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 1-6 of 6 results.

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

Original entry on oeis.org

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760
Offset: 0

Views

Author

Keywords

Comments

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with a
a(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

Examples

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a001654 n = a001654_list !! n
    a001654_list = zipWith (*) (tail a000045_list) a000045_list
    -- Reinhard Zumkeller, Jun 08 2013
    
  • Magma
    I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018
  • Maple
    with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
    seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007
  • Mathematica
    LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
    Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
    Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)
  • PARI
    A001654(n)=fibonacci(n)*fibonacci(n+1);
    
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016
    
  • Python
    from sympy import fibonacci as F
    def a(n): return F(n)*F(n + 1)
    [a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017
    
  • Python
    from math import prod
    from gmpy2 import fib2
    def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022
    

Formula

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Extended by Wolfdieter Lang, Jun 27 2000

A010048 Triangle of Fibonomial coefficients, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 6, 3, 1, 1, 5, 15, 15, 5, 1, 1, 8, 40, 60, 40, 8, 1, 1, 13, 104, 260, 260, 104, 13, 1, 1, 21, 273, 1092, 1820, 1092, 273, 21, 1, 1, 34, 714, 4641, 12376, 12376, 4641, 714, 34, 1, 1, 55, 1870, 19635, 85085, 136136, 85085, 19635, 1870, 55, 1
Offset: 0

Keywords

Comments

Conjecture: polynomials with (positive) Fibonomial coefficients are reducible iff n odd > 1. - Ralf Stephan, Oct 29 2004

Examples

			First few rows of the triangle T(n, k) are:
  n\k 0   1    2     3     4      5      6      7     8   9  10
   0: 1
   1: 1   1
   2: 1   1    1
   3: 1   2    2     1
   4: 1   3    6     3     1
   5: 1   5   15    15     5      1
   6: 1   8   40    60    40      8      1
   7: 1  13  104   260   260    104     13      1
   8: 1  21  273  1092  1820   1092    273     21     1
   9: 1  34  714  4641 12376  12376   4641    714    34   1
  10: 1  55 1870 19635 85085 136136  85085  19635  1870  55   1
... - Table extended and reformatted by _Wolfdieter Lang_, Oct 10 2012
For n=7 and k=3, n - k + 1 = 7 - 3 + 1 = 5, so T(7,3) = F(7)*F(6)*F(5)/( F(3)*F(2)*F(1)) = 13*8*5/(2*1*1) = 520/2 = 260. - _Michael B. Porter_, Sep 26 2016
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 15.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 84 and 492.

Crossrefs

Cf. A055870 (signed version of triangle).
Sums include: A056569 (row), A181926 (antidiagonal), A181927 (row square-sums).
Cf. A003267 and A003268 (central Fibonomial coefficients), A003150 (Fibonomial Catalan numbers), A144712, A099927, A385732/A385733 (Lucas).

Programs

  • Magma
    Fibonomial:= func< n,k | k eq 0 select 1 else (&*[Fibonacci(n-j+1)/Fibonacci(j): j in [1..k]]) >;
    [Fibonomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 20 2024
    
  • Maple
    A010048 := proc(n,k)
        mul(combinat[fibonacci](i),i=n-k+1..n)/mul(combinat[fibonacci](i),i=1..k) ;
    end proc:
    seq(seq(A010048(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 05 2015
  • Mathematica
    f[n_, k_] := Product[ Fibonacci[n - j + 1]/Fibonacci[j], {j, k}]; Table[ f[n, i], {n, 0, 10}, {i, 0, n}] (* Robert G. Wilson v, Dec 04 2009 *)
    Column[Round@Table[GoldenRatio^(k(n-k)) QBinomial[n, k, -1/GoldenRatio^2], {n, 0, 10}, {k, 0, n}], Center] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Sep 25 2016 *)
    T[n_, k_] := With[{c = ArcCsch[2] - I Pi/2}, Product[I^j Sinh[c j], {j, k + 1, n}] / Product[I^j Sinh[c j], {j, 1, n - k}]]; Table[Simplify[T[n, k]], {n, 0, 10}, {k, 0, n}] // Flatten  (* Peter Luschny, Jul 08 2025 *)
  • Maxima
    ffib(n):=prod(fib(k),k,1,n);
    fibonomial(n,k):=ffib(n)/(ffib(k)*ffib(n-k));
    create_list(fibonomial(n,k),n,0,20,k,0,n); /* Emanuele Munarini, Apr 02 2012 */
    
