A117202 Binomial transform of n*F(n).

0, 1, 4, 15, 52, 170, 534, 1631, 4880, 14373, 41810, 120406, 343884, 975325, 2749852, 7713435, 21540304, 59917826, 166094370, 458998523, 1264919720, 3477182961, 9536877614, 26102772910, 71309161752, 194468551225, 529490287924

Offset: 0

Author

Paul Barry, Mar 02 2006

Keywords

• easy
• nonn

Comments

Binomial transform of A045925.

Number of acyclic subgraphs of the wheel graph W_n (on n+1 vertices) with exactly n-1 edges. - Emil R. Vaughan, Jun 12 2007

Equivalently, number of two-component spanning forests of the wheel graph W_n (on n+1 vertices). - Harry Richman, Jul 31 2023

Starting (1, 4, 15, 52, ...) = binomial transform of A136376. - Gary W. Adamson, Sep 03 2008

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..1000
• Harry Richman, Farbod Shokrieh, and Chenxi Wu, Counting two-forests and random cut size via potential theory, arXiv:2308.03859 [math.CO], 2023. See p. 18.
• J. Salas and A. D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys. 135 (2009) 279-373; arXiv:0711.1738 [cond-mat.stat-mech], 2007-2009. Mentions this sequence. - _N. J. A. Sloane_, Mar 14 2014
• Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Crossrefs

Cf. A001519, A111262.

Cf. A136376.

Cf. A004146 (number of spanning trees of wheel graph).

Programs

Mathematica

Table[n Fibonacci[2n-1],{n,0,26}] (* or *) Table[Sum[Fibonacci[2k]*BernoulliB[2n-2k]*Binomial[2n,2k],{k,1,n}],{n,0,26}] (* or *) CoefficientList[Series[x(1-2x+2x^2)/(1-3x+x^2)^2 ,{x,0,26}],x] (* Indranil Ghosh, Feb 26 2017 *)

PARI

a(n) = n*fibonacci(2*n-1); \\ Indranil Ghosh, Feb 26 2017

PARI

concat(0, Vec(x*(1-2*x+2*x^2) / (1-3*x+x^2)^2 + O(x^30))) \\ Colin Barker, Feb 26 2017

Formula

G.f.: x*(1-2x+2x^2)/(1-3x+x^2)^2.

a(n) = 6*a(n-1)-11*a(n-2)+6*a(n-3)-a(n-4).

a(n) = Sum_{k=0..n} C(n,k)*k*F(k).

From Benoit Cloitre, Nov 29 2006: (Start)

a(n) = Sum_{k=1..n} F(2k)*B(2n-2k)*binomial(2n,2k) where F=Fibonacci numbers and B=Bernoulli numbers;

a(n) = n*F(2n-1). (End)

a(n) = (2^(-1-n)*(-(-5+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(5+sqrt(5)))*n) / 5. - Colin Barker, Feb 26 2017

a(n) = (1/sqrt(5)) * n * (((1 + sqrt(5)) / 2)^(2*n-1) - ((1 - sqrt(5)) / 2)^(2*n-1)). - Harry Richman, Jul 31 2023

a(n) = round((1/sqrt(5)) * n * phi^(2n-1)), where phi = (1+sqrt(5))/2 is the golden ratio A001622. - Harry Richman, Jul 31 2023