%I #26 May 19 2021 05:40:33
%S 1,0,3,0,1,9,0,0,6,27,0,0,1,27,81,0,0,0,9,108,243,0,0,0,1,54,405,729,
%T 0,0,0,0,12,270,1458,2187,0,0,0,0,1,90,1215,5103,6561,0,0,0,0,0,15,
%U 540,5103,17496,19683,0,0,0,0,0,1,135,2835,20412,59049,59049,0,0,0,0,0,0,18,945,13608,78732,196830,177147
%N Riordan array (1, 3+x).
%C Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).
%C Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Nov 10 2008
%H G. C. Greubel, <a href="/A099097/b099097.txt">Rows n = 0..50 of the triangle, flattened</a>
%F Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).
%F G.f. of column k: (3*x + x^2)^k.
%F G.f.: 1/(1 - 3*y*x - y*x^2). - _Philippe Deléham_, Nov 21 2011
%F Sum_{k=0..n} T(n,k)*x^k = A000007(n), A006190(n+1), A135030(n+1), A181353(n+1) for x = 0,1,2,3 respectively. - _Philippe Deléham_, Nov 21 2011
%e Triangle begins:
%e 1;
%e 0, 3;
%e 0, 1, 9;
%e 0, 0, 6, 27;
%e 0, 0, 1, 27, 81;
%e 0, 0, 0, 9, 108, 243;
%e ...
%t Table[3^(2*k-n)*Binomial[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, May 19 2021 *)
%o (Sage) flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 19 2021
%Y Cf. A027465.
%Y Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).
%K easy,nonn,tabl
%O 0,3
%A _Paul Barry_, Sep 25 2004