  • PARI
    T(n, k) = prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j));
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018
    
  • SageMath
    def fibonomial(n,k): return 1 if k==0 else product(fibonacci(n-j+1)/fibonacci(j) for j in range(1,k+1))
    flatten([[fibonomial(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 20 2024

Formula

T(n, k) = ((n, k)) = (F(n)*F(n-1)*...*F(n-k+1))/(F(k)*F(k-1)*...*F(1)), F(i) = Fibonacci numbers A000045.
T(n, k) = Fibonacci(n-k-1)*T(n-1, k-1) + Fibonacci(k+1)*T(n-1, k).
T(n, k) = phi^(k*(n-k)) * C(n, k)A001622%20is%20the%20golden%20ratio,%20and%20C(n,%20k)_q%20is%20the%20q-binomial%20coefficient.%20-%20_Vladimir%20Reshetnikov">{-1/phi^2}, where phi = (1+sqrt(5))/2 = A001622 is the golden ratio, and C(n, k)_q is the q-binomial coefficient. - _Vladimir Reshetnikov, Sep 26 2016
G.f. of column k: x^k * exp( Sum_{j>=1} Fibonacci((k+1)*j)/Fibonacci(j) * x^j/j ). - Seiichi Manyama, May 07 2025
T(n, k) = Product_{j=k+1..n} i^j*sinh(c*j) / Product_{j=1..n-k} i^j*sinh(c*j) where c = arccsch(2) - i*Pi/2 and i is the imaginary unit. If you substitute sinh by cosh you get the Lucas triangle A385732/A385733, which is a rational triangle. - Peter Luschny, Jul 08 2025

A055870 Signed Fibonomial triangle.

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -2, -2, 1, 1, -3, -6, 3, 1, 1, -5, -15, 15, 5, -1, 1, -8, -40, 60, 40, -8, -1, 1, -13, -104, 260, 260, -104, -13, 1, 1, -21, -273, 1092, 1820, -1092, -273, 21, 1, 1, -34, -714, 4641, 12376, -12376, -4641, 714, 34, -1, 1, -55, -1870, 19635, 85085, -136136, -85085, 19635, 1870, -55, -1
Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

Row n+1 (n >= 1) of the signed triangle lists the coefficients of the recursion relation for the n-th power of Fibonacci numbers A000045: Sum_{m=0..n+1} T(n+1,m)*(Fibonacci(k-m))^n = 0, k >= n+1; inputs: (Fibonacci(k))^n, k=0..n.
The inverse of the row polynomial p(n,x) := Sum_{m=0..n} T(n,m)*x^m is the g.f. for the column m=n-1 of the Fibonomial triangle A010048.
The row polynomials p(n,x) factorize according to p(n,x) = G(n-1)*p(n-2,-x), with inputs p(0,x)= 1, p(1,x)= 1-x and G(n):= 1 - A000032(n)*x + (-1)^n*x^2. (Derived from Riordan's result and Knuth's exercise).
The row polynomials are the characteristic polynomials of product of the binomial matrix binomial(i,j) and the exchange matrix J_n (matrix with 1's on the antidiagonal, 0 elsewhere). - Paul Barry, Oct 05 2004

Examples

			Row polynomial for n=4: p(4,x) = 1-3*x-6*x^2+3*x^3+x^4 = (1+x-x^2)*(1-4*x-x^2). 1/p(4,x) is G.f. for A010048(n+3,3), n >= 0: {1,3,15,60,...} = A001655(n).
For n=3: 1*(Fibonacci(k))^3 - 3*(Fibonacci(k-1))^3 - 6*(Fibonacci(k-2))^3 + 3*(Fibonacci(k-3))^3 + 1*(Fibonacci(k-4))^3 = 0, k >= 4; inputs: (Fibonacci(k))^3, k=0..3.
The triangle begins:
  n\m 0   1     2    3     4      5     6    7   8   9
  0   1
  1   1  -1
  2   1  -1    -1
  3   1  -2    -2    1
  4   1  -3    -6    3     1
  5   1  -5   -15   15     5     -1
  6   1  -8   -40   60    40     -8    -1
  7   1 -13  -104  260   260   -104   -13    1
  8   1 -21  -273 1092  1820  -1092  -273   21   1
  9   1 -34  -714 4641 12376 -12376 -4641  714  34  -1
  ... [_Wolfdieter Lang_, Aug 06 2012; a(7,1) corrected, Oct 10 2012]
		

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, pp. 84-5 and 492.

