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p always divides A316269(k, p-Kronecker(k^2-4, p)), so it's interesting to see when p^2 also divides A316269(k, p-Kronecker(k^2-4, p)).

In the following comments, let p = prime(n). Note that A316269(0, m) and A316269(1, m) is not defined, so here k must be understood as a remainder modulo p^2. because A316269(k+s*p^2, m) == A316269(k, m) (mod p^2).

Let p = prime(n). Every row contains 0, 1 and p^2 - 1. For n >= 3, the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 3, 4, ..., p - 3, p - 1}.

Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 3)/2.

Equivalently, for n >= 2, row n lists 0 <= k < p^2 such that p^2 divides A316269(k, (p-Kronecker(k^2-4, p))/2), p = prime(n).

The rank of {A316269(k,m)} modulo n is the smallest l such that n divides A316269(k,l).

Though {A316269(0,m)} is not defined, it can be understood as the sequence 0, 1, 0, -1, 0, 1, 0, -1, ... So the first column of each row (apart from the first one) is always 2.

Though {A316269(1,m)} is not defined, it can be understood as the sequence 0, 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, ... So the second column of each row is always 3.

Every row excluding the first term is antisymmetric, that is, T(n,k) = T(n,n-k) for 1 <= k <= n - 1.

T(n,k) is the multiplicative order of ((k + sqrt(k^2 - 4))/2)^2 modulo n*sqrt(k^2 - 4), where the multiplicative order of u modulo z is the smallest positive integer l such that (u^l - 1)/z is an algebraic integer.

The period of {A316269(k,m)} modulo n is the smallest l such that A316269(k,m) == A316269(k,m+l) (mod n) for every m >= 0. Clearly, T(n,k) is divisible by A321478(n,k). Actually, the ratio is always 1 or 2.

Though {A316269(0,m)} is not defined, it can be understood as the sequence 0, 1, 0, -1, 0, 1, 0, -1, ... So the first column of each row (apart from the first and second ones) is always 4.

Though {A316269(1,m)} is not defined, it can be understood as the sequence 0, 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, ... So the second column of each row (apart from the second one) is always 6.

T(n,k) is the LCM of A321478(n,k) and the multiplicative order of (k + sqrt(k^2 - 4))/2 modulo n, where the multiplicative order of u modulo z is the smallest positive integer l such that (u^l - 1)/z is an algebraic integer.

Equals A073133 with an additional column A(.,0).

If the first column and top row are deleted, antidiagonal reading yields A118243.

Adding a top row of 1's and antidiagonal reading downwards yields A157103.

Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....

From Jianing Song, Jul 14 2018: (Start)

All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.

Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.

E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.

Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.

pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.

If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.

Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.

Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:

((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4

......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*

......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*

......1.........3,7.............1...................0...................0

.....-1.........1,5.............0...................0...................1

.....-1.........3,7.............0...................1...................0

* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]

z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.

(End)

From Michael A. Allen, Mar 06 2023: (Start)

Removing the first (n=0) row of A352361 gives this sequence.

Row n is the n-metallonacci sequence.

A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

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].

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].

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].

For Lucas sequences, say, rows in A316269, we are mainly concerned about the periods, ranks and the ratios of the periods to the ranks of them modulo a given integer n. The period of {A316269(k,m) modulo m} is given as A321479(n,k), and the rank, which is defined as the smallest l > 0 such that n divides A316269(k,l), is given as A321478(n,k). T(n,k) is their ratio, which is the multiplicative order of A316269(k, A321478(n,k)+1) modulo n.

T(n,k) has value 1 or 2. This is because A316269(k,m+1)^2 == 1 (mod A316269(k,m)). See A172236 for some further properties.

It seems that the n-th row contains more 2's than 1's unless n is a power of 2, in which case the numbers of 1's and 2's are always the same if n >= 4.