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A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

%I A227074 #10 May 11 2025 23:47:51
%S A227074 1,4,4,16,8,16,64,24,24,64,256,88,48,88,256,1024,344,136,136,344,1024,
%T A227074 4096,1368,480,272,480,1368,4096,16384,5464,1848,752,752,1848,5464,
%U A227074 16384,65536,21848,7312,2600,1504,2600,7312,21848,65536,262144,87384,29160
%N A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.
%C A227074 All rows except the zeroth are divisible by 4. Is there a closed-form formula for these numbers, like for binomial coefficients?
%H A227074 T. D. Noe, <a href="/A227074/b227074.txt">Rows n = 0..50 of triangle, flattened</a>
%e A227074 Triangle begins:
%e A227074   1,
%e A227074   4, 4,
%e A227074   16, 8, 16,
%e A227074   64, 24, 24, 64,
%e A227074   256, 88, 48, 88, 256,
%e A227074   1024, 344, 136, 136, 344, 1024,
%e A227074   4096, 1368, 480, 272, 480, 1368, 4096,
%e A227074   16384, 5464, 1848, 752, 752, 1848, 5464, 16384,
%e A227074   65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536
%t A227074 t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 4^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 A227074 Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
%Y A227074 Cf. A051601 (n on the borders), A137688 (2^n on borders).
%Y A227074 Cf. A165665 (row sums: 3*4^n - 2*2^n), A227075 (3^n edges), A227076 (5^n edges).
%K A227074 nonn,tabl
%O A227074 0,2
%A A227074 _T. D. Noe_, Aug 06 2013