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%I #15 Aug 09 2018 04:54:18
%S 1,1,1,5,2,1,33,9,3,1,261,57,15,4,1,2361,441,99,23,5,1,23805,3933,783,
%T 165,33,6,1,263313,39249,7083,1383,261,45,7,1,3161781,430677,71415,
%U 13083,2361,393,59,8,1,40907241,5137641,789939,136863,23805,3861,567,75,9,1
%N Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p*(column p+3 of T), or [T^p](m,0) = p*T(p+m,p+3) for all m>=1 and p>=-3.
%C Column 0 equals A111530 (related to log of factorial series). Column 3 (A111547) equals SHIFT_LEFT(column 0 of log(T)), where the matrix logarithm, log(T), equals the integer matrix A111549.
%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+2, 2)*T(n, j+k+1) for n>k>0, with T(n, n) = 1, T(n+1, n) = n+1, T(n+3, 2) = 3*T(n+1, 0), T(n+4, 4) = T(n+1, 0), for n>=0.
%e SHIFT_LEFT(column 0 of T^-3) = -3*(column 0 of T);
%e SHIFT_LEFT(column 0 of T^-2) = -2*(column 1 of T);
%e SHIFT_LEFT(column 0 of T^-1) = -1*(column 2 of T);
%e SHIFT_LEFT(column 0 of log(T)) = column 3 of T;
%e SHIFT_LEFT(column 0 of T^1) = 1*(column 4 of T);
%e where SHIFT_LEFT of column sequence shifts 1 place left.
%e Triangle T begins:
%e 1;
%e 1,1;
%e 5,2,1;
%e 33,9,3,1;
%e 261,57,15,4,1;
%e 2361,441,99,23,5,1;
%e 23805,3933,783,165,33,6,1;
%e 263313,39249,7083,1383,261,45,7,1;
%e 3161781,430677,71415,13083,2361,393,59,8,1; ...
%e After initial term, column 2 is 3 times column 0.
%e Matrix inverse T^-1 = A111548 starts:
%e 1;
%e -1,1;
%e -3,-2,1;
%e -15,-3,-3,1;
%e -99,-15,-3,-4,1;
%e -783,-99,-15,-3,-5,1;
%e -7083,-783,-99,-15,-3,-6,1; ...
%e where columns are all equal after initial terms;
%e compare columns of T^-1 to column 2 of T.
%e Matrix logarithm log(T) = A111549 is:
%e 0;
%e 1,0;
%e 4,2,0;
%e 23,6,3,0;
%e 165,32,9,4,0;
%e 1383,222,47,13,5,0;
%e 13083,1824,321,70,18,6,0; ...
%e compare column 0 of log(T) to column 3 of T.
%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 + 2, 2] T[n, j + k + 1], {j, 0, n - k - 1}]]]];
%t Table[T[n, k], {n, 0, 9}, {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+2,2)*T(n,j+k+1)))))
%Y Cf. A111545 (column 1), A111546 (column 2), A111547 (column 3), A111552 (row sums), A111548 (matrix inverse), A111549 (matrix log); related tables: A111528, A104980, A111536, A111553.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Aug 07 2005
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