login
Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.
37

%I #22 Apr 17 2023 22:12:28

%S 1,1,1,1,1,1,1,2,2,1,1,3,5,3,1,1,5,12,10,4,1,1,8,29,33,17,5,1,1,13,70,

%T 109,72,26,6,1,1,21,169,360,305,135,37,7,1,1,34,408,1189,1292,701,228,

%U 50,8,1,1,55,985,3927,5473,3640,1405,357,65,9,1

%N Array A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1, read by antidiagonals.

%C From _Michael A. Allen_, Mar 30 2023: (Start)

%C Column k is the k-metallonacci sequence for k > 0.

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

%H G. C. Greubel, <a href="/A157103/b157103.txt">Antidiagonals n = 0..50, flattened</a>

%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.

%H Michelle Rudolph-Lilith, <a href="http://arxiv.org/abs/1508.07894">On the Product Representation of Number Sequences, with Application to the Fibonacci Family</a>, arXiv preprint arXiv:1508.07894 [math.NT], 2015. See Table 3.

%F A(n, k) = Fibonacci(n+1, k), with A(n, 0) = A(n, n) = 1 (array).

%F A(n, 1) = A000045(n+1).

%F T(n, k) = k*T(n-1, k) + T(n-2, k) with T(n, 0) = T(n, n) = 1 (triangle).

%F From _G. C. Greubel_, Jan 11 2022: (Start)

%F T(n, k) = Fibonacci(n-k+1, k), with T(n, 0) = T(n, n) = 1.

%F T(2*n, n) = A084845(n) for n >= 1, with T(0, 0) = 1.

%F T(2*n+1, n+1) = A084844(n). (End)

%e Array begins:

%e 1, 1, 1, 1, 1, 1, 1, 1, ... (A000012);

%e 1, 1, 2, 3, 4, 5, 6, 7, ... (A000027);

%e 1, 2, 5, 10, 17, 26, 37, 50, ... (A002522);

%e 1, 3, 12, 33, 72, 135, 228, 357, ...;

%e 1, 5, 29, 109, 305, 701, 1405, 2549, ...;

%e 1, 8, 70, 360, 1292, 3640, 8658, 18200, ...;

%e 1, 13, 169, 1189, 5473, 18901, 53353, 129949, ...;

%e 1, 21, 408, 3927, 23184, 98145, 328776, 927843, ...;

%e ...

%e First few rows of the triangle:

%e 1;

%e 1, 1;

%e 1, 1, 1;

%e 1, 2, 2, 1;

%e 1, 3, 5, 3, 1;

%e 1, 5, 12, 10, 4, 1;

%e 1, 8, 29, 33, 17, 5, 1;

%e 1, 13, 70, 109, 72, 26, 6, 1;

%e 1, 21, 169, 360, 305, 135, 37, 7, 1;

%e 1, 34, 408, 1189, 1292, 701, 228, 50, 8, 1;

%e 1, 55, 985, 3927, 5473, 3640, 1405, 357, 65, 9, 1;

%e 1, 89, 2378, 12970, 23184, 18901, 8658, 2549, 528, 82, 10, 1;

%e 1, 144, 5741, 42837, 98209, 98145, 53353, 18200, 4289, 747, 101, 11, 1;

%e ...

%e Example: Column 3 = (1, 3, 10, 33, 109, 360, ...) = A006190.

%p A157103 := proc(n,k)

%p if k = 0 then

%p 1;

%p else

%p mul(k-2*I*cos(l*Pi/(n+1)),l=1..n) ;

%p combine(%,trig) ;

%p round(%) ;

%p end if;

%p end proc:

%p seq( seq(A157103(d-k,k),k=0..d),d=0..12) ; # _R. J. Mathar_, Feb 27 2023

%t (* First program *)

%t T[_, 0]=1; T[n_, n_]=1; T[_, _]=0;

%t T[n_, k_] /; 0 <= k <= n := k T[n-1, k] + T[n-2, k];

%t Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* _Jean-François Alcover_, Aug 07 2018 *)

%t (* Second program *)

%t T[n_, k_]:= If[k==0 || k==n, 1, Fibonacci[n-k+1, k]];

%t Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 11 2022 *)

%o (Magma)

%o A157103:= func< n,k | k eq 0 or k eq n select 1 else Evaluate(DicksonSecond(n, -1), k) >;

%o [A157103(n-k, k): k in [0..n], n in [0..15]]; // _G. C. Greubel_, Jan 11 2022

%o (Sage)

%o def A157103(n,k): return 1 if (k==0 or k==n) else lucas_number1(n+1, k, -1)

%o flatten([[A157103(n-k, k) for k in (0..n)] for n in (0..10)]) # _G. C. Greubel_, Jan 11 2022

%Y Columns k=1 to 9: A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371.

%Y Cf. A084844, A084845.

%Y Essentially the transpose of A073133, A172236, A352361.

%K nonn,easy,tabl

%O 0,8

%A _Gary W. Adamson_, Feb 22 2009

%E Edited by _G. C. Greubel_, Jan 11 2022