Crossrefs

Sums include: A055871 (signed row), A056569 (row).
Central column: A003268.
Cf. A383715.

Programs

  • Magma
    Fibonomial:= func< n,k | k eq 0 select 1 else (&*[Fibonacci(n-j+1)/Fibonacci(j): j in [1..k]]) >;
    [(-1)^Floor((k+1)/2)*Fibonomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 20 2024
    
  • Maple
    A055870 := proc(n,k)
        (-1)^floor((k+1)/2)*A010048(n,k) ;
    end proc: # R. J. Mathar, Jun 14 2015
  • Mathematica
    T[n_, m_]:= {1,-1,-1,1}[[Mod[m,4] + 1]] * Product[ Fibonacci[n-j+1]/Fibonacci[j], {j, m}];
    Table[T[n, m], {n, 0, 10}, {m, 0, n}]//Flatten (* Jean-François Alcover, Jul 05 2013 *)
  • SageMath
    def fibonomial(n,k): return 1 if k==0 else product(fibonacci(n-j+1)/fibonacci(j) for j in range(1,k+1))
    flatten([[(-1)^((k+1)//2)*fibonomial(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 20 2024

Formula

T(n, m) = (-1)^floor((m+1)/2)*A010048(n, m), where A010048(n, m) := fibonomial(n, m).
G.f. for column m: (-1)^floor((m+1)/2)*x^m/p(m+1, x) with the row polynomial of the (signed) triangle: p(n, x) := Sum_{m=0..n} T(n, m)*x^m.
Sum_{k=0..n} T(n,k) * x^k = exp( -Sum_{k>=1} Fibonacci(n*k)/Fibonacci(k) * x^k/k ). - Seiichi Manyama, May 07 2025

A001656 Fibonomial coefficients.

Original entry on oeis.org

1, 5, 40, 260, 1820, 12376, 85085, 582505, 3994320, 27372840, 187628376, 1285992240, 8814405145, 60414613805, 414088493560, 2838203264876, 19453338487220, 133335155341960, 913892777190965, 6263914210945105
Offset: 0

Keywords

Examples

			G.f. = 1 + 5*x + 40*x^2 + 260*x^3 + 1820*x^4 + 12376*x^5 + 85085*x^6 + ... .
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Programs

  • Maple
    with (combinat): a:=n->1/6*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3): seq(a(n), n=1..18); # Zerinvary Lajos, Oct 07 2007
    A001656:=-1/(z-1)/(z**2-7*z+1)/(z**2+3*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1]*Fibonacci[n])/6,{n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
    LinearRecurrence[{5,15,-15,-5,1},{1,5,40,260,1820},20] (* Vincenzo Librandi, Aug 02 2012 *)
    Times@@@Partition[Fibonacci[Range[30]],4,1]/6 (* Harvey P. Dale, Oct 13 2016 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 4))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = ((4+n, 4)) (see A010048), or fibonomial(4+n, 4).
G.f.: 1/(1-5*x-15*x^2+15*x^3+5*x^4-x^5) = 1/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)) (see Comments to A055870). a(n)= 7*a(n-1)-a(n-2)+((-1)^n)*fibonomial(n+2, 2), n >= 2; a(0)=1, a(1)=5; fibonomial(n+2, 2)= A001654(n+1).
a(n) = Product_{k=1..n} Fibonacci(k+4)/Fibonacci(k). - Gary Detlefs, Feb 06 2011
a(n) = (F(n+3)^2-F(n+2)^2)*F(n+3)*F(n+2)/6, where F(n) is the n-th Fibonacci number. - Gary Detlefs, Oct 12 2011
a(n) = a(-5-n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - 2*a(n+2)) + a(n+1)*(-5*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
From Peter Bala, Mar 30 2015: (Start)
The o.g.f. A(x) = 1/(1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5). Hence A(x) (mod 25) = 1/(1 - 5*x + 10*x^2 - 10^x^3 + 5*x^4 - x^5) (mod 25) = 1/(1 - x)^5 (mod 25). It follows by Theorem 1 of Heninger et al. that A(x)^(1/5) = 1 + x + 6*x^2 + 26*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Fibonacci(5*n)/Fibonacci(n)*x^n/n ). Cf. A084175, A099930. (End)
Sum_{n>=0} 1/a(n) = 51/2 - 15*phi, where phi is the golden ratio (A001622) (Koshy, 2022, section 3.3, p. 9). - Amiram Eldar, Jan 23 2025

Extensions

Corrected and extended by Wolfdieter Lang, Jun 27 2000
More terms from Vladimir Joseph Stephan Orlovsky, Nov 23 2009

A001657 Fibonomial coefficients: column 5 of A010048.

