A318773 -id:A318773 - cp's OEIS frontend

A304249 Triangle T(n,k) = 3*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1 and T(n,k) = 0 for n < 0 or k < 0, read by rows.

1, 3, 9, 1, 27, 6, 81, 27, 1, 243, 108, 9, 729, 405, 54, 1, 2187, 1458, 270, 12, 6561, 5103, 1215, 90, 1, 19683, 17496, 5103, 540, 15, 59049, 59049, 20412, 2835, 135, 1, 177147, 196830, 78732, 13608, 945, 18, 531441, 649539, 295245, 61236, 5670, 189, 1

Offset: 0

Zagros Lalo, May 08 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n), cf. formula.
The coefficients in the expansion of 1/(1-3x-x^2) are given by the sequence generated by the row sums.
The sequence of the row sums are the "Bronze Fibonacci numbers" A006190, and the limit of their ratio is 3.30277563773... (Bronze ratio), see A098316.

			Triangle begins:
          1;
          3;
          9,         1;
         27,         6;
         81,        27,         1;
        243,       108,         9;
        729,       405,        54,        1;
       2187,      1458,       270,       12;
       6561,      5103,      1215,       90,        1;
      19683,     17496,      5103,      540,       15;
      59049,     59049,     20412,     2835,      135,       1;
     177147,    196830,     78732,    13608,      945,      18;
     531441,    649539,    295245,    61236,     5670,     189,      1;
    1594323,   2125764,   1082565,   262440,    30618,    1512,     21;
    4782969,   6908733,   3897234,  1082565,   153090,   10206,    252,    1;
   14348907,  22320522,  13817466,  4330260,   721710,   61236,   2268,   24;
   43046721,  71744535,  48361131, 16888014,  3247695,  336798,  17010,  324,  1;
  129140163, 229582512, 167403915, 64481508, 14073345, 1732104, 112266, 3240, 27;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 86, 363.

Zagros Lalo, Left justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1+3x)^n

Row sums give A006190.
Cf. A013610, A098316.
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), this sequence (q=2), A317497 (q=3), A318773 (q=4).

[3^(n-2*k)*Binomial(n-k,k): k in [0..Floor(n/2)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, 3T[n-1, k] + T[n-2, k-1]];
Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}]//Flatten
With[{q=2}, Table[3^(n-q*k)*Binomial[n-(q-1)*k, k], {n,0,24}, {k,0,Floor[n/q]}] ]//Flatten (* G. C. Greubel, May 12 2021 *)

T(n,k)=if( n>0 && k>0, 3*T(n-1, k) + T(n-2, k-1), !n && !k)
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

flatten([[3^(n-2*k)*binomial(n-k,k) for k in (0..n//2)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = A013610(n-k, n-2k). - M. F. Hasler, Jun 01 2018

A317497 Triangle T(n,k) = 3*T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 27, 1, 81, 6, 243, 27, 729, 108, 1, 2187, 405, 9, 6561, 1458, 54, 19683, 5103, 270, 1, 59049, 17496, 1215, 12, 177147, 59049, 5103, 90, 531441, 196830, 20412, 540, 1, 1594323, 649539, 78732, 2835, 15, 4782969, 2125764, 295245, 13608, 135, 14348907, 6908733, 1082565, 61236, 945, 1

Offset: 0

Zagros Lalo, Jul 31 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and  along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in 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/(1-3x-x^3) are given by the sequence generated by the row sums.
The row sums give A052541.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.1038034027355..., when n approaches infinity.

			Triangle begins:
         1;
         3;
         9;
        27,        1;
        81,        6;
       243,       27;
       729,      108,       1;
      2187,      405,       9;
      6561,     1458,      54;
     19683,     5103,     270,      1;
     59049,    17496,    1215,     12;
    177147,    59049,    5103,     90;
    531441,   196830,   20412,    540,    1;
   1594323,   649539,   78732,   2835,   15;
   4782969,  2125764,  295245,  13608,  135;
  14348907,  6908733, 1082565,  61236,  945,  1;
  43046721, 22320522, 3897234, 262440, 5670, 18;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366

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

Row sums give A052541.
Cf. A013610, A027465, A317496, A304236, A304249.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3), A036217 (column 4).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), this sequence (q=3), A318773 (q=4).

Flat(List([0..16],n->List([0..Int(n/3)],k->3^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018

[3^(n-3*k)*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-3k)(n-2k)!/((n-3k)! k!); Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3 T[n-1, k] + T[n-3, k-1]]; Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}]//Flatten

flatten([[3^(n-3*k)*binomial(n-2*k,k) for k in (0..n//3)] for n in (0..24)]) # G. C. Greubel, May 12 2021

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

A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 10, 13, 25, 46, 76, 115, 190, 328, 556, 901, 1471, 2455, 4123, 6826, 11239, 18604, 30973, 51451, 85168, 140980, 233899, 388252, 643756, 1066696, 1768393, 2933149, 4864417, 8064505, 13369684, 22169131, 36762382, 60955897, 101064949, 167572342, 277859488, 460727179, 763922026, 1266639052

Offset: 0

Zagros Lalo, Sep 04 2018

The coefficients in the expansion of 1/(1 - x - 3*x^4) are given by the sequence generated by the row sums in triangle A318772.

Coefficients in expansion of 1/(1 - x - 3*x^4) are given by the sum of numbers along "third Layer" skew diagonals pointing top-right in triangle A013610 ((1+3x)^n) and by the sum of numbers along "third Layer" skew diagonals pointing top-left in triangle A027465 ((3+x)^n), see links.

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3 x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3).

Cf. A013610, A027465, A318772, A318773.

Essentially a duplicate of A143454.

[n le 4 select 1 else Self(n-1) +3*Self(n-4): n in [1..51]]; // G. C. Greubel, May 08 2021

CoefficientList[Series[1/(1-x-3x^4), {x, 0, 50}], x]
a[n_]:= a[n]= If[n<4, 1, a[n-1] + 3*a[n-4]]; Table[a[n], {n,0,50}]
LinearRecurrence[{1,0,0,3}, {1,1,1,1}, 51]

my(p=Mod('x,x^4-'x^3-3)); a(n) = vecsum(Vec(lift(p^n))); \\ Kevin Ryde, May 11 2021

def a(n): return 1 if (n<4) else a(n-1) + 3*a(n-4)
[a(n) for n in (0..50)] # G. C. Greubel, May 08 2021

a(n) = a(n-1) + 3*a(n-4) for n >= 0, a(n)=0 for n < 0, with a(0) = a(1) = a(2) = a(3) = 1.

cp's OEIS Frontend

A304249 Triangle T(n,k) = 3*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1 and T(n,k) = 0 for n < 0 or k < 0, read by rows.

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A317497 Triangle T(n,k) = 3*T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

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A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

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