%I #4 Jun 13 2015 10:29:35
%S 1,1,1,1,1,1,3,2,1,1,19,9,3,1,1,310,105,25,5,1,1,10978,2702,480,68,8,
%T 1,1,868140,154609,20657,2184,182,13,1,1,149688297,19092682,1906051,
%U 152579,9562,483,21,1,1,57339888914,5161046609,378639419,22799907
%N Triangle T, read by rows, where row n of T = row n-1 of T^fibonacci(n) with appended '1' for n>=1 starting with a single '1' in row 0.
%F See example section for two different methods of generating this triangle.
%e Triangle T begins:
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 3, 2, 1, 1;
%e 19, 9, 3, 1, 1;
%e 310, 105, 25, 5, 1, 1;
%e 10978, 2702, 480, 68, 8, 1, 1;
%e 868140, 154609, 20657, 2184, 182, 13, 1, 1;
%e 149688297, 19092682, 1906051, 152579, 9562, 483, 21, 1, 1;
%e 57339888914, 5161046609, 378639419, 22799907, 1090125, 41480, 1275, 34, 1, 1; ...
%e GENERATE T FROM MATRIX POWERS OF T.
%e Row n of T = row n-1 of T^fibonacci(n) with appended '1'.
%e Examples.
%e Row 5 of T is given by row 4 of matrix power T^fibonacci(5) = T^5:
%e 1;
%e 5, 1;
%e 15, 5, 1;
%e 55, 20, 5, 1;
%e 310, 105, 25, 5, 1; <== row 5 of T
%e 3796, 1070, 215, 35, 5, 1; ...
%e Row 6 of T is given by row 5 of matrix power T^fibonacci(6) = T^8:
%e 1;
%e 8, 1;
%e 36, 8, 1;
%e 164, 44, 8, 1;
%e 978, 268, 52, 8, 1;
%e 10978, 2702, 480, 68, 8, 1; <== row 6 of T
%e 262838, 53648, 8082, 964, 92, 8, 1; ...
%e ALTERNATE GENERATING METHOD.
%e To obtain row n, start with a '1' repeated fibonacci(n) times,
%e and build a table where row k+1 equals the partial sums of row k
%e but with the last term appearing fibonacci(n-k) times, for k=1..n-1;
%e listing the final terms in each row forms row n of this triangle.
%e Example.
%e To obtain row 5, start with a '1' repeated fibonacci(5)=5 times:
%e (1,1,1,1,1);
%e take partial sums, writing the last term fibonacci(4)=3 times:
%e 1,2,3,4, (5,5,5);
%e take partial sums, writing the last term fibonacci(3)=2 times:
%e 1,3,6,10,15,20, (25,25);
%e take partial sums, writing the last term fibonacci(2)=1 times:
%e 1,4,10,20,35,55,80, (105);
%e take partial sums, writing the last term fibonacci(1)=1 times:
%e 1,5,15,35,70,125,205, (310).
%e Final terms in the above partial sums forms row 5: [310,105,25,5,1];
%e repeating this process will generate all the rows of this triangle.
%o (PARI) /* Generate using matrix power method: */ T(n,k)=local(A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, B[i, j]=(A^(fibonacci(i-1)))[i-1, j]); )); A=B); return( ((A)[n+1, k+1]))
%o (PARI) /* Generate using partial sums method (faster) */ T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k, p=fibonacci(n+2)-fibonacci(n-j+2)-j; A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A[p+1]
%Y Cf. columns: A136171, A136172, A136173; variants: A101479, A132610, A132615.
%K nonn,tabl
%O 0,7
%A _Paul D. Hanna_, Dec 17 2007