Original entry on oeis.org

1, 8, 104, 1092, 12376, 136136, 1514513, 16776144, 186135312, 2063912136, 22890661872, 253854868176, 2815321003313, 31222272414424, 346260798314872, 3840089017377228, 42587248616222024, 472299787252290712, 5237885063192296801, 58089034826620525728
Offset: 0

Keywords

Examples

			G.f. = 1 + 8*x + 104*x^2 + 1092*x^3 + 12376*x^4 + 136136*x^5 + 1514513*x^6 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(combinat) : a:=n-> 1/30*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3)*fibonacci(n+4): seq(a(n), n=1..19); # Zerinvary Lajos, Oct 07 2007
    A001657:=-1/(z**2+11*z-1)/(z**2-4*z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_] := Times @@ Fibonacci[Range[n + 1, n + 5]]/30; t = Table[f[n], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
    LinearRecurrence[{8,40,-60,-40,8,1},{1,8,104,1092,12376,136136},20] (* Harvey P. Dale, Nov 30 2019 *)
  • PARI
    a(n)=(n->(n^5-n)/30)(fibonacci(n+3)) \\ Charles R Greathouse IV, Apr 24 2012
    
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 5))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(5+n, 5) (or fibonomial(5+n, 5)).
G.f.: 1/(1-8*x-40*x^2+60*x^3+40*x^4-8*x^5-x^6) = 1/((1-x-x^2)*(1+4*x-x^2)*(1-11*x-x^2)) (see Comments to A055870).
a(n) = 11*a(n-1) + a(n-2) + ((-1)^n)*fibonomial(n+3, 3), n >= 2; a(0)=1, a(1)=8; fibonomial(n+3, 3)= A001655(n).
a(n) = Fibonacci(n+3)*(Fibonacci(n+3)^4-1)/30. - Gary Detlefs, Apr 24 2012
a(n) = (A049666(n+3) + 2*(-1)^n*A001076(n+3) - 3*A000045(n+3))/150, n >= 0, with A049666(n) = F(5*n)/5, A001076(n) = F(3*n)/2 and A000045(n) = F(n). From the partial fraction decomposition of the o.g.f. and recurrences. - Wolfdieter Lang, Aug 23 2012
a(n) = a(-6-n) * (-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - 3*a(n+2)) + a(n+1)*(-8*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
G.f.: exp( Sum_{k>=1} F(6*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extensions

Corrected and extended by Wolfdieter Lang, Jun 27 2000

A056566 Fibonomial coefficients.

Original entry on oeis.org

1, 34, 1870, 83215, 3994320, 186135312, 8771626578, 411591708660, 19344810307020, 908637119420910, 42689423937884208, 2005443612183077232, 94214069697350815795, 4426039514623184676790, 207929935924379904006970, 9768275694729434277258589, 458901121999204061365680096
Offset: 0

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Crossrefs

Cf. A010048, A000045, A001654-8, A056565, A001906 (signed), A004187, A049660 (signed), A049668.

Programs

  • Mathematica
    a[n_] := (1/65520) Times @@ Fibonacci[n + Range[8]]; Array[a, 20, 0] (* Giovanni Resta, May 08 2016 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
    vector(20, n, b(n-1, 8)) \\ Joerg Arndt, May 08 2016

Formula

a(n) = A010048(n+8, 8) = Fibonomial(n+8, 8).
G.f.: 1/p(9, n) with p(9, n)= 1 - 34*x - 714*x^2 + 4641*x^3 + 12376*x^4 - 12376*x^5 - 4641*x^6 + 714*x^7 + 34*x^8 - x^9 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)* (1 + 18*x + x^2)*(1 - 47*x + x^2) (n=9 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references).
Recursion: a(n) = 47*a(n-1) - a(n-2) + ((-1)^n)*A001658(n), n >= 2, a(0)=1, a(1)=34.
G.f.: exp( Sum_{k>=1} F(9*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025
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