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 A101479 Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal. 25
 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 19, 9, 3, 1, 1, 191, 70, 18, 4, 1, 1, 2646, 795, 170, 30, 5, 1, 1, 46737, 11961, 2220, 335, 45, 6, 1, 1, 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1, 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1, 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,7 COMMENTS Remarkably, T equals the product of these triangular matrices: T = A107867*A107862^-1 = A107870*A107867^-1 = A107873*A107870^-1; reversing the order of these products yields triangle A107876. LINKS Alois P. Heinz, Rows n = 0..140, flattened (first 31 rows from Paul D. Hanna) EXAMPLE Triangle begins: 1; 1, 1; 1, 1, 1; 3, 2, 1, 1; 19, 9, 3, 1, 1; 191, 70, 18, 4, 1, 1; 2646, 795, 170, 30, 5, 1, 1; 46737, 11961, 2220, 335, 45, 6, 1, 1; 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1; 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1; ... Row 4 starts with row 3 of T^3 which begins: 1; 3, 1; 6, 3, 1; 19, 9, 3, 1; ... row 5 starts with row 4 of T^4 which begins: 1; 4, 1; 10, 4, 1; 34, 14, 4, 1; 191, 70, 18, 4, 1; ... An ALTERNATE GENERATING METHOD is illustrated as follows. For row 4: Start with a '1' and append 2 zeros, take partial sums and append 1 zero, take partial sums thrice more, resulting in: 1, 0, 0; 1, 1, 1, 0; 1, 2, 3, 3; 1, 3, 6, 9; 1, 4,10,19. Final nonzero terms form row 4: [19,9,3,1,1]. For row 5: Start with a '1' and append 3 zeros, take partial sums and append 2 zeros, take partial sums and append 1 zero, take partial sums thrice more, resulting in: 1, 0, 0, 0; 1, 1, 1, 1, 0,  0; 1, 2, 3, 4, 4,  4,  0; 1, 3, 6,10,14, 18, 18; 1, 4,10,20,34, 52, 70; 1, 5,15,35,69,121,191; where the final nonzero terms form row 5: [191,70,18,4,1,1]. Likewise, for row 6: 1, 0, 0, 0,  0; 1, 1, 1, 1,  1,  0,  0,  0; 1, 2, 3, 4,  5,  5,  5,  5,   0,   0; 1, 3, 6,10, 15, 20, 25, 30,  30,  30,   0; 1, 4,10,20, 35, 55, 80,110, 140, 170, 170; 1, 5,15,35, 70,125,205,315, 455, 625, 795; 1, 6,21,56,126,251,456,771,1226,1851,2646; where the final nonzero terms form row 6: [2646,795,170,30,5,1,1]. Continuing in this way generates all rows of this triangle. MAPLE b:= proc(n) option remember;       Matrix(n, (i, j)-> T(i-1, j-1))^(n-1)     end: T:= proc(n, k) option remember;      `if`(n=k, 1, `if`(k>n, 0, b(n)[n, k+1]))     end: seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 13 2020 MATHEMATICA b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], n-1]; T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]]; Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 25 2020, after Alois P. Heinz *) PROG (PARI) {T(n, k) = my(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^(i-2))[i-1, j]); )); A=B); return(A[n+1, k+1])} for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print("")) (PARI) {T(n, k) = my(A=vector(n+1), p); A[1]=1; for(j=1, n-k-1, p=(n-1)*(n-2)/2-(n-j-1)*(n-j-2)/2; A = Vec((Polrev(A)+x*O(x^p))/(1-x))); A = Vec((Polrev(A) +x*O(x^p)) / (1-x) ); A[p+1]} for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print("")) CROSSREFS Columns are A101481, A101482, A101483, row sums form A101484. Cf. A107876 (dual triangle). Cf. A304184, A304185, A304186, A304187. Sequence in context: A214742 A204124 A316674 * A136170 A245188 A137241 Adjacent sequences:  A101476 A101477 A101478 * A101480 A101481 A101482 KEYWORD nonn,tabl AUTHOR Paul D. Hanna, Jan 21 2005, Jul 26 2006, May 27 2007 STATUS approved

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Last modified December 3 20:35 EST 2021. Contains 349468 sequences. (Running on oeis4.)