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Triangle T(n,k) = T(n-1,k) + 3*T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1, T(n,k) = 0 for n or k < 0, read by rows.
5

%I #14 Sep 08 2022 08:46:22

%S 1,1,1,1,3,1,6,1,9,1,12,9,1,15,27,1,18,54,1,21,90,27,1,24,135,108,1,

%T 27,189,270,1,30,252,540,81,1,33,324,945,405,1,36,405,1512,1215,1,39,

%U 495,2268,2835,243,1,42,594,3240,5670,1458,1,45,702,4455,10206,5103,1,48,819,5940,17010,13608,729

%N Triangle T(n,k) = T(n-1,k) + 3*T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1, T(n,k) = 0 for n or k < 0, read by rows.

%C The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n) and along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)

%C The coefficients in the expansion of 1/(1-x-3x^3) 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 1.863706527819..., 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. 364-366.

%H G. C. Greubel, <a href="/A317496/b317496.txt">Rows n = 0..120 of the irregular triangle, flattened</a>

%H Zagros Lalo, <a href="/A317496/a317496.pdf">Second layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n</a>

%H Zagros Lalo, <a href="/A317496/a317496_1.pdf">Second layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n</a>

%F T(n,k) = 3^k * (n-2*k)!/ (k! * (n-3*k)!) where n is a nonnegative integer and k = 0..floor(n/3).

%e Triangle begins:

%e 1;

%e 1;

%e 1;

%e 1, 3;

%e 1, 6;

%e 1, 9;

%e 1, 12, 9;

%e 1, 15, 27;

%e 1, 18, 54;

%e 1, 21, 90, 27;

%e 1, 24, 135, 108;

%e 1, 27, 189, 270;

%e 1, 30, 252, 540, 81;

%e 1, 33, 324, 945, 405;

%e 1, 36, 405, 1512, 1215;

%e 1, 39, 495, 2268, 2835, 243;

%e 1, 42, 594, 3240, 5670, 1458;

%e 1, 45, 702, 4455, 10206, 5103;

%e 1, 48, 819, 5940, 17010, 13608, 729;

%t T[n_, k_]:= T[n, k] = 3^k*(n-2*k)!/((n-3*k)!*k!); Table[T[n, k], {n, 0, 18}, {k, 0, Floor[n/3]} ]//Flatten

%t T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, T[n-1, k] + 3T[n-3, k-1]]; Table[T[n, k], {n, 0, 18}, {k, 0, Floor[n/3]}]//Flatten

%o (GAP) Flat(List([0..20],n->List([0..Int(n/3)],k->3^k/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # _Muniru A Asiru_, Aug 01 2018

%o (Magma) [3^k*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // _G. C. Greubel_, May 12 2021

%o (Sage) flatten([[3^k*binomial(n-2*k,k) for k in (0..n//3)] for n in (0..24)]) # _G. C. Greubel_, May 12 2021

%Y Row sums give A084386.

%Y Cf. A013610, A027465, A304236, A304249, A317497.

%Y Sequences of the form 3^k*binomial(n-(q-1)*k, k): A013610 (q=1), A304236 (q=2), this sequence (q=3), A318772 (q=4).

%K tabf,nonn,easy

%O 0,5

%A _Zagros Lalo_, Jul 31 2018