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%I #9 Dec 14 2015 11:49:44
%S 1,1,1,3,2,1,15,10,3,1,108,75,21,4,1,1036,753,208,36,5,1,12569,9534,
%T 2637,442,55,6,1,185704,146353,40731,6742,805,78,7,1,3247546,2647628,
%U 742620,122350,14330,1325,105,8,1,65762269,55251994,15624420,2571620
%N Triangle P, read by rows, where column k of P^3 equals column 0 of P^(3k+3) such that column 0 of P^3 equals column 0 of P shift one row up, with P(0,0)=1.
%F Denote this triangle by P and define as follows.
%F Let [P^m]_k denote column k of matrix power P^m,
%F so that triangular matrix W = A136231 may be defined by
%F [W]_k = [P^(3k+3)]_0, for k>=0, such that
%F (1) W = P^3 and (2) [W]_0 = [P]_0 shift up one row.
%F Define the triangular matrix U = A136228 by
%F [U]_k = [P^(3k+1)]_0, for k>=0,
%F and the triangular matrix V = A136230 by
%F [V]_k = [P^(3k+2)]_0, for k>=0.
%F Then columns of P may be formed from powers of U:
%F [P]_k = [U^(k+1)]_0, for k>=0,
%F and columns of P^2 may be formed from powers of V:
%F [P^2]_k = [V^(k+1)]_0, for k>=0.
%F Further, columns of powers of P, U, V and W satisfy:
%F [U^(j+1)]_k = [P^(3k+1)]_j,
%F [V^(j+1)]_k = [P^(3k+2)]_j,
%F [W^(j+1)]_k = [P^(3k+3)]_j,
%F [W^(j+1)]_k = [W^(k+1)]_j,
%F [P^(3j+3)]_k = [P^(3k+3)]_j, for all j>=0, k>=0.
%F Also, we have the column transformations:
%F U * [P]_k = [P]_{k+1},
%F V * [P^2]_k = [P^2]_{k+1},
%F W * [P^3]_k = [P^3]_{k+1},
%F W * [U]_k = [U]_{k+1},
%F W * [V]_k = [V]_{k+1},
%F W * [W]_k = [W]_{k+1}, for all k>=0.
%F Other identities include the matrix products:
%F U = P * [P^2 shift right one column];
%F V = P^2 * [P shift right one column];
%F V = U * [U shift down one row];
%F W = V * [V shift down one row];
%F where the triangle transformations "shift right" and "shift down" are illustrated in examples of entries A136228 (U) and A136230 (V).
%e Triangle P begins:
%e 1;
%e 1, 1;
%e 3, 2, 1;
%e 15, 10, 3, 1;
%e 108, 75, 21, 4, 1;
%e 1036, 753, 208, 36, 5, 1;
%e 12569, 9534, 2637, 442, 55, 6, 1;
%e 185704, 146353, 40731, 6742, 805, 78, 7, 1;
%e 3247546, 2647628, 742620, 122350, 14330, 1325, 105, 8, 1;
%e 65762269, 55251994, 15624420, 2571620, 298240, 26943, 2030, 136, 9, 1; ...
%e where column k of P = column 0 of U^(k+1) and U = A136228.
%e Matrix cube, W = P^3 (A136231), begins:
%e 1;
%e 3, 1;
%e 15, 6, 1;
%e 108, 48, 9, 1;
%e 1036, 495, 99, 12, 1;
%e 12569, 6338, 1323, 168, 15, 1;
%e 185704, 97681, 21036, 2754, 255, 18, 1; ...
%e where column k of P^3 = column 0 of P^(3k+3) such that
%e column 0 of P^3 = column 0 of P shift one row up.
%e Matrix square, P^2 (A136225), begins:
%e 1;
%e 2, 1;
%e 8, 4, 1;
%e 49, 26, 6, 1;
%e 414, 232, 54, 8, 1;
%e 4529, 2657, 629, 92, 10, 1;
%e 61369, 37405, 9003, 1320, 140, 12, 1; ...
%e where column k of P^2 = column 0 of V^(k+1) and
%e triangle V = A136230 begins:
%e 1;
%e 2, 1;
%e 8, 5, 1;
%e 49, 35, 8, 1;
%e 414, 325, 80, 11, 1;
%e 4529, 3820, 988, 143, 14, 1;
%e 61369, 54800, 14696, 2200, 224, 17, 1; ...
%e where column k of V = column 0 of P^(3k+2).
%e Related triangle U = A136228 begins:
%e 1;
%e 1, 1;
%e 3, 4, 1;
%e 15, 24, 7, 1;
%e 108, 198, 63, 10, 1;
%e 1036, 2116, 714, 120, 13, 1;
%e 12569, 28052, 9884, 1725, 195, 16, 1; ...
%e where column k of U = column 0 of P^(3k+1)
%e and column k of P = column 0 of U^(k+1).
%e Surprisingly, column 0 of P is also found in square A136217:
%e (1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;
%e (1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),...;
%e (3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),...;
%e (15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),...;
%e (108),(414),1036,(2116),3493,(5555),8040,(11477),15483,...;
%e (1036),(4529),12569,(28052),48800,(82328),124335,(186261),...;
%e (12569),(61369),185704,(446560),811111,(1438447),2250731,...;
%e ...
%e and has a recurrence similar to that of square array A136212
%e which generates the triple factorials.
%o (PARI) {T(n,k)=local(P=Mat(1),U,PShR);if(n>0,for(i=0,n, PShR=matrix(#P,#P, r,c,
%o if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));U=P*PShR^2; U=matrix(#P+1,
%o #P+1, r,c, if(r>=c, if(r<#P+1,U[r,c], if(c==1,(P^3)[ #P,1],(P^(3*c-1))[r-c+1,
%o 1])))); P=matrix(#U, #U, r,c, if(r>=c, if(r<#R,P[r,c], (U^c)[r-c+1,
%o 1])))));P[n+1,k+1]}
%Y Columns: A136221, A136222, A136223, A136224.
%Y Related tables: A136228 (U), A136230 (V), A136231 (W=P^3), A136217, A136218.
%Y Variants: A091351, A135880.
%K nice,nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Dec 25 2007, corrected Jan 24 2008
%E Typo in example corrected by _Paul D. Hanna_, Mar 27 2010
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