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.

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A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

Original entry on oeis.org

1, 3, 15, 60, 260, 1092, 4641, 19635, 83215, 352440, 1493064, 6324552, 26791505, 113490195, 480752895, 2036500788, 8626757644, 36543528780, 154800876945, 655747029795, 2777789007071, 11766903040368, 49845401197200, 211148507782800, 894439432403425
Offset: 0

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Author

Keywords

Comments

In a triangle having sides of F(n+1), 2*F(n+2) and F(n+3), the product of the area and circumradius will be a(n). For example: a triangle having sides of 5, 16 and 13 will have an area of 4*sqrt(51), a circumradius of 65*sqrt(51)/51, and the product is 4*65 = 260. - Gary Detlefs, Dec 14 2010
Explanation of this comment: if a triangle with sides (a, b, c) has a circumradius R and an area A, then A*R = abc/4; here, with a = F(n+1), b=2*F(n+2) and c=F(n+3), this gives a(n)= A*R. - Bernard Schott, Jan 26 2023

Examples

			G.f. = 1 + 3*x + 15*x^2 + 60*x^3 + 260*x^4 + 1092*x^5 + 4641*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

Cf. A066258 (first differences), A215037 (partial sums), A363753 (alternating sums).

Programs

  • Magma
    [Fibonacci(n+3)*Fibonacci(n+2)*Fibonacci(n+1)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2016
  • Maple
    A001655:=1/(z**2-z-1)/(z**2+4*z-1); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1])/2, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
    LinearRecurrence[{3, 6, -3, -1}, {1, 3, 15, 60}, 25] (* Jean-François Alcover, Sep 23 2017 *)
  • PARI
    b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(20, n, b(n-1, 3))  \\ Joerg Arndt, May 08 2016
    

Formula

G.f.: 1/(1-3*x-6*x^2+3*x^3+x^4) = 1/((1+x-x^2)*(1-4*x-x^2)) (see Comments to A055870).
a(n) = A010048(n+3, 3) = fibonomial(n+3, 3).
a(n) = (1/2) * A065563(n).
a(n) = 4*a(n-1) + a(n-2) + ((-1)^n)*F(n+1), n >= 2; a(0)=1, a(1)=3.
a(n) = (F(n+3)^3 - F(n+2)^3 - F(n+1)^3)/6. - Gary Detlefs, Dec 24 2010
a(n-1) = Sum_{k=0..n} F(k+1)*F(k)^2, n >= 1. - Wolfdieter Lang, Aug 01 2012
From Wolfdieter Lang, Aug 09 2012: (Start)
a(n-1)*(-1)^n = Sum_{k=0..n} (-1)^k*F(k+1)^2*F(k), n >= 1. See the link under A215037, eq. (25).
a(n) = (F(3*(n+2)) + 2*(-1)^n*F(n+2))/10, n >= 0. See the same link, eq. (32). (End)
a(n) = -a(-4-n)*(-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - a(n+2)) + a(n+1)*(-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
O.g.f.: exp( Sum_{n >= 1} L(n)*L(2*n)*x^n/n ), where L(n) = A000032(n) is a Lucas number. Cf. A114525, A256178. - Peter Bala, Mar 18 2015
Sum_{n>=0} (-1)^n/a(n) = 2 * A079586 - 6. - Amiram Eldar, Oct 04 2020
The formula by Gary Detlefs above is valid for all sequences of the Fibonacci type f(n) = f(n-1) + f(n-2): 3*f(n+2)*f(n+1)*f(n) = f(n+2)^3 - f(n+1)^3 - f(n)^3. - Klaus Purath, Mar 25 2021
a(n) = sqrt(Sum_{j=1..n+1} F(j)^3*F(j+1)^3). See Treeby link. - Michel Marcus, Apr 10 2022
a(n) = Sum_{k=1..n+1} A000129(k)*A056570(n+2-k). - Michael A. Allen, Jan 25 2023
G.f.: exp( Sum_{k>=1} F(4*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A215037 a(n) = Sum_{k=0..n} fibonomial(k+3,3), n >= 0.

