1,9

EqualsA073133with an additional column A(.,0).

If the first column and top row are deleted, antidiagonal reading yieldsA118243.

Adding a top row of 1's and antidiagonal reading downwards yieldsA157103.

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

FromJianing 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 byJianing 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)

FromMichael A. Allen,Mar 06 2023: (Start)

Removing the first (n=0) row ofA352361gives 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)

Jianing Song,Antidiagonals n = 1..100, flattened(the first 30 antidiagonals from Vincenzo Librandi)

Michael A. Allen and Kenneth Edwards,Fence tiling derived identities involving the metallonacci numbers squared or cubed,Fib. Q. 60:5 (2022) 5-17.

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. -Jianing Song,Jun 27 2018

For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). -Jianing Song,Apr 21 2019

FromG. C. Greubel,Sep 29 2024: (Start)

A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).

T(n, k) = F_{k}(n-k) (antidiagonal triangle).

Sum_{k=0..n-1} T(n, k) =A304357(n) - (1-(-1)^n)/2.

Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.

T(2*n, n) =A084844(n).

T(2*n+1, n+1) =A084845(n). (End)

The array, A(n, k), starts in row n = 1 with columns k >= 0 as

0 1 1 2 3 5 8

0 1 2 5 12 29 70

0 1 3 10 33 109 360

0 1 4 17 72 305 1292

0 1 5 26 135 701 3640

0 1 6 37 228 1405 8658

0 1 7 50 357 2549 18200

0 1 8 65 528 4289 34840

0 1 9 82 747 6805 61992

0 1 10 101 1020 10301 104030

0 1 11 122 1353 15005 166408

Antidiagonal triangle, T(n, k), begins as:

0;

0, 1;

0, 1, 1;

0, 1, 2, 2;

0, 1, 3, 5, 3;

0, 1, 4, 10, 12, 5;

0, 1, 5, 17, 33, 29, 8;

0, 1, 6, 26, 72, 109, 70, 13;

0, 1, 7, 37, 135, 305, 360, 169, 21;

0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

A172236[n_, k_]:=Fibonacci[k, n-k];

Table[A172236[n, k], {n, 15}, {k, 0, n-1}]//Flatten

(PARI) A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));

tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), "," )); print); \\Jianing Song,Jul 14 2018 (program fromMichel Marcus;see alsoA316269)

(PARI) A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\Jianing Song,Nov 10 2018

(Magma)

A172236:= func< n, k | k le 1 select k else Evaluate(DicksonSecond(k-1, -1), n-k) >;

[A172236(n, k): k in [0..n-1], n in [1..13]]; //G. C. Greubel,Sep 29 2024

(SageMath)

defA172236(n, k): return sum(binomial(k-j-1, j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))

flatten([[A172236(n, k) for k in range(n)] for n in range(1, 14)]) #G. C. Greubel,Sep 29 2024

Rows n include:A000045(n=1),A000129(n=2),A006190(n=3),A001076(n=4),A052918(n=5),A005668(n=6),A054413(n=7),A041025(n=8),A099371(n=9),A041041(n=10),A049666(n=11),A041061(n=12),A140455(n=13),A041085(n=14),A154597(n=15),A041113(n=16),A178765(n=17),A041145(n=18),A243399(n=19),A041181(n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)

Columns k include:A000004(k=0),A000012(k=1),A000027(k=2),A002522(k=3),A054602(k=4),A057721(k=5),A124152(k=6).

Entry points for A(n,k) modulo m:A001177(n=1),A214028(n=2),A322907(n=3).

Pisano period for A(n,k) modulo m:A001175(n=1),A175181(n=2),A175182(n=3),A175183(n=4),A175184(n=5),A175185(n=6).

Number of zeros in a period for A(n,k) modulo m:A001176(n=1),A214027(n=2),A322906(n=3).

Cf.A084844,A084845,A118243,A157103,A271782,A316269.

Sums include:A304357,A304359.

Similar to:A073133.

Sequence in context:A317205A296068A144064*A191646A297321A277938

Adjacent sequences:A172233A172234A172235*A172237A172238A172239

nonn,tabl,easy

Roger L. Bagula,Jan 29 2010

More terms fromJianing Song,Jul 14 2018

approved