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A364367
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An irregular triangle read by rows, the 4th row-symmetric Fibonaccian triangle: T(n,k) is the Whitney number of level k of the (4,n)-th symmetric Fibonaccian lattice (0 <= n, 0 <= k <= 3*n).
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2
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1, 1, 1, 1, 1, 1, 2, 3, 3, 3, 2, 1, 1, 3, 6, 8, 10, 10, 8, 6, 3, 1, 1, 4, 10, 17, 25, 31, 33, 31, 25, 17, 10, 4, 1, 1, 5, 15, 31, 53, 77, 98, 110, 110, 98, 77, 53, 31, 15, 5, 1, 1, 6, 21, 51, 100, 166, 242, 313, 364, 383, 364, 313, 242, 166, 100, 51, 21, 6, 1
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OFFSET
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0,7
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COMMENTS
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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=4. Let L be the (4,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) = 1 and p(1,x) = 1+x+x^2+x^3, since L is a chain with 4 elements. For n >= 2, we have, by definition, p(n,x) = Sum_{k=0..3*n} T(n,k)*x^k. The Whitney number T(n,k) is the number of (4,n) Fibonaccian strings S=(S(1),...,S(n)) whose coordinate sum S(1)+...+S(n) is equal to 4*(n*(n+1)/2)-k.
For m=3, see A364366. For m=5, see A364368. 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].
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LINKS
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FORMULA
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With T(0,0)=1, then T(n,k) = T(n-1,k-3) + T(n-1,k-2) + T(n-1,k-1) + T(n-1,k) - T(n-2,k-3) for n >= 1 and 0 <= k <= 3*n, understanding T(i,j) to be zero when j < 0 or j > 3*i. That the preceding recurrence holds is equivalent to the identity p(n,x) = (1+x+x^2+x^3)*p(n-1,x) - x^3*p(n-2,x) for n >= 1, where p(0,x)=1 and p(-1,x) is taken to be 0.
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EXAMPLE
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Triangle T(n,k) (with rows n >= 0 and columns k >= 0) starts as follows:
1;
1, 1, 1, 1;
1, 2, 3, 3, 3, 2, 1;
1, 3, 6, 8, 10, 10, 8, 6, 3, 1;
1, 4, 10, 17, 25, 31, 33, 31, 25, 17, 10, 4, 1;
1, 5, 15, 31, 53, 77, 98, 110, 110, 98, 77, 53, 31, 15, 5, 1;
...
Below are the 15 (4,2) Fibonaccian strings (organized by rank level) that comprise the (4,2)nd symmetric Fibonaccian lattice:
rank=6: (1,5)
rank=5: (1,6) (2,5)
rank=4: (1,7) (2,6) (3,5)
rank=3: (1,8) (2,7) (3,6)
rank=2: (2,8) (3,7) (4,6)
rank=1: (3,8) (4,7)
rank=0: (4,8)
The pair (4,5) is disallowed as a (4,2) Fibonaccian string since it contains consecutive integers.
In the (4,3)rd symmetric Fibonaccian lattice, rank level 5 consists of exactly the (4,3) Fibonaccian strings whose coordinate sum is 4*(3*(3+1)/2)-5=19: (1,6,12), (1,7,11), (1,8,10), (2,5,12), (2,6,11), (2,7,10), (3,5,11), (3,6,10), (3,7,9), and (4,6,9), confirming that T(3,5)=10.
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CROSSREFS
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Sum of row n (n >= 0) is A001353(n+1), cf. row n=4 of the array A316269.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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