A099920 -id:A099920 - cp's OEIS frontend

A099924 Self-convolution of Lucas numbers.

4, 4, 13, 22, 45, 82, 152, 274, 491, 870, 1531, 2676, 4652, 8048, 13865, 23798, 40713, 69446, 118144, 200510, 339559, 573894, 968183, 1630632, 2742100, 4604572, 7721797, 12933334, 21637221, 36159610, 60367976, 100687786

Offset: 0

Ralf Stephan, Nov 01 2004

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 57.

Michael De Vlieger, Table of n, a(n) for n = 0..4767
É. Czabarka, R. Flórez, and L. Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), #15.1.6.
Sergio Falcon, Half self-convolution of the k-Fibonacci sequence, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 96-106.
Tamás Szakács, Convolution of second order linear recursive sequences. II. Commun. Math. 25, No. 2, 137-148 (2017). See remark 4.
Tamás Szakács, Linear recursive sequences and factorials, Ph. D. Thesis, Univ. Debrecen (Hungary, 2024). See p. 37.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A001629, A000032. Bisection: A203573 (even), 2*A203574 (odd).

Table[Sum[LucasL[k]LucasL[n-k],{k,0,n}],{n,0,40}] (* or *) LinearRecurrence[ {2,1,-2,-1},{4,4,13,22},40] (* Harvey P. Dale, Mar 06 2012 *)

a(n) = (n+1)*L(n) + 2F(n+1) = Sum_{k=0..n} L(k)*L(n-k).
G.f.: (2-x)^2/(1-x-x^2)^2, corrected Aug 23 2022
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), a(0)=4, a(1)=4, a(2)=13, a(3)=22. - Harvey P. Dale, Mar 06 2012
a(n) = 2*A099920(n+1)-A099920(n). - R. J. Mathar, Aug 23 2022

A264147 a(n) = nF(n+1) - (n+1)F(n), where F = A000045.

Original entry on oeis.org

0, -1, 1, 1, 5, 10, 22, 43, 83, 155, 285, 516, 924, 1639, 2885, 5045, 8773, 15182, 26162, 44915, 76855, 131119, 223101, 378696, 641400, 1084175, 1829257, 3081193, 5181893, 8702290, 14594830, 24446971, 40902299, 68359619, 114132765, 190373580, 317258388, 528265207

Offset: 0

Bruno Berselli, Nov 04 2015

a(n) is prime for n = 4, 7, 8, 26, 28, 52, 86, 87, 93, 97, 158, 196, 303, 2908, 3412, 4111, 4208, 6183, 6337, 9878, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A001629, A007502, A354044.
Cf. A178521: n*F(n+1) + (n+1)*F(n).
Cf. A094588: n*F(n-1) + F(n).
Cf. A099920: Sum_{i=0..n} F(i)*L(n-i).
Cf. A023607: Sum_{i=0..n} F(i)*L(n+1-i).

# The function 'fibrec' is defined in A354044.
function A264147(n)
    n == 0 && return BigInt(0)
    a, b = fibrec(n)
    n*b - a*(n + 1)
end # Peter Luschny, May 16 2022

[n*Fibonacci(n+1)-(n+1)*Fibonacci(n): n in [0..40]];

A264147 := proc(n)
    n*combinat[fibonacci](n+1)-(n+1)*combinat[fibonacci](n) ;
end proc:
seq(A264147(n),n=0..10) ; # R. J. Mathar, Jun 02 2022

Table[n Fibonacci[n + 1] - (n + 1) Fibonacci[n], {n, 0, 40}]

makelist(n*fib(n+1)-(n+1)*fib(n), n, 0, 40);

for(n=0, 40, print1(n*fibonacci(n+1)-(n+1)*fibonacci(n)", "));

concat(0, Vec(-x*(1 - 3*x) / (1 - x - x^2)^2 + O(x^50))) \\ Colin Barker, Jul 27 2017

