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

Author

Paul Barry, Mar 22 2003

Keywords

• easy
• nonn

Comments

Binomial transform of F(2*n-1), index shifted by 1, where F is A000045. - corrected by Richard R. Forberg, Aug 12 2013

Case k=2 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=1, a(1)=k+1.

Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n+1, s(0) = 3, s(2*n+1) = 4.

a(n+1) gives the number of periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008

a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Sep 14 2008

Counts all paths of length (2*n+1), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010

Given the 3 X 3 matrix M = [1,1,1; 1,1,0; 1,1,3], a(n) = term (1,1) in M^(n+1). - Gary W. Adamson, Aug 06 2010

Number of nonisomorphic graded posets with 0 and 1 of rank n+2, with exactly 2 elements of each rank level between 0 and 1. Also the number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. (This is by Stanley's definition of graded, that all maximal chains have the same length.) - David Nacin, Feb 26 2012

a(n) = 3^n a(n;1/3) = Sum_{k=0..n} C(n,k) * F(k-1) * (-1)^k * 3^(n-k), which also implies the Deleham formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012

The limiting ratio a(n)/a(n-1) is 1 + phi^2. - Bob Selcoe, Mar 17 2014

a(n) counts closed walks on K_2 containing 3 loops on the index vertex and 2 loops on the other. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(3,1; 1,2). - David Neil McGrath, Nov 18 2014

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

a(n) = 5*A052936(n-1), n > 1.

Row sums of A114164.

Cf. A000045, A007051 (INVERTi transform), A007598, A028387, A030191, A039717, A049310, A081568 (binomial transform), A086351 (INVERT transform), A090041, A093129, A094441, A111776, A147748, A178381, A189315.

Programs

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

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

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 *)

PARI

Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014

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

a(n) = A030191(n) - 2*A030191(n-1). - R. J. Mathar, Jul 19 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

a(n) = A093129(n) + A030191(n-1). - Gary W. Adamson, Apr 24 2023

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