%I #6 Apr 12 2013 01:40:35
%S 1,1,1,2,1,1,4,2,2,1,9,5,5,3,1,1,24,15,15,10,5,5,2,1,77,53,53,38,23,
%T 23,13,8,3,3,1,295,218,218,165,112,112,74,51,28,28,15,7,4,1,1,1329,
%U 1034,1034,816,598,598,433,321,209,209,135,84,56,28,28,13,6,2,1,6934,5605
%N Triangle, read by rows, where row n+1 is generated from row n by first inserting zeros at positions {(m+2)*(m+3)/2, m>=0} in row n and then taking the partial sums in reverse order, for n>=2, starting with 1's in the initial two rows.
%C Column 0 forms A091352, which also equals column 1 of table A125781, where table A125781 is generated by a complementary recurrence of this triangle. The number of terms in row n is A127419(n).
%e To generate row 6, start with row 5:
%e 24, 15, 15, 10, 5, 5, 2, 1;
%e insert zeros at positions [1,4,8,13,..., (m+2)*(m+3)/2 - 2,...]:
%e 24, 0, 15, 15, 0, 10, 5, 5, 0, 2, 1;
%e then row 6 equals the partial sums of row 5 taken in reverse order:
%e 24, _0, 15, 15, _0, 10, _5, 5, 0, 2, 1;
%e 77, 53, 53, 38, 23, 23, 13, 8, 3, 3, 1.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 4, 2, 2, 1;
%e 9, 5, 5, 3, 1, 1;
%e 24, 15, 15, 10, 5, 5, 2, 1;
%e 77, 53, 53, 38, 23, 23, 13, 8, 3, 3, 1;
%e 295, 218, 218, 165, 112, 112, 74, 51, 28, 28, 15, 7, 4, 1, 1;
%e 1329, 1034, 1034, 816, 598, 598, 433, 321, 209, 209, 135, 84, 56, 28, 28, 13, 6, 2, 1;
%e Column 0 of this triangle equals column 1 of triangle A091351, where triangle A091351 begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 4, 3, 1;
%e 1, 9, 9, 4, 1;
%e 1, 24, 30, 16, 5, 1;
%e 1, 77, 115, 70, 25, 6, 1;
%e 1, 295, 510, 344, 135, 36, 7, 1; ...
%e and column k of A091351 = row sums of matrix power A091351^k for k>=0.
%Y Cf. A091352, A091351, A125781, A127419.
%K nonn,tabf
%O 0,4
%A _Paul D. Hanna_, Jan 14 2007