[n*fibonacci(n+1)-(n+1)*fibonacci(n) for n in (0..40)]

G.f.:  x*(-1 + 3*x)/(1 - x - x^2)^2.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4).
a(n) = n*F(n-1) - F(n).
a(n) = Sum_{i=0..n} F(i)*L(n-1-i), where L() is a Lucas number (A000032).
a(n) = 3*A001629(n) - A001629(n+1).
a(n) = -(-1)^n*A178521(-n).
a(n+2) - a(n) = A007502(n+1).
Sum_{i>0} 1/a(i) = 1.39516607051636028893879220294180374...
a(n) = (-((1+sqrt(5))/2)^n*(2*sqrt(5) + (-5+sqrt(5))*n) + ((1-sqrt(5))/2)^n*(2*sqrt(5) + (5+sqrt(5))*n)) / 10. - Colin Barker, Jul 27 2017
a(n) = (-i)^n*(n*sin(c*(n+1)) - (n+1)*sin(c*n)*i)/sqrt(5/4) where c = arccos(i/2). - Peter Luschny, May 16 2022

A122076 Coefficients of a generalized Jaco-Lucas polynomial (even indices) read by rows.

Original entry on oeis.org

2, 3, 2, 7, 8, 2, 18, 30, 15, 2, 47, 104, 80, 24, 2, 123, 340, 355, 170, 35, 2, 322, 1068, 1410, 932, 315, 48, 2, 843, 3262, 5208, 4396, 2079, 532, 63, 2, 2207, 9760, 18280, 18784, 11440, 4144, 840, 80, 2, 5778, 28746, 61785, 74838, 55809, 26226, 7602, 1260

Offset: 0

R. J. Mathar, Oct 16 2006

Row sums give A052539. - Franck Maminirina Ramaharo, Jul 09 2018

Alternating row sums seem to be 1, except when n=0. - F. Chapoton, Nov 09 2021

			The triangle T(n,k) begins:
n\k:    0      1       2       3       4       5      6      7     8    9 10
0:      2
1:      3      2
2:      7      8       2
3:     18     30      15       2
4:     47    104      80      24       2
5:    123    340     355     170      35       2
6:    322   1068    1410     932     315      48      2
7:    843   3262    5208    4396    2079     532     63      2
8:   2207   9760   18280   18784   11440    4144    840     80     2
9:   5778  28746   61785   74838   55809   26226   7602   1260    99    2
10: 15127  83620  202840  282980  249815  144488  54690  13080  1815  120  2
... reformatted and extended. - _Franck Maminirina Ramaharo_, Jul 09 2018

Muniru A Asiru, Table of n, a(n) for n = 0..1325 (offset adapted by _Georg Fischer_, Jan 31 2019).
Franck Ramaharo, A one-variable bracket polynomial for some Turk's head knots, arXiv:1807.05256 [math.CO], 2018.
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, (2005) 359-370, Table 3.3.

Cf. A200073.

Concatenation([2],Flat(List([1..10],n->List([0..n],k->Sum([0..n],j->2*n*Binomial(2*n-j,j)*Binomial(j,k)/(2*n-j)))))); # Muniru A Asiru, Jul 27 2018

T[n_, k_] := Sum[ 2n*Binomial[2n - j, j]*Binomial[j, k]/(2n - j), {j, 0, n}]; T[0, 0] = 2; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Jul 23 2018 *)

t(n,k)={if(n>=1, sum(j=0,n/2, n*binomial(n-j,j)*binomial(j,k)/(n-j)), 2 );}
T(n,k) = t(2*n, k);
{ nmax=10 ; for(n=0,nmax, for(k=0,n, print1(T(n,k),",") ; ); ); }

T(n,k) = Sum_(j=0..n) 2n*binomial(2n-j,j)*binomial(j,k)/(2n-j).
From Franck Maminirina Ramaharo, Jul 09 2018: (Start)
T(n,0) = A005248(n).
T(n,1) = A099920(2*n-1).
T(n,n-1) = A005563(n).
(End)