Original entry on oeis.org

1, 4, 19, 79, 339, 1431, 6072, 25707, 108922, 461362, 1954426, 8278978, 35070483, 148560678, 629313573, 2665814361, 11292572005, 47836100785, 202636977730, 858384007525, 3636173014596, 15403076054964, 65248477252164
Offset: 0

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Author

Wolfdieter Lang, Aug 09 2012

Keywords

Comments

This sum is obtained from the m=2 member of the m-family of sums s(m;n) := Sum_{k=0..n} F(k+m)*F(k+1)*F(k), n>=0, given by
(F(n+m)*F(n+2)*F(n+1) - (-1)^n*F(m)*A008346(n))/2 with A008346(n) = (F(n) + (-1)^n), where F = A000045.
The formula for s(m;n), m>=0, n>=0, follows by induction on m, using the sums for m=0 and m=1. s(0,n) = F(n+1)*(F(n+1)^2 - (-1)^n)/2 = F(n+2)*F(n+1)*F(n)/2 (see A001655(n-1)), and s(1,n) = (F(n+2)*F(n+1)^2 - (-1)^n*A008346(n))/2 (see A215038). For the formulas for s(0,n) and s(1,n) see also the link on partial summation, eqs. (6) and (7). There Sum_{k=0..n} fibonomial(k+2,k) is obtained more directly in eq. (5) with the help of the partial summation formula.

Examples

			a(3) = 2*1*1/2 + 3*2*1/2 + 5*3*2/2 + 8*5*3/2 = 1 + 3 + 15 + 60 = 79.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{4, 3, -9, 2, 1}, {1, 4, 19, 79, 339}, 23] (* Hans J. H. Tuenter, Jun 26 2023 *)

Formula

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=0..n} F(k+3)*F(k+2)*F(k+1)/2.
a(n) = (F(n+3)^2*F(n+2) + (-1)^n*A008346(n+1))/4, n>=0, with A008346(n) = F(n) + (-1)^n. See a comment above.
G.f.: 1/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from the g.f. of the Fibonomials A001655).
From Hans J. H. Tuenter, Jun 26 2023: (Start)
a(n) = (F(n+3)^2*F(n+2) + (-1)^n*F(n+1)-1)/4.
a(n) = (F(n+3)^3 + F(n+2)^3 + (-1)^n*F(n+1) - 2)/8.
a(n) = (F(3*n+8) + 4*(-1)^n*F(n+1) - 5)/20.
a(n) = 4*a(n-1) + 3*a(n-2) - 9*a(n-3) + 2*a(n-4) + a(n-5).
a(-n) = A363753(n-3).
(End)

A363754 a(n) = Sum_{k=0..n} F(2k-1)*F(2k)*F(2k+1)/2, where F(n) is the Fibonacci number A000045(n).

Original entry on oeis.org

0, 1, 16, 276, 4917, 88132, 1581196, 28372701, 509125596, 9135883240, 163936760185, 2941725767256, 52787126964456, 947226559367881, 16997290941068152, 305004010378316172, 5473074895864584141, 98210344115173624636, 1762313119177232976916, 31623425801074947486405
Offset: 0

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Author

Hans J. H. Tuenter, Jun 19 2023

Keywords

Comments

This is one of the triple Fibonacci sums that were considered by Subba Rao (1953).
Taking any of the given closed-form expressions for a(n) with Fibonacci numbers, one can extend a(n) to negative indices by using the property F(-n)=(-1)^(n+1). This gives a(-n)=a(n-1).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{22, -77, 77, -22, 1}, {0, 1, 16, 276, 4917}]

Formula

a(n) = (F(2n+1)^3 + F(2n+1) - 2)/8.
a(n) = (F(6*n+3)+8*F(2*n+1)-10)/40.
a(n) = 22*a(n-1) - 77*a(n-2) + 77*a(n-3) - 22*a(n-4) + a(n-5).
G.f.: x*(1 - 6*x + x^2)/((1 - x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).
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