OFFSET
0,4
COMMENTS
The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n).
The coefficients in the expansion of 1/(1-x-3x^2) are given by the sequence generated by the row sums.
REFERENCES
Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 88, 363.
LINKS
G. C. Greubel, Rows n = 0..100 of the irregular triangle, flattened
Zagros Lalo, Left-justified triangle
FORMULA
T(n,k) = 3^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).
EXAMPLE
Triangle begins:
1;
1;
1, 3;
1, 6;
1, 9, 9;
1, 12, 27;
1, 15, 54, 27;
1, 18, 90, 108;
1, 21, 135, 270, 81;
1, 24, 189, 540, 405;
1, 27, 252, 945, 1215, 243;
1, 30, 324, 1512, 2835, 1458;
1, 33, 405, 2268, 5670, 5103, 729;
1, 36, 495, 3240, 10206, 13608, 5103;
1, 39, 594, 4455, 17010, 30618, 20412, 2187;
1, 42, 702, 5940, 26730, 61236, 61236, 17496;
1, 45, 819, 7722, 40095, 112266, 153090, 78732, 6561;
1, 48, 945, 9828, 57915, 192456, 336798, 262440, 59049;
MAPLE
seq(seq( 3^k*binomial(n-k, k), k=0..floor(n/2)), n=0..24); # G. C. Greubel, May 12 2021
MATHEMATICA
T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, T[n-1, k] + 3*T[n-2, k-1]]; Table[T[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}]//Flatten.
Table[3^k Binomial[n-k, k], {n, 0, 17}, {k, 0, Floor[n/2]}]//Flatten.
PROG
(PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1, k) + 3*T(n-2, k-1)));
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 10 2018
(Magma) /* As triangle */ [[3^k*Binomial(n-k, k): k in [0..Floor(n/2)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018
(Sage) flatten([[3^k*binomial(n-k, k) for k in (0..n//2)] for n in (0..24)]) # G. C. Greubel, May 12 2021
CROSSREFS
KEYWORD
tabf,nonn,easy
AUTHOR
Zagros Lalo, May 08 2018
STATUS
approved