A111262 -id:A111262 - cp's OEIS frontend

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

Paul Barry, Mar 02 2006

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

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).

Cf. A001519, A111262.
Cf. A136376.
Cf. A004146 (number of spanning trees of wheel graph).

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 *)

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

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

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

A124125 a(n)=(1/(4n))sum(k=1,n,F(6k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

Original entry on oeis.org

2, 19, 245, 3631, 58121, 973843, 16773677, 293759095, 5196109073, 92455824667, 1650850175669, 29537478199039, 529130102195225, 9485447592486691, 170110949757514301, 3051485664370912903, 54745886982174938657

Offset: 1

Benoit Cloitre, Nov 29 2006

nonn

Linear recurrence and empirical g.f. confirmed by more terms. - Ray Chandler, Mar 07 2024

Ray Chandler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (35, -383, 1465, -1516, 80).

Cf. A111262.

a(n)=(1/4)*(fibonacci(6*n-3)+4^n*fibonacci(2*n-1)+2*5^(n-1))

a(n)=(1/4)*(F(6n-3)+4^n*F(2n-1)+2*5^(n-1))

Empirical G.f.: -x*(68*x^4-597*x^3+346*x^2-51*x+2) / ((5*x-1)*(x^2-18*x+1)*(16*x^2-12*x+1)). [Colin Barker, Dec 01 2012]

A124126 a(n)=(1/(3n))sum(k=1,n,F(8k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

Original entry on oeis.org

7, 168, 5425, 199367, 7890120, 327681361, 14071534535, 618924449640, 27702229113265, 1255905441590279, 57477374413516680, 2648841480448502353, 122698149590393354375, 5704992303566275023912, 265994788806640480586545

Offset: 1

Benoit Cloitre, Nov 29 2006

nonn

Linear recurrence and empirical g.f. confirmed with more terms. - Ray Chandler, Mar 07 2024

Ray Chandler, Table of n, a(n) for n = 1..599
Index entries for linear recurrences with constant coefficients, signature (132, -6594, 157376, -1866219, 10372992, -22605346, 11987460, -245025).

Cf. A111262.

a(n)=(1/3/n)*sum(k=1,n,fibonacci(8*k)*bernfrac(2*n-2*k)*binomial(2*n,2*k))

a(n)=(1/(3n))*(F(8n-4)+2*L(4n-2)*5^(n-1)+2*F(2n-1)*3^(2n-1)+U(n)) where L=Lucas numbers and U(n) satisfies the order 2 recursion : U(1)=2, U(2)=24, U(n)=23U(n-1)-121U(n-2).

Empirical g.f.: x*(48015*x^7 +9278012*x^6 -12039433*x^5 +3970491*x^4 -510573*x^3 +29407*x^2 -756*x +7) / ((x^2 -47*x +1)*(25*x^2 -35*x +1)*(81*x^2 -27*x +1)*(121*x^2 -23*x +1)). - Colin Barker, Jun 28 2013

cp's OEIS Frontend

A117202 Binomial transform of n*F(n).

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A124125 a(n)=(1/(4n))sum(k=1,n,F(6k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

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A124126 a(n)=(1/(3n))sum(k=1,n,F(8k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

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cp's OEIS Frontend

A117202 Binomial transform of n*F(n).

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Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A124125 a(n)=(1/(4n))*sum(k=1,n,F(6k)*B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

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Author

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Crossrefs

Programs

PARI

Formula

A124126 a(n)=(1/(3n))*sum(k=1,n,F(8k)*B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

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Author

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Comments

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Programs

PARI

Formula

A124125 a(n)=(1/(4n))sum(k=1,n,F(6k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.

A124126 a(n)=(1/(3n))sum(k=1,n,F(8k)B(2n-2k)*binomial(2n,2k)) where F=Fibonacci numbers and B=Bernoulli numbers.