cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

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Author

Ralf Stephan, Feb 02 2005

Keywords

Comments

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).

Examples

			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, ...
  ...
		

References

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

Crossrefs

Main diagonal gives A100399.
Cf. A244003.

Programs

  • Maple
    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
  • Mathematica
    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

Formula

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}) ] ... ]].

A105317 Powers of Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 9, 13, 16, 21, 25, 27, 32, 34, 55, 64, 81, 89, 125, 128, 144, 169, 233, 243, 256, 377, 441, 512, 610, 625, 729, 987, 1024, 1156, 1597, 2048, 2187, 2197, 2584, 3025, 3125, 4096, 4181, 6561, 6765, 7921, 8192, 9261, 10946, 15625, 16384, 17711
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 25 2005

Keywords

Comments

The subset of nontrivial Fibonacci powers [numbers A000045(k)^n which are not in A000045] starts 4, 9, 16, 25, 27, 32, 64, 81, 125, 128, 169, 243, 256, 441, 512, 625, 729, 1024, 1156... - R. J. Mathar, Jan 26 2015. These are the initial terms of A254719. - Reinhard Zumkeller, Feb 06 2015

Examples

			2197 = 13^3 = A000045(7)^3, therefore 2197 is a term.
		

Crossrefs

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a105317 n = a105317_list !! (n-1)
    a105317_list = 0 : 1 : h 1 (drop 4 a000045_list) (singleton (2, 2)) where
      h y xs'@(x:xs) s
        | x < ff    = h y xs (insert (x, x) s)
        | ff == y   = h y xs' s'
        | otherwise = ff : h ff xs' (insert (f * ff, f) s')
        where ((ff, f), s') = deleteFindMin s
    -- Reinhard Zumkeller, Feb 06 2015
    
  • Maple
    N:= 10^6: # to get all terms <= N
    select(`<=`,{0,1,seq(seq(combinat:-fibonacci(i)^j, i = 3 ..floor(log[phi](sqrt(5)*N^(1/j)+1))),j=1..ilog2(N))},N);
    # if using Maple 11 or earlier, uncomment the next line
    # sort(convert(%,list)); # Robert Israel, Jan 26 2015
  • Mathematica
    lim = 10^5; t = Table[f = Fibonacci[n]; f^Range[Floor[Log[lim]/Log[f]]], {n, 3, Ceiling[Log[GoldenRatio, lim] + 1]}]; Union[{0, 1}, Flatten[t]] (* T. D. Noe, Sep 27 2011 *)
  • PARI
    list(lim)=my(v=List([0]),k=1,f,t); while(k<=lim, listput(v,k); k*=2); k=3; while(k<=lim, listput(v,k); k*=3); k=5; while(k<=lim, listput(v,k); k*=5); k=6; while((f=fibonacci(k++))<=lim, t=1; while((t*=f)<=lim, listput(v,t))); Set(v) \\ Charles R Greathouse IV, Oct 03 2016

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

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

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).

Crossrefs

Programs

Formula

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

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

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

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).

Crossrefs

Programs

Formula

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

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

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).

Crossrefs

Programs

Formula

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).

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jul 17 2001
Showing 1-5 of 5 results.