A081567 Second binomial transform of F(n+1).
1, 3, 10, 35, 125, 450, 1625, 5875, 21250, 76875, 278125, 1006250, 3640625, 13171875, 47656250, 172421875, 623828125, 2257031250, 8166015625, 29544921875, 106894531250, 386748046875, 1399267578125, 5062597656250, 18316650390625, 66270263671875, 239768066406250
Offset: 0
Examples
a(4)=125: 35*(3 + (35 mod 10 - 10 mod 3)/(10-3)) = 35*(3 + 4/7) = 125. - _Bob Selcoe_, Mar 17 2014
References
- R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
- Carolina Benedetti, Christopher R. H. Hanusa, Pamela E. Harris, Alejandro H. Morales, and Anthony Simpson, Kostant's partition function and magic multiplex juggling sequences, arXiv:2001.03219 [math.CO], 2020. See Table 1 p. 12.
- S. Butler and R. Graham, Enumerating (multiplex) juggling sequences, arXiv:0801.2597 [math.CO], 2008.
- P. E. Harris, E. Insko, and L. K. Williams, The adjoint representation of a Lie algebra and the support of Kostant's weight multiplicity formula, arXiv preprint arXiv:1401.0055 [math.RT], 2013.
- Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
- A. Laradji and A. Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
- A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), #04.3.8.
- Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
- D. Nacin, The Minimal Non-Koszul A(Gamma), arXiv preprint arXiv:1204.1534 [math.QA], 2012. - From _N. J. A. Sloane_, Oct 05 2012
- Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042.
- Index entries for linear recurrences with constant coefficients, signature (5,-5).
Crossrefs
Programs
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Magma
I:=[1, 3]; [n le 2 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 27 2012
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Maple
with(GraphTheory):G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=23; n2:=nmax*2+2: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..9); od: seq(a(2*n+1),n=0..nmax); # Johannes W. Meijer, May 29 2010
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Mathematica
Table[MatrixPower[{{2,1},{1,3}},n][[2]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *) LinearRecurrence[{5,-5},{1,3},30] (* Vincenzo Librandi, Feb 27 2012 *)
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PARI
Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014
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Python
def a(n, adict={0:1, 1:3}): if n in adict: return adict[n] adict[n]=5*a(n-1) - 5*a(n-2) return adict[n] # David Nacin, Mar 04 2012
Formula
a(n) = 5*a(n-1) - 5*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 3.
a(n) = (1/2 - sqrt(5)/10) * (5/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2) * (sqrt(5)/2 + 5/2)^n.
G.f.: (1 - 2*x)/(1 - 5*x + 5*x^2).
a(n-1) = Sum_{k=1..n} binomial(n, k)*F(k)^2. - Benoit Cloitre, Oct 26 2003
a(n) = A090041(n)/2^n. - Paul Barry, Mar 23 2004
The sequence 0, 1, 3, 10, ... with a(n) = (5/2 - sqrt(5)/2)^n/5 + (5/2 + sqrt(5)/2)^n/5 - 2(0)^n/5 is the binomial transform of F(n)^2 (A007598). - Paul Barry, Apr 27 2004
From Paul Barry, Nov 15 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, j)*binomial(j+k, 2k);
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, k+j)*binomial(k, k-j)2^(n-k-j);
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(n+k-j, n-k-j)*binomial(k, j)(-1)^j*2^(n-k-j). (End)
a(n) = A111776(n, n). - Abdullahi Umar, Sep 14 2008
a(n) = Sum_{k=0..n} A094441(n,k)*2^k. - Philippe Deléham, Dec 14 2009
a(n+1) = Sum_{k=-floor(n/5)..floor(n/5)} ((-1)^k*binomial(2*n, n+5*k)/2). -Mircea Merca, Jan 28 2012
G.f.: Q(0,u)/x - 1/x, where u=x/(1-2*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
For n>=3: a(n) = a(n-1)*(3+(a(n-1) mod a(n-2) - a(n-2) mod a(n-3))/(a(n-2) - a(n-3))). - Bob Selcoe, Mar 17 2014
a(n) = sqrt(5)^(n-1)*(3*S(n-1, sqrt(5)) - sqrt(5)*S(n-2, sqrt(5))) with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0 and S(-2, x) = -1. This is the (1,1) entry of A^n with the matrix A=(3,1;1,2). See the comment by David Neil McGrath, Nov 18 2014. - Wolfdieter Lang, Dec 04 2014
Conjecture: a(n) = 2*a(n-1) + A039717(n). - Benito van der Zander, Nov 20 2015
a(n) = A189315(n+1) / 10. - Tom Copeland, Dec 08 2015
E.g.f.: exp(5*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024
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