A193673 Triangle given by p(n,k)=(coefficient of x^(n-k) in (1/2) ((x+3)^n+(x+1)^n)), 0<=k<=n.

1, 2, 1, 5, 4, 1, 14, 15, 6, 1, 41, 56, 30, 8, 1, 122, 205, 140, 50, 10, 1, 365, 732, 615, 280, 75, 12, 1, 1094, 2555, 2562, 1435, 490, 105, 14, 1, 3281, 8752, 10220, 6832, 2870, 784, 140, 16, 1, 9842, 29529, 39384, 30660, 15372, 5166, 1176, 180, 18, 1, 29525

Offset

0,2

Links

Table of n, a(n) for n=0..55.

Formula

From Mélika Tebni, Dec 09 2023: (Start)

T(n,k) = binomial(n,k)*(1 + 3^(n-k)) / 2.

E.g.f. of column k: exp(2*x)*cosh(x)*x^k / k!. (End)

From Peter Bala, Mar 07 2024: (Start)

Exponential Riordan array (exp(2*x)*cosh(x), x).

The zeros of the n-th row polynomial R(n,x) = ((1 + x)^n + (3 + x)^n)/2 lie on the vertical line Re(x) = -2 in the complex plane.

Triangle equals P * (I + P^2)/2 = P * A119468 = P^2 * A119467, where P denotes Pascal's triangle A007318. (End)

Example

First five rows:
   1
   2   1
   5   4   1
  14  15   6   1
  41  56  30   8   1

Mathematica

q[n_, k_] := 1; r[0] = 1;
 r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
p[n_, k_] := Coefficient[(1/2) ((x + 3)^n + (x + 1)^n), x, k] (* A193673 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 20}]    (* A193661 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 10}, {k, 0, n}]] (* A193673 as a triangle *)
Flatten[%] (* A193673 as a sequence *)

Crossrefs

Cf. A119467, A119468, A193661.

Sequence in context: A104710 A039598 A128738 * A126181 A362924 A154930

Adjacent sequences: A193670 A193671 A193672 * A193674 A193675 A193676

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Aug 02 2011

Status

approved