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URL: https://oeis.org/A227075

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A227075
A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.
6
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
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
KEYWORD
nonn,tabl
AUTHOR
T. D. Noe, Aug 01 2013
STATUS
approved