%I #28 Mar 05 2021 19:51:16
%S 2,3,4,5,7,10,9,13,19,28,17,25,37,55,82,33,49,73,109,163,244,65,97,
%T 145,217,325,487,730,129,193,289,433,649,973,1459,2188,257,385,577,
%U 865,1297,1945,2917,4375,6562,513,769,1153,1729,2593,3889,5833,8749,13123,19684
%N Triangular array T of numbers generated by these rules: 2 is in T; and if x is in T, then 2x-1 and 3x-2 are in T.
%C To obtain row n from row n-1, apply 2x-1 to each x in row n-1 and then put 1+3^n at the end. Or, instead, apply 3x-2 to each x in row n-1 and then put 1+2^n at the beginning.
%C From _Lamine Ngom_, Feb 10 2021: (Start)
%C Triangle read by diagonals provides all the sequences of the form 1+2^(k-1)*3^n, where k is the k-th diagonal.
%C For instance, the terms of the first diagonal form the sequence 2, 4, 10, 28, ..., i.e., 1+3^n (A034472).
%C The 2nd diagonal leads to the sequence 3, 7, 19, 55, ..., i.e., 1+2*3^n (A052919).
%C The 3rd diagonal is the sequence 5, 13, 37, 109, ..., i.e., 1+4*3^n (A199108).
%C And for k = 4, we obtain the sequence 9, 25, 73, 217, ..., i.e., 1+8*3^n (A199111). (End)
%H Ivan Neretin, <a href="/A094617/b094617.txt">Table of n, a(n) for n = 1..5151</a>
%F When offset is zero, then the first term is T(0,0) = 2, and
%F T(n,0) = 1 + 2^n = A000051(n),
%F T(n,n) = 1 + 3^n = A048473(n),
%F T(2n,n) = 1 + 6^n = A062394(n).
%F Row sums = A094618.
%F a(n) = A036561(n-1) + 1. - _Filip Zaludek_, Nov 19 2016
%e Rows of this triangle begin:
%e 2;
%e 3, 4;
%e 5, 7, 10;
%e 9, 13, 19, 28;
%e 17, 25, 37, 55, 82;
%e 33, 49, 73, 109, 163, 244;
%e 65, 97, 145, 217, 325, 487, 730;
%e 129, 193, 289, 433, 649, 973, 1459, 2188;
%e 257, 385, 577, 865, 1297, 1945, 2917, 4375, 6562;
%e 513, 769, 1153, 1729, 2593, 3889, 5833, 8749, 13123, 19684;
%e ...
%t FoldList[Append[2 #1 - 1, 1 + 3^#2] &, {2}, Range[9]] // Flatten (* _Ivan Neretin_, Mar 30 2016 *)
%Y Cf. A094616, A094618, A138247.
%Y Cf. A034472, A052919, A199108, A199111.
%K nonn,tabl
%O 1,1
%A _Clark Kimberling_, May 14 2004