Offset changed from 1 to 0 by Franck Maminirina Ramaharo, Jul 30 2018

A240847 a(n) = 2a(n-1) + a(n-2) - 2a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, -2, -5, -12, -25, -50, -96, -180, -331, -600, -1075, -1908, -3360, -5878, -10225, -17700, -30509, -52390, -89664, -153000, -260375, -442032, -748775, -1265832, -2136000, -3598250, -6052061, -10164540

Offset: 0

Paul Curtz, Apr 13 2014

F1(m, n) is the difference table of a(n):
    0,   0,   1,   0,   1,   0,   0,  -2, ...
    0,   1,  -1,   1,  -1,   0,  -2,  -3, ...
    1,  -2,   2,  -2,   1,  -2,  -1,  -4, ...
   -3,   4,  -4,   3,  -3,   1,  -3,  -2, ...
    7,  -8,   7,  -6,   4,  -4,   1,  -4, ...
  -15,  15, -13,  10,  -8,   5,  -5,   1, ...
   30, -28,  23, -18,  13, -10,   6,  -6, ...
The recurrence holds for every row and every signed column.
Main diagonal: F1(n, n) = A001477(n).
First upper diagonal: F1(n, n+1) = -A001477(n).
F1(m, n) = F1(m, n-1) + F1(m+1, n-1).
Inverse binomial transform: 0, 0, 1, -3, 7, -15, 30, ... = 0, 0, followed by (-1)^n*A023610(n). Without signs: F2(0, n) = 0, 0, 1, 3, 7, 15, 30, ... = b(n) has the same recurrence.
F1(0, n) + F2(0, n) = 0, followed by A099920(n).
a(n) and b(n) are reciprocal by their inverse binomial transform.
0, followed by A001629(n) is an autosequence.
F1(m, 1) = (-1)^n*A029907(n).
F1(1, n) = 0, 1, -1, 1, -1, followed by -A226432(n+3).
F1(m, 2) = (-1)^n*A208354(n).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000032, A000045, A001629 (main sequence for the recurrence), A067331.

List([0..40], n-> (6*Fibonacci(n-3) - (n-3)*Lucas(1,-1,n-3)[2])/5 ); # G. C. Greubel, Feb 06 2020

[(6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5: n in [0..40]]; // G. C. Greubel, Feb 06 2020

with(combinat): seq( ((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5, n=0..40); # G. C. Greubel, Feb 06 2020

a[n_]:= a[n]= 2*a[n-1] +a[n-2] -2*a[n-3] -a[n-4]; a[0]= a[1]= a[3]= 0; a[2]= 1; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Apr 17 2014 *)
CoefficientList[Series[x^2*(1-2*x)/(1-x-x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+c-2b-a}; NestList[nxt,{0,0,1,0},40][[All,1]] (* Harvey P. Dale, Sep 17 2022 *)

Vec(x^2*(1-2*x)/(1-x-x^2)^2 + O(x^100)) \\ Colin Barker, Apr 13 2014

vector(41, n, my(m=n-1); ((m+3)*fibonacci(m-3) - 2*(m-3)*fibonacci(m-2) )/5 ) \\ G. C. Greubel, Feb 06 2020

[((n+3)*fibonacci(n-3) - 2*(n-3)*fibonacci(n-2))/5 for n in (0..40)] # G. C. Greubel, Feb 06 2020

a(n) = 0, 0, 1, 0, 1, 0, 0, followed by -A067331.
G.f.: x^2*(1-2*x)/(1-x-x^2)^2. - Colin Barker, Apr 13 2014
a(n) = ( (10*n + (3-5*n)*t)*(1+t)^n + (10*n-(3-5*n)*t)*(1-t)^n )/(25*2^n), where t=sqrt(5). - Bruno Berselli, Apr 17 2014
a(n) = (6*Fibonacci(n-3) - (n-3)*Lucas(n-3))/5 = ((n+3)*Fibonacci(n-3) - 2*(n-3)*Fibonacci(n-2))/5. - G. C. Greubel, Feb 06 2020

A245960 Triangle read by rows: T(n,k) is the number of the vertices of the Lucas cube H_n having degree k (0<=k<=n).

