A305837 - OEIS

%I #13 Sep 05 2018 02:27:19

%S 1,5,25,1,125,10,625,75,1,3125,500,15,15625,3125,150,1,78125,18750,

%T 1250,20,390625,109375,9375,250,1,1953125,625000,65625,2500,25,

%U 9765625,3515625,437500,21875,375,1,48828125,19531250,2812500,175000,4375,30,244140625,107421875,17578125,1312500,43750,525,1

%N Triangle read by rows: T(0,0) = 1; T(n,k) = 5*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 A013612 ((1+5*x)^n).

%C The coefficients in the expansion of 1/(1-5x-x^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 5.1925824035..., a metallic mean (see A098318), 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, 72, 92, 380, 382.

%H Shara Lalo, <a href="/A305837/a305837.pdf">Left-justified triangle </a>

%H Shara Lalo, <a href="/A305837/a305837_1.pdf">Skew diagonals in triangle A013612</a>

%F G.f.: 1/(1 - 5*t*x - t^2).

%e Triangle begins:

%e             1;

%e             5;

%e            25,           1;

%e           125,          10;

%e           625,          75,          1;

%e          3125,         500,         15;

%e         15625,        3125,        150,         1;

%e         78125,       18750,       1250,        20;

%e        390625,      109375,       9375,       250,        1;

%e       1953125,      625000,      65625,      2500,       25;

%e       9765625,     3515625,     437500,     21875,      375,      1;

%e      48828125,    19531250,    2812500,    175000,     4375,     30;

%e     244140625,   107421875,   17578125,   1312500,    43750,    525,     1;

%e    1220703125,   585937500,  107421875,   9375000,   393750,   7000,    35;

%e    6103515625,  3173828125,  644531250,  64453125,  3281250,  78750,   700,  1;

%e   30517578125, 17089843750, 3808593750, 429687500, 25781250, 787500, 10500, 40;

%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 5 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

%Y Row sums give A052918.

%Y Cf. A000351 (column 0), A053464 (column 1), A081135 (column 2), A081143 (column 3), A036071 (column 4).

%Y Cf. A013612.

%Y Cf. A098318.

%K tabf,nonn,easy

%O 0,2

%A _Shara Lalo_, Jun 11 2018