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A364366
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An irregular triangle read by rows, the 3rd row-symmetric Fibonaccian triangle: T(n,k) is the Whitney number of level k of the (3,n)-th symmetric Fibonaccian lattice (0 <= n, 0 <= k <= 2*n).
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2
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1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 5, 4, 3, 1, 1, 4, 7, 10, 11, 10, 7, 4, 1, 1, 5, 11, 18, 24, 26, 24, 18, 11, 5, 1, 1, 6, 16, 30, 46, 58, 63, 58, 46, 30, 16, 6, 1, 1, 7, 22, 47, 81, 116, 143, 158, 143, 116, 81, 47, 22, 7, 1
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OFFSET
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0,6
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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=3. Let L be the (3,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) = 1 and p(1,x) = 1+x+x^2, since L is a chain with 3 elements. For n >= 2, we have, by definition, p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k. The Whitney number T(n,k) is the number of (3,n) Fibonaccian strings S=(S(1),...,S(n)) whose coordinate sum S(1)+...+S(n) is equal to 3*(n*(n+1)/2)-k. This irregular triangle of T(n,k)'s is obtained from the regular triangles in entries A079487 and A123245 by removing all even length rows (with the exception of the row of length two, all such even length rows are 'asymmetric').
For m=4, see A364367. 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-2) + T(n-1,k-1) + T(n-1,k) - T(n-2,k-2) for n >= 1 and 0 <= k <= 2*n, understanding T(i,j) to be zero when j < 0 or j > 2*i. That the preceding recurrence holds is equivalent to the identity p(n,x) = (1+x+x^2)*p(n-1,x) - x^2*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, 2, 2, 2, 1;
1, 3, 4, 5, 4, 3, 1;
1, 4, 7, 10, 11, 10, 7, 4, 1;
1, 5, 11, 18, 24, 26, 24, 18, 11, 5, 1;
1, 6, 16, 30, 46, 58, 63, 58, 46, 30, 16, 6, 1;
1, 7, 22, 47, 81, 116, 143, 158, 143, 116, 81, 47, 22, 7, 1;
...
Below are the 21 (3,3) Fibonaccian strings (organized by rank level) that comprise the (3,3)rd symmetric Fibonaccian lattice:
rank=6: (1,4,7)
rank=5: (1,4,8) (1,5,7) (2,4,7)
rank=4: (1,4,9) (1,5,8) (2,4,8) (2,5,7)
rank=3: (1,5,9) (1,6,8) (2,4,9) (2,5,8) (3,5,7)
rank=2: (1,6,9) (2,5,9) (2,6,8) (3,5,8)
rank=1: (2,6,9) (3,5,9) (3,6,8)
rank=0: (3,6,9)
The triples (3,4,7), (3,4,8), (3,4,9), (1,6,7), (2,6,7), and (3,6,7) are disallowed as (3,3) Fibonaccian strings since each contains consecutive integers.
In the (3,5)th symmetric Fibonaccian lattice, rank level 8 consists of exactly the (3,5) Fibonaccian strings whose coordinate sum is 3*(5*(5+1)/2)-8=37: (1,4,7,10,15), (1,4,7,11,14), (1,4,8,10,14), (1,4,8,11,13), (1,5,7,10,14), (1,5,7,11,13), (1,5,8,10,13), (2,4,7,10,14), (2,4,7,11,13), (2,4,8,10,13), and (2,5,7,10,13), confirming that T(5,8)=11.
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CROSSREFS
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Sum of row n (n >= 0) is A001906(n+1), cf. row n=3 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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