%I
%S 1,1,1,1,2,1,2,2,1,3,3,1,4,5,2,5,7,3,6,10,5,1,8,14,7,1,10,19,11,2,12,
%T 26,15,3,15,35,22,5,18,46,30,7,22,60,42,11,27,78,56,15,32,10,76,22,1,
%U 38,128,100,30,1,46,162,133,42,2,54,204,173,56,3
%N Triangle read by rows, generated from an array of terms in powers of triangle A173305.
%C Row sums = A000041, the partition numbers.
%F Given triangle A173305 in which every column >0 = A000009 shifted down twice.
%F We create an array in which nth row = columns in (n1)th power of triangle
%F A173305. Finite differences of successive columns of the array become row terms
%F of A173306.
%e Given triangle A173305, we create an array by extracting terms in powers of A173305:
%e 1, 1, 1, 2, 2, 3, .4, .5, .6, .8, 10, 12, 15,...; = column terms of A173305
%e 1, 1, 2, 3, 4, 6, .9, 12, 16, 22, 29, 38, 50,...; = terms of A173305^2
%e 1, 1, 2, 3, 5, 7, 11, 15, 21, 29, 40, 53, 72,...; = terms of A173305^3
%e 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77,...; = terms of A173305^4
%e ...
%e (rows quickly converge to A000041, the partition numbers).
%e Taking finite difference terms from the top, we obtain the array:
%e 1, 1, 1, 2, 2, 3, .4, .5, .6,..8, 10, 12, 15,...;
%e ......1, 1, 2, 3, .5, .7, 10, 14, 19, 26, 35,...;
%e ............1, 1, .2, .3, .5, .7, 11, 15, 22,...;
%e ...........................1, .1, .2, .3, .5,...;
%e ...
%e Finally, columns of the above array become rows of A173306:
%e 1;
%e 1;
%e 1, 1;
%e 2, 1;
%e 2, 2, 1;
%e 3, 3, 1;
%e 4, 5, 2;
%e 5, 7, 3;
%e 6, 10, 5, 1;
%e 8, 14, 7, 1;
%e 10, 19, 11, 2;
%e 12, 26, 15, 3;
%e 15, 35, 22, 5;
%e 18, 46, 30, 7;
%e 22, 60, 42, 11;
%e 27, 78, 56, 15;
%e 32, 100, 76, 22, 1;
%e 38, 128, 100, 30, 1;
%e 46, 162, 133, 42, 2;
%e 54, 204, 173, 56, 3;
%e ...
%Y Cf. A000009, A000041, A173305
%K nonn,tabl
%O 0,5
%A _Gary W. Adamson_, Feb 15 2010
