%I #14 Sep 05 2018 02:40:01
%S 1,10,100,1,1000,20,10000,300,1,100000,4000,30,1000000,50000,600,1,
%T 10000000,600000,10000,40,100000000,7000000,150000,1000,1,1000000000,
%U 80000000,2100000,20000,50,10000000000,900000000,28000000,350000,1500,1,100000000000,10000000000,360000000,5600000,35000,60
%N Triangle read by rows: T(0,0) = 1; T(n,k) = 10 T(n-1,k) + 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-left in center-justified triangle given in A013617 ((1+10*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038303 ((10+x)^n).
%C The coefficients in the expansion of 1/(1-10x-x^2) are given by the sequence generated by the row sums.
%C The row sums are Denominators of continued fraction convergents to sqrt(26), see A041041.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 10.09901951359278483002... (a metallic mean) when n approaches infinity (see A176537: (5+sqrt(26))).
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 102
%H Zagros Lalo, <a href="/A317055/a317055.pdf">Left-justified triangle</a>
%H Zagros Lalo, <a href="/A317055/a317055_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 10x)^n</a>
%H Zagros Lalo, <a href="/A317055/a317055_2.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (10 + x)^n</a>
%e Triangle begins:
%e 1;
%e 10;
%e 100, 1;
%e 1000, 20;
%e 10000, 300, 1;
%e 100000, 4000, 30;
%e 1000000, 50000, 600, 1;
%e 10000000, 600000, 10000, 40;
%e 100000000, 7000000, 150000, 1000, 1;
%e 1000000000, 80000000, 2100000, 20000, 50;
%e 10000000000, 900000000, 28000000, 350000, 1500, 1;
%e 100000000000, 10000000000, 360000000, 5600000, 35000, 60;
%e 1000000000000, 110000000000, 4500000000, 84000000, 700000, 2100, 1;
%e 10000000000000, 1200000000000, 55000000000, 1200000000, 12600000, 56000, 70;
%e 100000000000000, 13000000000000, 660000000000, 16500000000, 210000000, 1260000, 2800, 1;
%t t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 10 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten.
%Y Row sums give A041041.
%Y Cf. A013617
%Y Cf. A038303
%Y Cf. A176537
%Y Cf. A011557 (column 0), A053541 (column 1), A081140 (column 2).
%K tabf,nonn,easy
%O 0,2
%A _Zagros Lalo_, Jul 21 2018