Original entry on oeis.org

1, 1, 0, 0, 2, 1, 0, 3, 0, 1, 0, 0, 6, 0, 1, 0, 0, 5, 5, 0, 1, 0, 0, 3, 8, 6, 0, 1, 0, 0, 0, 14, 7, 7, 0, 1, 0, 0, 0, 8, 22, 8, 8, 0, 1, 0, 0, 0, 3, 27, 27, 9, 9, 0, 1, 0, 0, 0, 0, 25, 42, 35, 10, 10, 0, 1, 0, 0, 0, 0, 11, 66, 55, 44, 11, 11, 0, 1

Offset: 0

Emeric Deutsch, Aug 08 2014

The vertex set of the Lucas cube H_n is the set of all binary strings of length n without consecutive 1's and without strings that start and end with 1. Two vertices are adjacent if their strings differ in exactly one bit.
Sum of entries in row n (n>=1) is the Lucas number L(n) = F(n-1)+F(n+1), where F(n) = A000045(n) are the Fibonacci numbers.
Sum(k*T(n,k), k=0..n) = 2*n*F(n-1) = 2*A099920(n-1).

			Row 2 is 0,2,1 because the Lucas cube H_2 is  the path-tree P_3 having 2 vertices of degree 1 and 1 vertex of degree 2.
Row 3 is 0,3,0,1 because the Lucas cube H_3 is the star tree with 4 vertices; the vertex degrees are 1, 1, 1, and 3.
Triangle starts:
  1;
  1,0;
  0,2,1;
  0,3,0,1;
  0,0,6,0,1;
  0,0,5,5,0,1;

S. Klavzar, M. Mollard, M. Petkovsek, The degree sequence of Fibonacci and Lucas cubes, Discrete Math., 311, 2011, 1310-1322.
S. Klavzar, Structure of Fibonacci cubes: a survey, preprint.
S. Klavzar, Structure of Fibonacci cubes: a survey, J. Comb. Optim., 25, 2013, 505-522

Cf. A000045, A099920.

G := (1+(1-t)*z+z^2*t^2+(1-t)*z^3*t-(1-t)^2*z^4*t)/((1-t*z)*(1-z^2*t)-z^3*t): Gserz := simplify(series(G, z = 0, 20)): T := proc (n, k) options operator, arrow: coeff(coeff(Gserz, z, n), t, k) end proc: for n from 0 to 15 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form

G.f.: (1+(1-t)*z+t^2*z^2+(1-t)*t*z^3-t*(1-t)^2*z^4)/((1-t*z)*(1-t*z^2)-t*z^3).

If n>=2 then T(n,k) = sum(2*binomial(i,2i+k-n)*binomial(n-2i-1,k-i)+binomial(i-1,2i+k-n)*binomial(n-2i,k-i), i=0..k). Recall that binomial(m,k)=0 if k<0.

A324128 a(n) = 2nFibonacci(n) + (-1)^n + 1.

Original entry on oeis.org

2, 2, 6, 12, 26, 50, 98, 182, 338, 612, 1102, 1958, 3458, 6058, 10558, 18300, 31586, 54298, 93026, 158878, 270602, 459732, 779286, 1318222, 2225666, 3751250, 6312438, 10606572, 17797418, 29825282, 49922402, 83468678, 139411778, 232622148, 387796318, 645922550, 1074985346, 1787678458, 2970700846

Offset: 0

N. J. A. Sloane, Feb 20 2019

This sequence is distantly related to the number of losing strings using a binary alphabet in the "same game" described by Burns and Purcell (2007) and Kurz (2001). - Petros Hadjicostas, Sep 01 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.
Sascha Kurz, Polynomials in "same game", 2001.
Index entries for linear recurrences with constant coefficients, signature (2,2,-4,-2,2,1).

