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%I #23 Jun 07 2021 15:45:29
%S 1,1,1,3,2,1,13,7,3,1,71,33,13,4,1,461,191,71,21,5,1,3447,1297,461,
%T 133,31,6,1,29093,10063,3447,977,225,43,7,1,273343,87669,29093,8135,
%U 1859,353,57,8,1,2829325,847015,273343,75609,17185,3251,523,73,9,1
%N Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p*(column p+1 of T), or [T^p](m,0) = p*T(p+m,p+1) for all m>=1 and p>=-1.
%C Column 0 equals A003319 (indecomposable permutations). Amazingly, column 1 (A104981) equals SHIFT_LEFT(column 0 of log(T)), where the matrix logarithm, log(T), equals the integer matrix A104986.
%C From _Paul D. Hanna_, Feb 17 2009: (Start)
%C Square array A156628 has columns found in this triangle T:
%C Column 0 of A156628 = column 0 of T = A003319;
%C Column 1 of A156628 = column 1 of T = A104981;
%C Column 2 of A156628 = column 2 of T = A003319 shifted;
%C Column 3 of A156628 = column 1 of T^2 (A104988);
%C Column 5 of A156628 = column 2 of T^2 (A104988). (End)
%H G. C. Greubel, <a href="/A104980/b104980.txt">Rows n = 0..50 of the triangle, flattened</a>
%H Paul Barry, <a href="https://arxiv.org/abs/1804.06801">A note on number triangles that are almost their own production matrix</a>, arXiv:1804.06801 [math.CO], 2018.
%F T(n, k) = k*T(n, k+1) + Sum_{j=0..n-k-1} T(j, 0)*T(n, j+k+1) for n>k>0, with T(n, n) = 1, T(n+1, n) = n+1, T(n+1, 2) = T(n, 0) for n>=0.
%e SHIFT_LEFT(column 0 of T^-1) = -1*(column 0 of T);
%e SHIFT_LEFT(column 0 of T^1) = 1*(column 2 of T);
%e SHIFT_LEFT(column 0 of T^2) = 2*(column 3 of T);
%e where SHIFT_LEFT of column sequence shifts 1 place left.
%e Triangle T begins:
%e 1;
%e 1, 1;
%e 3, 2, 1;
%e 13, 7, 3, 1;
%e 71, 33, 13, 4, 1;
%e 461, 191, 71, 21, 5, 1;
%e 3447, 1297, 461, 133, 31, 6, 1;
%e 29093, 10063, 3447, 977, 225, 43, 7, 1;
%e 273343, 87669, 29093, 8135, 1859, 353, 57, 8, 1;
%e 2829325, 847015, 273343, 75609, 17185, 3251, 523, 73, 9, 1; ...
%e Matrix inverse T^-1 is A104984 which begins:
%e 1;
%e -1, 1;
%e -1, -2, 1;
%e -3, -1, -3, 1;
%e -13, -3, -1, -4, 1;
%e -71, -13, -3, -1, -5, 1;
%e -461, -71, -13, -3, -1, -6, 1; ...
%e Matrix T also satisfies:
%e [I + SHIFT_LEFT(T)] = [I - SHIFT_DOWN(T)]^-1, which starts:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 7, 3, 1, 1;
%e 33, 13, 4, 1, 1;
%e 191, 71, 21, 5, 1, 1; ...
%e where SHIFT_DOWN(T) shifts columns of T down 1 row,
%e and SHIFT_LEFT(T) shifts rows of T left 1 column,
%e with both operations leaving zeros in the diagonal.
%t T[n_, k_]:= T[n, k]= If[n<k || k<0, 0, If[n==k, 1, If[n==k+1, n, k T[n, k+1] + Sum[T[j, 0] T[n, j+k+1], {j, 0, n-k-1}]]]];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* _Jean-François Alcover_, Aug 09 2018, from PARI *)
%o (PARI) {T(n,k) = if(n<k||k<0, 0, if(n==k, 1, if(n==k+1, n, k*T(n,k+1) + sum(j=0,n-k-1,T(j,0)*T(n,j+k+1)))))}
%o for(n=0, 10, for(k=0, n, print1(T(n,k),", ")); print(""))
%o (PARI) {T(n,k) = if(n<k||k<0, 0, (matrix(n+1, n+1, m, j, if(m==j, 1, if(m==j+1, -m+1, -polcoeff((1-1/sum(i=0,m,i!*x^i))/x+O(x^m),m-j-1))))^-1)[n+1,k+1])}
%o for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))
%o (Sage)
%o @CachedFunction
%o def T(n,k):
%o if (k<0 or k>n): return 0
%o elif (k==n): return 1
%o elif (k==n-1): return n
%o else: return k*T(n, k+1) + sum( T(j, 0)*T(n, j+k+1) for j in (0..n-k-1) )
%o flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 07 2021
%Y Cf. A003319 (column 0), A104981 (column 1), A104983 (row sums), A104984 (matrix inverse), A104988 (matrix square), A104990 (matrix cube), A104986 (matrix log), A156628.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Apr 10 2005
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