory):with(combinat, fibonacci):nn:=30:for i from 1 to nn do:x:=fibonacci(i)^4-1: printf(`%d, `, x):od:

Table[(Fibonacci[n]^4 - 1), {n, 40}] (* Vincenzo Librandi, Jul 26 2014 *)
LinearRecurrence[{5,15,-15,-5,1},{0,0,15,80,624},30] (* Harvey P. Dale, Dec 01 2019 *)

a(n) = fibonacci(n)^4-1; \\ Michel Marcus, Oct 20 2020

From R. J. Mathar, Nov 02 2014: (Start)
G.f.: x^3*(-15-5*x+x^2) / ( (x-1)*(x^2-7*x+1)*(x^2+3*x+1) ).
a(n) = A056571(n)-1. (End)
Sum_{n>=3} 1/a(n) = 35/18 - 5*sqrt(5)/6 = 25/9 - 5*phi/3, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 20 2020
Sum_{n>=3} (-1)^(n+1)/a(n) = 1/18 (Ohtsuka, 2015). - Amiram Eldar, Dec 09 2021

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).

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

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A005969 Sum of fourth powers of Fibonacci numbers.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A056573 Sixth power of Fibonacci numbers A000045.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A031923 Let r and s be consecutive Fibonacci numbers. Sequence is r^4, r^3 s, r^2 s^2, and r s^3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

Extensions

A056574 Seventh power of Fibonacci numbers A000045.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A248822 Number of integers k^4 that divide 1!*2!*3!*...*n!.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

A248822 Number of integers k^4 that divide 1!2!3!...n!.