Cf. A000045, A035615, A099920, A309874, A324129.

A324128[n_]:=Fibonacci[n]2n+(-1)^n+1;Array[A324128,50,0] (* Paolo Xausa, Nov 15 2023 *)

Vec(2*(1 - x - x^2 + 2*x^3 + x^4 - x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)^2) + O(x^40)) \\ Colin Barker, Mar 03 2019

From Chai Wah Wu, Feb 20 2019: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) + 2*a(n-5) + a(n-6) for n > 5.
G.f.: (2*x^5 - 2*x^4 - 4*x^3 + 2*x^2 + 2*x - 2)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)
From Petros Hadjicostas, Sep 01 2019: (Start)
a(n) = 2*A324129(n) for n >= 0.
a(n) = A309874(n) + 2*A099920(n-1) = 2^n - A035615(n) + 2*A099920(n-1) for n >= 2.[Here A309874 counts the losing strings while A035615 counts the winning strings using a binary alphabet in the "same game". See Burns and Purcell (2007) and Kurz (2001).]
(End)

A324129 a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.

Original entry on oeis.org

1, 1, 3, 6, 13, 25, 49, 91, 169, 306, 551, 979, 1729, 3029, 5279, 9150, 15793, 27149, 46513, 79439, 135301, 229866, 389643, 659111, 1112833, 1875625, 3156219, 5303286, 8898709, 14912641, 24961201, 41734339, 69705889, 116311074, 193898159, 322961275

Offset: 0

N. J. A. Sloane, Feb 20 2019

Equals A324128(n)/2.

This sequence is distantly related to (one-half) the number of losing strings using a binary alphabet in the "same game" described by Burns and Purcell (2007) and Kurz (2001). - Petros Hadjicostas, Sep 01 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.
Sascha Kurz, Polynomials in "same game", 2001.
Index entries for linear recurrences with constant coefficients, signature (2,2,-4,-2,2,1).

Cf. A000045, A035615, A099920, A323812, A324128.

[n*Fibonacci(n)+((-1)^n+1)/2:n in [0..35]]; // Marius A. Burtea, Aug 29 2019

A324129[n_]:=Fibonacci[n]n+((-1)^n+1)/2;Array[A324129,50,0] (* Paolo Xausa, Nov 15 2023 *)

Vec((1 - x - x^2 + 2*x^3 + x^4 - x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)^2) + O(x^40)) \\ Colin Barker, Mar 03 2019

From Chai Wah Wu, Feb 20 2019: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) + 2*a(n-5) + a(n-6) for n > 5.
G.f.: (x^5 - x^4 - 2*x^3 + x^2 + x - 1)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)
a(n) = A309874(n)/2 + A099920(n-1) = 2^(n-1) - A035615(n)/2 + A099920(n-1) = A323812(n) + A099920(n-1) for n >= 2. [Sequence A309874 counts the losing strings while A035615 counts the winning strings using a binary alphabet in the "same game". See Burns and Purcell (2007) and Kurz (2001).] - Petros Hadjicostas, Sep 01 2019

A123265 Fibonacci-Lucas triangle read by rows.

Original entry on oeis.org

2, 1, 2, 3, 3, 2, 4, 8, 5, 2, 7, 15, 15, 7, 2, 11, 30, 35, 24, 9, 2, 18, 56, 80, 66, 35, 11, 2, 29, 104, 171, 170, 110, 48, 13, 2, 47, 189, 355, 407, 315, 169, 63, 15, 2, 76, 340, 715, 932, 832, 532, 245, 80, 17, 2

Offset: 0

Philippe Deléham, Nov 07 2006

As a Riordan array, this is ((2-x)/(1-x-x^2),x/(1-x-x^2)) . Row sums form A078343, diagonals sums form A014551.

