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%I #15 Nov 13 2021 05:49:39
%S 1,1,1,10,1,20,1,30,100,1,40,300,1,50,600,1000,1,60,1000,4000,1,70,
%T 1500,10000,10000,1,80,2100,20000,50000,1,90,2800,35000,150000,100000,
%U 1,100,3600,56000,350000,600000,1,110,4500,84000,700000,2100000,1000000,1,120,5500,120000,1260000,5600000,7000000
%N Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 10 * 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 A013617 ((1+10x)^n) and along skew diagonals pointing top-left in center-justified triangle given in A038303 ((10+x)^n).
%C The coefficients in the expansion of 1/(1-x-10x^2) are given by the sequence generated by the row sums.
%C The row sums are Generalized Fibonacci numbers (see A015446).
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.701562118716424343244... ((1+sqrt(41))/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, 102.
%H Zagros Lalo, <a href="/A317054/a317054.pdf">Left-justified triangle</a>
%H Zagros Lalo, <a href="/A317054/a317054_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 10x)^n</a>
%H Zagros Lalo, <a href="/A317054/a317054_2.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (10 + x)^n</a>
%e Triangle begins:
%e 1;
%e 1;
%e 1, 10;
%e 1, 20;
%e 1, 30, 100;
%e 1, 40, 300;
%e 1, 50, 600, 1000;
%e 1, 60, 1000, 4000;
%e 1, 70, 1500, 10000, 10000;
%e 1, 80, 2100, 20000, 50000;
%e 1, 90, 2800, 35000, 150000, 100000;
%e 1, 100, 3600, 56000, 350000, 600000;
%e 1, 110, 4500, 84000, 700000, 2100000, 1000000;
%e 1, 120, 5500, 120000, 1260000, 5600000, 7000000;
%e 1, 130, 6600, 165000, 2100000, 12600000, 28000000, 10000000;
%e 1, 140, 7800, 220000, 3300000, 25200000, 84000000, 80000000;
%e 1, 150, 9100, 286000, 4950000, 46200000, 210000000, 360000000, 100000000;
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 10 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten.
%t Table[10^k Binomial[n - k, k], {n, 0, 15}, {k, 0, Floor[n/2]}].
%o (PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1, k)+10*T(n-2, k-1)));
%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Jul 20 2018
%Y Row sums give A015446.
%Y Cf. A013617
%Y Cf. A038303
%K tabf,nonn,easy
%O 0,4
%A _Zagros Lalo_, Jul 20 2018
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