A111536 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p*(column p+2 of T), or [T^p](m,0) = p*T(p+m,p+2) for all m>=1 and p>=-2.
1, 1, 1, 4, 2, 1, 22, 8, 3, 1, 148, 44, 14, 4, 1, 1156, 296, 84, 22, 5, 1, 10192, 2312, 600, 148, 32, 6, 1, 99688, 20384, 4908, 1156, 242, 44, 7, 1, 1069168, 199376, 44952, 10192, 2084, 372, 58, 8, 1, 12468208, 2138336, 454344, 99688, 20012, 3528, 544, 74, 9, 1
Offset: 0
Examples
SHIFT_LEFT(column 0 of T^-2) = -2*(column 0 of T); SHIFT_LEFT(column 0 of T^-1) = -1*(column 1 of T); SHIFT_LEFT(column 0 of log(T)) = column 2 of T; SHIFT_LEFT(column 0 of T^1) = 1*(column 3 of T); SHIFT_LEFT(column 0 of T^2) = 2*(column 4 of T); where SHIFT_LEFT of column sequence shifts 1 place left. Triangle T begins: 1; 1, 1; 4, 2, 1; 22, 8, 3, 1; 148, 44, 14, 4, 1; 1156, 296, 84, 22, 5, 1; 10192, 2312, 600, 148, 32, 6, 1; 99688, 20384, 4908, 1156, 242, 44, 7, 1; 1069168, 199376, 44952, 10192, 2084, 372, 58, 8, 1; 12468208, 2138336, 454344, 99688, 20012, 3528, 544, 74, 9, 1; ... ... After initial term, column 1 is twice column 0. Matrix inverse T^-1 = A111540 starts: 1; -1, 1; -2, -2, 1; -8, -2, -3, 1; -44, -8, -2, -4, 1; -296, -44, -8, -2, -5, 1; -2312, -296, -44, -8, -2, -6, 1; -20384, -2312, -296, -44, -8, -2, -7, 1; -199376, -20384, -2312, -296, -44, -8, -2, -8, 1; ... where columns are all equal after initial terms; compare columns of T^-1 to column 1 of T. Matrix logarithm log(T) = A111541 is: 0; 1, 0; 3, 2, 0; 14, 5, 3, 0; 84, 22, 8, 4, 0; 600, 128, 36, 12, 5, 0; 4908, 896, 212, 58, 17, 6, 0; 44952, 7220, 1496, 360, 90, 23, 7, 0; 454344, 65336, 12128, 2652, 602, 134, 30, 8, 0; 5016768, 653720, 110288, 22320, 4736, 974, 192, 38, 9, 0; ... compare column 0 of log(T) to column 2 of T.
Links
- Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
Crossrefs
Programs
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Mathematica
T[n_, k_] := T[n, k] = If[n
Jean-François Alcover, Jan 24 2017, adapted from PARI *) -
PARI
T(n,k)=if(n
Formula
T(n, k) = k*T(n, k+1) + Sum_{j=0..n-k-1} T(j+1, 1)*T(n, j+k+1) for n>k>0, with T(n, n) = 1, T(n+1, n) = n+1, T(n+2, 1) = 2*T(n+1, 0), T(n+3, 3) = T(n+1, 0), for n>=0.
Comments