%I #28 Sep 08 2022 08:46:22
%S 1,1,1,9,1,18,1,27,81,1,36,243,1,45,486,729,1,54,810,2916,1,63,1215,
%T 7290,6561,1,72,1701,14580,32805,1,81,2268,25515,98415,59049,1,90,
%U 2916,40824,229635,354294,1,99,3645,61236,459270,1240029,531441,1,108,4455,87480,826686,3306744,3720087
%N Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 9 * T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.
%C The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013616 ((1+9*x)^n) and along skew diagonals pointing top-left in center-justified triangle given in A038291 ((9+x)^n).
%C The coefficients in the expansion of 1/(1-x-9x^2) are given by the sequence generated by the row sums.
%C The row sums are Generalized Fibonacci numbers (see A015445).
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.5413812651491... ((1+sqrt(37))/2), when n approaches infinity.
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 100
%H Zagros Lalo, <a href="/A317051/a317051.pdf">Left-justified triangle</a>
%H Zagros Lalo, <a href="/A317051/a317051_3.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 9x)^n</a>
%H Zagros Lalo, <a href="/A317051/a317051_4.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (9 + x)^n</a>
%F T(n,k) = 9^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).
%e Triangle begins:
%e 1;
%e 1;
%e 1, 9;
%e 1, 18;
%e 1, 27, 81;
%e 1, 36, 243;
%e 1, 45, 486, 729;
%e 1, 54, 810, 2916;
%e 1, 63, 1215, 7290, 6561;
%e 1, 72, 1701, 14580, 32805;
%e 1, 81, 2268, 25515, 98415, 59049;
%e 1, 90, 2916, 40824, 229635, 354294;
%e 1, 99, 3645, 61236, 459270, 1240029, 531441;
%e 1, 108, 4455, 87480, 826686, 3306744, 3720087;
%e 1, 117, 5346, 120285, 1377810, 7440174, 14880348, 4782969;
%e 1, 126, 6318, 160380, 2165130, 14880348, 44641044, 38263752;
%e 1, 135, 7371, 208494, 3247695, 27280638, 111602610, 172186884, 43046721;
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 9 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten.
%t Table[9^k Binomial[n - k, k], {n, 0, 15}, {k, 0, Floor[n/2]}].
%o (GAP) Flat(List([0..13],n->List([0..Int(n/2)],k->9^k*Binomial(n-k,k)))); # _Muniru A Asiru_, Jul 20 2018
%o (PARI) T(n, k) = 9^k*binomial(n-k,k);
%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Jul 20 2018
%o (Magma) /* As triangle */ [[9^k*Binomial(n-k,k): k in [0..Floor(n/2)]]: n in [0.. 15]]; // _Vincenzo Librandi_, Sep 05 2018
%Y Row sums give A015445.
%Y Cf. A013616
%Y Cf. A038291
%K tabf,nonn,easy
%O 0,4
%A _Zagros Lalo_, Jul 20 2018