%I #15 Sep 05 2018 02:27:23
%S 1,1,1,5,1,10,1,15,25,1,20,75,1,25,150,125,1,30,250,500,1,35,375,1250,
%T 625,1,40,525,2500,3125,1,45,700,4375,9375,3125,1,50,900,7000,21875,
%U 18750,1,55,1125,10500,43750,65625,15625,1,60,1375,15000,78750,175000,109375
%N Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 5*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 A013612 ((1+5*x)^n). The coefficients in the expansion of 1/(1-x-5x^2) are given by the sequence generated by the row sums.
%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.7913327..., when n approaches infinity. The row sums are A015440 (generalized Fibonacci numbers).
%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 380, 381.
%H Shara Lalo, <a href="/A305838/a305838.pdf">Skew diagonals in triangle A013612</a>
%H Shara Lalo, <a href="/A305838/a305838_1.pdf">Right-justified triangle</a>
%F G.f.: 1/(1 - t*x - 5*t^2).
%e Triangle begins:
%e 1;
%e 1;
%e 1, 5;
%e 1, 10;
%e 1, 15, 25;
%e 1, 20, 75;
%e 1, 25, 150, 125;
%e 1, 30, 250, 500;
%e 1, 35, 375, 1250, 625;
%e 1, 40, 525, 2500, 3125;
%e 1, 45, 700, 4375, 9375, 3125;
%e 1, 50, 900, 7000, 21875, 18750;
%e 1, 55, 1125, 10500, 43750, 65625, 15625;
%e 1, 60, 1375, 15000, 78750, 175000, 109375;
%e 1, 65, 1650, 20625, 131250, 393750, 437500, 78125;
%e 1, 70, 1950, 27500, 206250, 787500, 1312500, 625000;
%e 1, 75, 2275, 35750, 309375, 1443750, 3281250, 2812500, 390625;
%e 1, 80, 2625, 45500, 446875, 2475000, 7218750, 9375000, 3515625;
%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 5 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten
%Y Row sums give A015440.
%Y Cf. A000012 (column 0), A008587 (column 1), A123296 (column 2), A141480 (column 3).
%Y Cf. A013612.
%K tabf,nonn,easy
%O 0,4
%A _Shara Lalo_, Jun 11 2018