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Because (p(n,x)) is a strong divisibility sequence, for each integer k, the sequence (p(n,k)) is a strong divisibility sequence of integers.

From Klaus Purath, Mar 07 2023: (Start)

For any two terms (a(n), a(n+1)) = (x, y), x^2 - 5*x*y + y^2 = 37 = A082111(4). This is valid in general for all recursive sequences (t) with constant coefficients (5,-1) and t(0) = 1: x^2 - 5*x*y + y^2 = A082111(t(1)-5). This includes and interprets the Feb 04 2014 comment in A004253 by Colin Barker.

By analogy to all this, for three consecutive terms (x, y, z) of any sequence (t) of the form (5,-1) with t(0) = 1: y^2 - x*z = A082111(t(1)-5). (End)

Proposed by R. Guy in the seqfan list, Mar 29 2009.

The sequence is the case P1 = 13, P2 = 34, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

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].

Proposed by R. Guy in the seqfan list, Mar 29 2009.

From Klaus Purath, Aug 18 2024: (Start)

For any two consecutive terms (x,y), x^2 - 5xy + y^2 = -17 = A127147(6) always applies. In general, the following applies to all recurrences (t) with constant coefficients (5,-1) and t(0) = 2 and two consecutive terms (x,y): x^2 - 5xy + y^2 = A127147(t(1)+3) for any integer t(1). This includes and interprets the Feb 08 2014 comment on A003501 by Colin Barker.

By analogy to this, for three consecutive terms (x,y,z) of any recurrence (t) of the form (5,-1) with t(0) = 2: y^2 - xz = A127147(t(1)+3).

a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/6, where (t) is any third order recurrence with constant coefficients (6,-6,1) and initial values t(0) = x, t(1) = x + 2, t(2) = x + 5 for any integer x.

a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/4, where (t) is any third order recurrence with constant coefficients (4,4,-1) and initial values t(0) = x, t(1) = 2 - x, t(2) = x + 1 for any integer x. (End)