A364368 An irregular triangle read by rows, the 5th row-symmetric Fibonaccian triangle: T(n,k) is the Whitney number of level k of the (5,n)-th symmetric Fibonaccian lattice (0 <= n, 0 <= k <= 4*n).

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 3, 2, 1, 1, 3, 6, 10, 13, 16, 17, 16, 13, 10, 6, 3, 1, 1, 4, 10, 20, 32, 46, 59, 68, 71, 68, 59, 46, 32, 20, 10, 4, 1, 1, 5, 15, 35, 66, 109, 161, 215, 263, 296, 308, 296, 263, 215, 161, 109, 66, 35, 15, 5, 1

Offset: 0

Author

Robert G. Donnelly and Molly W. Dunkum, Jul 20 2023

Keywords

• nonn
• tabf

Comments

For integers m and n (m >= 2, n > 0), let L be the set of n-tuples S=(S(1),...,S(n)) with each S(j) in {(j-1)*m+1,(j-1)*m+2,...,j*m} and such that S has no consecutive integers. Partially order these '(m,n) Fibonaccian strings' comprising L by the rule R <= S iff R(j) >= S(j) for 1 <= j <= n (so, 'lightest' n-tuples are at the top of the Hasse diagram for L). Then L is a self-dual distributive lattice, the '(m,n)-th symmetric Fibonaccian lattice'. When n=1, L is a chain with m elements. Now allow n=0; in this case, regard L to be a singleton set. Let p(n,x) be the rank generating function of L, so p(n,1)=|L|, p(0,x)=1, and p(1,x)=1+x+...+x^(m-1). For n >= 2, the fact that p(n,x) = p(1,x)*p(n-1,x) - x^(m-1)*p(n-2,x) can be deduced from a recurrence of Whitney numbers of symmetric Fibonaccian lattices proved in Proposition 2.1 of [Donnelly, Dunkum, Lišková, and Nance, 2023].

The (m,n)-th symmetric Fibonaccian lattice realizes a p(n,1)-dimensional representation of the special linear Lie algebra sl(m,C). The representation is reducible exactly when m >= 3 and n >= 3. The polynomial p(n,x) is a natural specialization of the character of this representation, where the latter can be identified as a certain skew Schur function. In [Donnelly, Dunkum, Lišková, and Nance, 2023], these representations are uniformly constructed (as an application of [Donnelly and Dunkum, 2022]) and explicit formulas for p(n,x) are given.

In [Donnelly, Dunkum, Lišková, and Nance, 2023], the (m,n)-th symmetric Fibonaccian lattice L is also described using semistandard tableaux of a specific ribbon shape; the irreducible components of the associated sl(m,C)-representation are in one-to-one correspondence with what are called the 'ballot-admissible' (aka Littlewood-Richardson) tableaux. In terms of Fibonaccian strings, an element S = (S(1),...,S(n)) in L is ballot-admissible iff for any integer q between 1 and n (inclusive) and any integer r between 1 and m-1 (inclusive), the following integer quantity is nonnegative: Sum_{k=n+1-q..n}([n+1-k is odd]*([r+(k-1)*m = S(k)] - [r+(k-1)*m+1 = S(k)]) + [n+1-k is even]*([k*m-r = S(k)]-[k*m+1-r = S(k)])), where '[]' denotes the Iverson bracket. Enumerating the ballot-admissible tableaux or Fibonaccian strings in L seems to be an interesting problem when m >= 3; when m=3, the sizes of the sets of ballot-admissible tableaux conjecturally agree with A004148.

In this OEIS entry, we have m=5. Let L be the (5,n)-th symmetric Fibonaccian lattice. When n=0, we have T(0,0) = |L| = 1. When n=1, we have T(1,0) = T(1,1) = T(1,2) = T(1,3) = T(1,4) = 1 and p(1,x) = 1+x+x^2+x^3+x^4, since L is a chain with 5 elements. For n >= 2, we have, by definition, p(n,x) = Sum_{k=0..4*n} T(n,k)*x^k. The Whitney number T(n,k) is the number of (5,n) Fibonaccian strings S=(S(1),...,S(n)) whose coordinate sum S(1)+...+S(n) is equal to 5*(n*(n+1)/2)-k.