			Triangle begins:
2;
1, 2;
3, 3, 2;
4, 8, 5, 2;
7, 15, 15, 7, 2;
11, 30, 35, 24, 9, 2;
18, 56, 80, 66, 35, 11, 2;
29, 104, 171, 170, 110, 48, 13, 2;

Cf. A000032, A099920 ; A007395, A005408, A005563.

T(n,k)=T(n-1,k-1)+T(n-1,k)+T(n-2,k), T(0,0)=2, T(1,0)=1, T(n,k)=0 if k<0 or if k>n . T(n,0)=L(n)=A000032(n),Lucas numbers.

A245962 Triangle read by rows: T(n,k) is the number of induced subgraphs of the Lucas cube Lambda(n) that are isomorphic to the hypercube Q(k).

Original entry on oeis.org

1, 1, 3, 2, 4, 3, 7, 8, 2, 11, 15, 5, 18, 30, 15, 2, 29, 56, 35, 7, 47, 104, 80, 24, 2, 76, 189, 171, 66, 9, 123, 340, 355, 170, 35, 2, 199, 605, 715, 407, 110, 11, 322, 1068, 1410, 932, 315, 48, 2, 521, 1872, 2730, 2054, 832, 169, 13, 843, 3262, 5208, 4396, 2079, 532, 63, 2

Offset: 0

Emeric Deutsch, Aug 14 2014

Number of entries in row n is 1 + floor(n/2).
The entries in row n are the coefficients of the cube polynomial of the Lucas cube Lambda(n).
For n >= 1, sum of entries in row n = A014551(n) = 2^n + (-1)^n (the Jacobsthal-Lucas numbers).
Sum_{k >= 0} k*T(n,k) = A099429(n).
T(n,0) = A000032(n) (n >= 1; the Lucas numbers); T(n,1) = A099920(n-1); T(n,2) = A245961(n).
As communicated by the authors, Theorem 5.2 and Corollary 5.3 of the Klavzar et al. paper contains a typo: 2nd binomial should be binomial(n - a - 1, a) resp. binomial(n - i - 1, i).

			Row 4 is 7, 8, 2. Indeed, the Lucas cube Lambda(4) is the graph <><> obtained by identifying a vertex of a square with a vertex of another square; it has 7 vertices (i.e., Q(0)s), 8 edges (i.e., Q(1)s), and 2 squares (i.e., Q(2)s).
Triangle starts:
   1;
   1;
   3,  2;
   4,  3;
   7,  8,  2;
  11, 15,  5;

Paolo Xausa, Table of n, a(n) for n = 0..10200 (rows 0..200 of the triangle, flattened).
Sandi Klavzar and Michel Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
Sandi Klavzar and Michel Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105.
Jun Wan and Zuo-Ru Zhang, A proof of the only mode of a unimodal sequence, arXiv:2402.12858 [math.CO], 2024.

Cf. A000032, A014551, A099429, A099920, A245961.

T := proc (n, k) options operator, arrow: add((2*binomial(n-i, i)-binomial(n-i-1, i))*binomial(i, k), i = k .. floor((1/2)*n)) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. (1/2)*n) end do; # yields sequence in triangular form

A245962[n_, k_] := Sum[(2*Binomial[n-i, i]-Binomial[n-i-1, i])*Binomial[i, k], {i, k, n/2}]; Table[A245962[n, k], {n, 0, 15}, {k, 0, n/2}] (* Paolo Xausa, Feb 29 2024 *)

T(n,k) = Sum_{i = k..floor(n/2)} (2*binomial(n - i, i) - binomial(n - i - 1, i))*binomial(i, k).
G.f.: (1+(1+t)*z^2)/(1-z-(1+t)*z^2).
The generating polynomial of row n (i.e., the cube polynomial of Lambda(n)) is Sum_{i = 0..floor(n/2)} (2*binomial(n - i, i) - binomial(n - i - 1))(1+x)^i.
The generating polynomial of row n (i.e., the cube polynomial of Lambda(n)) is ((1+w)/2)^n + ((1-w)/2)^n, where w = sqrt(5 + 4x).
The generating function of column k (k >= 1) is z^(2k)(2-z)/(1-z-z^2)^(k+1).

