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%I #27 Feb 01 2023 08:19:20
%S 1,2,4,3,7,19,4,10,28,78,5,13,37,105,301,6,16,46,132,382,1108,7,19,55,
%T 159,463,1351,3951,8,22,64,186,544,1594,4680,13758,9,25,73,213,625,
%U 1837,5409,15945,47049,10,28,82,240,706,2080,6138,18132,53610,158616,11,31,91,267,787,2323,6867,20319,60171,178299,528619
%N Triangle read by rows where row n is the largest (or middle or n-th) column of the reverse pyramid summation of order n described in A359087.
%C The integer that is at the k-th row of the middle column of this pyramid of order n will be noted T(n,k).
%C Each row has n terms.
%F T(n,1) = n.
%F T(n,2) = 3n - 2.
%F T(n,3) = 9n - 8.
%F T(n,4) = 27n - 30.
%F T(n,5) = 81n - 104.
%F T(n,n) = A359087(n).
%F T(n,k) = 3^(k-1)*n - 2*A132894(k-1) for 1 <= k <= n (conjectured).
%e Triangle begins:
%e n=1: 1;
%e n=2: 2, 4;
%e n=3: 3, 7, 19;
%e n=4: 4, 10, 28, 78;
%e n=5: 5, 13, 37, 105, 301;
%e n=6: 6, 16, 46, 132, 382, 1108;
%e ...
%e For n=5, the reverse pyramid summation is as follows and row 5 here is the middle column 5,13,37,...
%e 1 2 3 4 5 4 3 2 1
%e 6 9 12 13 12 9 6
%e 27 34 37 34 27
%e 98 105 98
%e 301
%o (PARI) f(v) = if (#v == 1, v, vector(#v-2, i, v[i]+v[i+1]+v[i+2]));
%o row(n) = my(u = concat([1..n], Vecrev([1..n-1])), v=u, w = vector(n)); for (i=1, n, w[i] = v[#v\2+1]; v = f(v);); w; \\ _Michel Marcus_, Jan 30 2023
%Y Cf. A132894, A359087 (right diagonal).
%Y Columns k=1..3: A000027, A016777, A017173.
%K nonn,tabl
%O 1,2
%A _Bernard Schott_, Jan 27 2023