A318774 -id:A318774 - cp's OEIS frontend

A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

1, 1, 1, 1, 1, 3, 1, 6, 1, 9, 1, 12, 1, 15, 9, 1, 18, 27, 1, 21, 54, 1, 24, 90, 1, 27, 135, 27, 1, 30, 189, 108, 1, 33, 252, 270, 1, 36, 324, 540, 1, 39, 405, 945, 81, 1, 42, 495, 1512, 405, 1, 45, 594, 2268, 1215, 1, 48, 702, 3240, 2835, 1, 51, 819, 4455, 5670, 243, 1, 54, 945, 5940, 10206, 1458

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" 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.)
The coefficients in the expansion of 1/(1-x-3*x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.6580980673722..., when n approaches infinity.

			Triangle begins:
  1;
  1;
  1;
  1;
  1,  3;
  1,  6;
  1,  9;
  1, 12;
  1, 15,   9;
  1, 18,  27;
  1, 21,  54;
  1, 24,  90;
  1, 27, 135,   27;
  1, 30, 189,  108;
  1, 33, 252,  270;
  1, 36, 324,  540;
  1, 39, 405,  945,   81;
  1, 42, 495, 1512,  405;
  1, 45, 594, 2268, 1215;
  ...

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A318774.
Cf. A013610, A027465.
Cf. A304236, A304249.
Cf. A317496, A317497.

[3^k*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k]= 3^k(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]} ] // Flatten
T[0, 0] = 1; T[n_, k_] := T[n, k] = If[n<0 || k<0, 0, T[n-1, k] + 3T[n-4, k-1]]; Table[T[n, k], {n, 0, 21}, {k, 0, Floor[n/4]}] // Flatten

flatten([[3^k*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^k * (n - 3*k)!/ ((n - 4*k)! k!), where n >= 0 and 0 <= k <= floor(n/4).

A172369 Triangle read by rows: T(n,k,q) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a sequence defined in comments.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 4, 1, 1, 7, 28, 28, 28, 7, 1, 1, 10, 70, 280, 280, 70, 10, 1, 1, 13, 130, 910, 3640, 910, 130, 13, 1, 1, 25, 325, 3250, 22750, 22750, 3250, 325, 25, 1, 1, 46, 1150, 14950, 149500, 261625, 149500, 14950, 1150, 46, 1

Offset: 0

Roger L. Bagula, Feb 01 2010

Start from A143454 and its partial products c(n) = 1, 1, 1, 1, 1, 4, 28, 280, 3640, 91000, 4186000, ... . Then T(n,k) = round(c(n)/(c(k)*c(n-k))).

			Triangle begins as:
  1;
  1,  1;
  1,  1,    1;
  1,  1,    1,     1;
  1,  1,    1,     1,      1;
  1,  4,    4,     4,      4,      1;
  1,  7,   28,    28,     28,      7,      1;
  1, 10,   70,   280,    280,     70,     10,     1;
  1, 13,  130,   910,   3640,    910,    130,    13,    1;
  1, 25,  325,  3250,  22750,  22750,   3250,   325,   25,  1;
  1, 46, 1150, 14950, 149500, 261625, 149500, 14950, 1150, 46, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A172363 (q=1), A172368 (q=2), this sequence (q=3), A318774.

f[n_, q_]:= f[n, q]= If[n==0, 0, If[n<4, 1, f[n-1, q] +q*f[n-4, q]]];
c[n_, q_]:= Product[f[j, q], {j,n}];
T[n_, k_, q_]:= Round[c[n, q]/(c[k, q]*c[n - k, q])];
Table[T[n, k, 3], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, May 08 2021 *)

@CachedFunction
def f(n,q): return 0 if (n==0) else 1 if (n<4) else f(n-1, q) + q*f(n-4, q)
def c(n,q): return product( f(j,q) for j in (1..n) )
def T(n,k,q): return round(c(n, q)/(c(k, q)*c(n-k, q)))
flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 08 2021

T(n, k, q) = round( c(n,q)/(c(k,q)*c(n-k,q)) ), where c(n, q) = Product_{j=1..n} f(j, q), f(n, q) = f(n-1, q) + q*f(n-4, q), f(0, q) = 0, f(1, q) = f(2, q) = f(3, q) = 1, and q = 3. - G. C. Greubel, May 08 2021

Definition corrected to give integral terms, G. C. Greubel, May 08 2021

cp's OEIS Frontend

A318772 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 3 * T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

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