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%I #23 Aug 18 2013 08:47:27
%S 1,3,3,9,6,9,27,15,15,27,81,42,30,42,81,243,123,72,72,123,243,729,366,
%T 195,144,195,366,729,2187,1095,561,339,339,561,1095,2187,6561,3282,
%U 1656,900,678,900,1656,3282,6561,19683,9843,4938,2556,1578,1578,2556,4938
%N A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.
%C All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?
%C 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
%H T. D. Noe, <a href="/A227075/b227075.txt">Rows n = 0..50 of triangle, flattened</a>
%e Triangle:
%e 1,
%e 3, 3,
%e 9, 6, 9,
%e 27, 15, 15, 27,
%e 81, 42, 30, 42, 81,
%e 243, 123, 72, 72, 123, 243,
%e 729, 366, 195, 144, 195, 366, 729,
%e 2187, 1095, 561, 339, 339, 561, 1095, 2187,
%e 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561
%t 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]
%Y Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
%Y Cf. A051601 (n on the borders), A137688 (2^n on borders).
%Y Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).
%K nonn,tabl
%O 0,2
%A _T. D. Noe_, Aug 01 2013
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