For m=3, see A364366. For m=4, see A364367. When m=2, the (2,n)-th symmetric Fibonaccian lattice is a chain with n+1 elements and rank generating function 1+x+...+x^(n-1)+x^n. Therefore, the 2nd row-symmetric Fibonaccian triangle is a regular triangle of 1's. The 1st row-symmetric Fibonaccian 'triangle' is regarded to be the signed sequence 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0, ... (A010892). 'Gibonaccian' versions of such triangles are considered in [Donnelly, Dunkum, Huber, and Knupp, 2021].

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k >= 0) starts as follows:
  1;
  1,  1,  1,  1,  1;
  1,  2,  3,  4,  4,  4,  3,  2,  1;
  1,  3,  6, 10, 13, 16, 17, 16, 13, 10,  6,  3,  1;
  1,  4, 10, 20, 32, 46, 59, 68, 71, 68, 59, 46, 32, 20, 10,  4,  1;
...
Below are the 24 (5,2) Fibonaccian strings (organized by rank level) that comprise the (5,2)nd symmetric Fibonaccian lattice:
rank=8: (1,6)
rank=7: (1,7)   (2,6)
rank=6: (1,8)   (2,7)   (3,6)
rank=5: (1,9)   (2,8)   (3,7)   (4,6)
rank=4: (1,10)  (2,9)   (3,8)   (4,7)
rank=3: (2,10)  (3,9)   (4,8)   (5,7)
rank=2: (3,10)  (4,9)   (5,8)
rank=1: (4,10)  (5,9)
rank=0: (5,10)
The pair (5,6) is disallowed as a (5,2) Fibonaccian string since it contains consecutive integers.
In the (5,3)rd symmetric Fibonaccian lattice, rank level 9 consists of exactly the (5,3) Fibonaccian strings whose coordinate sum is 5*(3*(3+1)/2)-9=21: (1,6,14), (1,7,13), (1,8,12), (1,9,11), (2,6,13), (2,7,12), (2,8,11), (3,6,12), (3,7,11), and (4,6,11), confirming that T(3,9)=10.
		

Links

• R. G. Donnelly, M. W. Dunkum, M. L. Huber, and L. Knupp, Sign-alternating Gibonacci polynomials, arXiv:2012.14993 [math.CO], 2020.
• R. G. Donnelly, M. W. Dunkum, M. L. Huber, and L. Knupp, "Sign-alternating Gibonacci polynomials", Enumer. Comb. Appl. 1:2 (2021), art. id. S2R15.
• R. G. Donnelly and M. W. Dunkum, Gelfand--Tsetlin-type weight bases for all special linear Lie algebra representations corresponding to skew Schur functions, arXiv:2012.14986 [math.CO], 2020-2022.
• R. G. Donnelly and M. W. Dunkum, "Gelfand--Tsetlin-type weight bases for all special linear Lie algebra representations corresponding to skew Schur functions", Adv. Appl. Math. 139 (2022), art. id. 102356.
• R. G. Donnelly, M. W. Dunkum, S. V. Lišková, and A. Nance, Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras, arXiv:2012.14991 [math.CO], 2020-2022.
• R. G. Donnelly, M. W. Dunkum, S. V. Lišková, and A. Nance, "Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras", Involve 16:2 (2023), 201-226.

Crossrefs

Sum of row n (n >= 0) is A004254(n+1), cf. row n=5 of the array A316269.

Formula

With T(0,0)=1, then T(n,k) = T(n-1,k-4) + T(n-1,k-3) + T(n-1,k-2) + T(n-1,k-1) + T(n-1,k) - T(n-2,k-4) for n >= 1 and 0 <= k <= 4*n, understanding T(i,j) to be zero when j < 0 or j > 4*i. That the preceding recurrence holds is equivalent to the identity p(n,x) = (1+x+x^2+x^3+x^4)*p(n-1,x) - x^4*p(n-2,x) for n >= 1, where p(0,x)=1 and p(-1,x) is taken to be 0.