A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938

Offset

0,2

Comments

All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?

Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013

Links

T. D. Noe, Rows n = 0..50 of triangle, flattened

Example

Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561

Mathematica

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

Crossrefs

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).

Cf. A051601 (n on the borders), A137688 (2^n on borders).

Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).

Sequence in context: A197414 A247571 A292885 * A165351 A215665 A200494

Adjacent sequences: A227072 A227073 A227074 * A227076 A227077 A227078

Keyword

nonn,tabl

Author

T. D. Noe, Aug 01 2013

Status

approved