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).
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
Examples
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;
Links
- 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.
Programs
-
Maple
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
-
Mathematica
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 *)
Formula
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).
Comments