A246174 Triangle read by rows: T(n,k) is the number of vertex pairs at distance k of the Lucas cube Lambda(n) (1<=k<=n).

Original entry on oeis.org

2, 1, 3, 3, 8, 8, 4, 1, 15, 20, 15, 5, 30, 48, 44, 24, 6, 1, 56, 105, 119, 84, 35, 7, 104, 224, 296, 256, 144, 48, 8, 1, 189, 459, 696, 711, 495, 228, 63, 9, 340, 920, 1570, 1840, 1522, 880, 340, 80, 10, 1, 605, 1804, 3421, 4521, 4312, 2981, 1463, 484, 99, 11

Offset: 2

Emeric Deutsch, Aug 18 2014

The Lucas cube Lambda(n) can be defined as the graph whose vertices are the binary strings of length n without either two consecutive 1's or a 1 in the first and in the last position, and in which two vertices are adjacent when their Hamming distance is exactly 1.
The number of entries in row n is equal to n if n is even and equal to n-1 if n is odd.
The entries in row n are the coefficients of the Hosoya polynomial of the Lucas cube Lambda(n).
T(n,1) = A099920(n-1) = number of edges in Lambda(n).
Sum(kT(n,k), k>=1) = A238420(n) = the Wiener index of Lambda(n).

			Row 2 is 2,1. Indeed, Lambda(2)  is the path-tree P(3) having vertex-pair distances 1,1, and 2.
Triangle starts:
2,1;
3,3;
8,8,4,1;
15,20,15,5;
30,48,44,24,6,1;

S. Klavzar, M. Mollard, Wiener index and Hosoya polynomial of Fibonacci and Lucas cubes, MATCH Commun. Math. Comput. Chem., 68, 2012, 311-324.
E. Munarini, C. P. Cippo, N. Z. Salvi, On the Lucas cubes, The Fibonacci Quarterly, 39, No. 1, 2001, 12-21.

Cf. A099920, A238420, A246173.

g := t*z^2*(2+t-z+t*z-3*t*z^2+t*z^3+t*z^4)/((1+t*z)*(1-z-t*z-z^2-t*z^2+t*z^3)*(1-z-z^2)): gserz := simplify(series(g, z = 0, 20)): for j from 2 to 18 do H[j] := sort(coeff(gserz, z, j)) end do: for j from 2 to 13 do seq(coeff(H[j], t, k), k = 1 .. 2*floor((1/2)*j)) end do; # yields sequence in triangular form

G.f.: tz^2(2+t-z+tz-3tz^2+tz^3+tz^4)/((1+tz)(1-z-tz-z^2-tz^2+tz^3)(1-z-z^2)). Derived from Theorem 4.3 of the Klavzar-Mollard reference in which the g.f. of the ordered Hosoya polynomials is given.

cp's OEIS Frontend

A099924 Self-convolution of Lucas numbers.

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A264147 a(n) = n*F(n+1) - (n+1)*F(n), where F = A000045.

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A122076 Coefficients of a generalized Jaco-Lucas polynomial (even indices) read by rows.

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A240847 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

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A245960 Triangle read by rows: T(n,k) is the number of the vertices of the Lucas cube H_n having degree k (0<=k<=n).

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A324128 a(n) = 2*n*Fibonacci(n) + (-1)^n + 1.

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A264147 a(n) = nF(n+1) - (n+1)F(n), where F = A000045.

A240847 a(n) = 2a(n-1) + a(n-2) - 2a(n-3) - a(n-4) for n>3, a(0)=a(1)=a(3)=0, a(2)=1.

A324128 a(n) = 2nFibonacci(n) + (-1)^n + 1.