%I #70 Sep 30 2024 12:51:52
%S 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,
%T 26,72,109,70,13,0,1,7,37,135,305,360,169,21,0,1,8,50,228,701,1292,
%U 1189,408,34,0,1,9,65,357,1405,3640,5473,3927,985,55,0,1,10,82,528,2549,8658,18901,23184,12970,2378,89,0,1,11,101,747,4289,18200,53353,98145,98209,42837,5741,144
%N Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.
%C Equals A073133 with an additional column A(.,0).
%C If the first column and top row are deleted, antidiagonal reading yields A118243.
%C Adding a top row of 1's and antidiagonal reading downwards yields A157103.
%C Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
%C From _Jianing Song_, Jul 14 2018: (Start)
%C 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.
%C 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.
%C 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.
%C 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.
%C 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.
%C 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.
%C 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.
%C Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
%C ((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
%C ......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
%C ......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
%C ......1.........3,7.............1...................0...................0
%C .....-1.........1,5.............0...................0...................1
%C .....-1.........3,7.............0...................1...................0
%C * 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]
%C 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.
%C (End)
%C From _Michael A. Allen_, Mar 06 2023: (Start)
%C Removing the first (n=0) row of A352361 gives this sequence.
%C Row n is the n-metallonacci sequence.
%C 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)
%H Jianing Song, <a href="/A172236/b172236.txt">Antidiagonals n = 1..100, flattened</a> (the first 30 antidiagonals from Vincenzo Librandi)
%H Michael A. Allen and Kenneth Edwards, <a href="https://www.fq.math.ca/Papers1/60-5/allen.pdf">Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.
%F 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
%F 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
%F From _G. C. Greubel_, Sep 29 2024: (Start)
%F A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
%F T(n, k) = F_{k}(n-k) (antidiagonal triangle).
%F Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
%F Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
%F T(2*n, n) = A084844(n).
%F T(2*n+1, n+1) = A084845(n). (End)
%e The array, A(n, k), starts in row n = 1 with columns k >= 0 as
%e 0 1 1 2 3 5 8
%e 0 1 2 5 12 29 70
%e 0 1 3 10 33 109 360
%e 0 1 4 17 72 305 1292
%e 0 1 5 26 135 701 3640
%e 0 1 6 37 228 1405 8658
%e 0 1 7 50 357 2549 18200
%e 0 1 8 65 528 4289 34840
%e 0 1 9 82 747 6805 61992
%e 0 1 10 101 1020 10301 104030
%e 0 1 11 122 1353 15005 166408
%e Antidiagonal triangle, T(n, k), begins as:
%e 0;
%e 0, 1;
%e 0, 1, 1;
%e 0, 1, 2, 2;
%e 0, 1, 3, 5, 3;
%e 0, 1, 4, 10, 12, 5;
%e 0, 1, 5, 17, 33, 29, 8;
%e 0, 1, 6, 26, 72, 109, 70, 13;
%e 0, 1, 7, 37, 135, 305, 360, 169, 21;
%e 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;
%t A172236[n_,k_]:=Fibonacci[k, n-k];
%t Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten
%o (PARI) A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
%o tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ _Jianing Song_, Jul 14 2018 (program from _Michel Marcus_; see also A316269)
%o (PARI) A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ _Jianing Song_, Nov 10 2018
%o (Magma)
%o A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
%o [A172236(n,k): k in [0..n-1], n in [1..13]]; // _G. C. Greubel_, Sep 29 2024
%o (SageMath)
%o def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
%o flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # _G. C. Greubel_, Sep 29 2024
%Y 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.)
%Y Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
%Y Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
%Y 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).
%Y Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
%Y Cf. A084844, A084845, A118243, A157103, A271782, A316269.
%Y Sums include: A304357, A304359.
%Y Similar to: A073133.
%K nonn,tabl,easy
%O 1,9
%A _Roger L. Bagula_, Jan 29 2010
%E More terms from _Jianing Song_, Jul 14 2018