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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.
+10
19
1, 1, 1, 3, 2, 1, 15, 10, 3, 1, 108, 75, 21, 4, 1, 1036, 753, 208, 36, 5, 1, 12569, 9534, 2637, 442, 55, 6, 1, 185704, 146353, 40731, 6742, 805, 78, 7, 1, 3247546, 2647628, 742620, 122350, 14330, 1325, 105, 8, 1, 65762269, 55251994, 15624420, 2571620
FORMULA
Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix W = A136231 may be defined by [W]_k = [P^(3k+3)]_0, for k>=0, such that
(1) W = P^3 and (2) [W]_0 = [P]_0 shift up one row.
Define the triangular matrix U = A136228 by [U]_k = [P^(3k+1)]_0, for k>=0,
and the triangular matrix V = A136230 by [V]_k = [P^(3k+2)]_0, for k>=0.
Then columns of P may be formed from powers of U:
[P]_k = [U^(k+1)]_0, for k>=0,
and columns of P^2 may be formed from powers of V:
[P^2]_k = [V^(k+1)]_0, for k>=0.
Further, columns of powers of P, U, V and W satisfy:
[U^(j+1)]_k = [P^(3k+1)]_j,
[V^(j+1)]_k = [P^(3k+2)]_j,
[W^(j+1)]_k = [P^(3k+3)]_j,
[W^(j+1)]_k = [W^(k+1)]_j,
[P^(3j+3)]_k = [P^(3k+3)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
U * [P]_k = [P]_{k+1},
V * [P^2]_k = [P^2]_{k+1},
W * [P^3]_k = [P^3]_{k+1},
W * [U]_k = [U]_{k+1},
W * [V]_k = [V]_{k+1},
W * [W]_k = [W]_{k+1}, for all k>=0.
Other identities include the matrix products:
U = P * [P^2 shift right one column];
V = P^2 * [P shift right one column];
V = U * [U shift down one row];
W = V * [V shift down one row];
where the triangle transformations "shift right" and "shift down" are illustrated in examples of entries A136228 (U) and A136230 (V).
EXAMPLE
Triangle P begins:
1;
1, 1;
3, 2, 1;
15, 10, 3, 1;
108, 75, 21, 4, 1;
1036, 753, 208, 36, 5, 1;
12569, 9534, 2637, 442, 55, 6, 1;
185704, 146353, 40731, 6742, 805, 78, 7, 1;
3247546, 2647628, 742620, 122350, 14330, 1325, 105, 8, 1;
65762269, 55251994, 15624420, 2571620, 298240, 26943, 2030, 136, 9, 1; ...
where column k of P = column 0 of U^(k+1) and U = A136228. Matrix cube, W = P^3 ( A136231), begins: 1;
3, 1;
15, 6, 1;
108, 48, 9, 1;
1036, 495, 99, 12, 1;
12569, 6338, 1323, 168, 15, 1;
185704, 97681, 21036, 2754, 255, 18, 1; ...
where column k of P^3 = column 0 of P^(3k+3) such that
column 0 of P^3 = column 0 of P shift one row up.
Matrix square, P^2 ( A136225), begins: 1;
2, 1;
8, 4, 1;
49, 26, 6, 1;
414, 232, 54, 8, 1;
4529, 2657, 629, 92, 10, 1;
61369, 37405, 9003, 1320, 140, 12, 1; ...
where column k of P^2 = column 0 of V^(k+1) and
1;
2, 1;
8, 5, 1;
49, 35, 8, 1;
414, 325, 80, 11, 1;
4529, 3820, 988, 143, 14, 1;
61369, 54800, 14696, 2200, 224, 17, 1; ...
where column k of V = column 0 of P^(3k+2).
Related triangle U = A136228 begins: 1;
1, 1;
3, 4, 1;
15, 24, 7, 1;
108, 198, 63, 10, 1;
1036, 2116, 714, 120, 13, 1;
12569, 28052, 9884, 1725, 195, 16, 1; ...
where column k of U = column 0 of P^(3k+1)
and column k of P = column 0 of U^(k+1).
Surprisingly, column 0 of P is also found in square A136217: (1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),...;
(3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),...;
(15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),...;
(108),(414),1036,(2116),3493,(5555),8040,(11477),15483,...;
(1036),(4529),12569,(28052),48800,(82328),124335,(186261),...;
(12569),(61369),185704,(446560),811111,(1438447),2250731,...;
...
and has a recurrence similar to that of square array A136212which generates the triple factorials.
PROG
(PARI) {T(n, k)=local(P=Mat(1), U, PShR); if(n>0, for(i=0, n, PShR=matrix(#P, #P, r, c,
if(r>=c, if(r==c, 1, if(c==1, 0, P[r-1, c-1])))); U=P*PShR^2; U=matrix(#P+1,
#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,
1])))); P=matrix(#U, #U, r, c, if(r>=c, if(r<#R, P[r, c], (U^c)[r-c+1,
1]))))); P[n+1, k+1]}
Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {floor(m*(m+7)/6), m>=0} and then taking partial sums, starting with all 1's in row 0.
+10
13
1, 1, 1, 3, 2, 1, 15, 8, 3, 1, 108, 49, 15, 4, 1, 1036, 414, 108, 24, 5, 1, 12569, 4529, 1036, 198, 34, 6, 1, 185704, 61369, 12569, 2116, 306, 46, 7, 1, 3247546, 996815, 185704, 28052, 3493, 453, 59, 8, 1, 65762269, 18931547, 3247546, 446560, 48800, 5555, 622, 74, 9, 1
COMMENTS
A variant of the triple factorial array A136212. Compare to triangle array A136218, which is generated by a complementary process.
FORMULA
Let triangular matrix P = A136220, then: column 0 ( A136221) = column 0 of P; column 1 ( A136226) = column 0 of P^2; column 3 ( A136229) = column 0 of P^4.
EXAMPLE
Square array begins:
(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),18,19,20,(21),..;
(3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),216,242,(269),..;
(15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),2062,2485,(2943),..;
(108),(414),1036,(2116),3493,(5555),8040,(11477),15483,20748,(26748),33528,..;
(1036),(4529),12569,(28052),48800,(82328),124335,(186261),260856,364551,..;
(12569),(61369),185704,(446560),811111,(1438447),2250731,(3513569),5078154,..;
(185704),(996815),3247546,(8325700),15684001,(29039188),46830722,...;
(3247546),(18931547),65762269,(178284892),346583419,...;
(65762269),(412345688),1515642725,(4317391240),...; ...
where terms in parenthesis are at positions {floor(m*(m+7)/6), m>=0} and are removed before taking partial sums to obtain the next row.
To generate the array, start with all 1's in row 0; from then on, obtain row n+1 from row n by first removing terms in row n at positions {floor(m*(m+7)/6), m>=0} and then taking partial sums.
For example, to generate row 2 from row 1:
[(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),18,...],
remove terms at positions [0,1,3,5,7,10,13,16,20,...] to get:
[3, 5, 7, 9,10, 12,13, 15,16, 18,19,20, 22,23,24, 26,27,28,...]
then take partial sums to obtain row 2:
[3,8,15,24,34,46,59,74,90,108,127,147,169,192,216,242,269,...].
Continuing in this way will generate all the rows of this array.
Amazingly, column 0 of this array = column 0 of triangle P= A136220: 1;
1, 1;
3, 2, 1;
15, 10, 3, 1;
108, 75, 21, 4, 1;
1036, 753, 208, 36, 5, 1;
12569, 9534, 2637, 442, 55, 6, 1;
185704, 146353, 40731, 6742, 805, 78, 7, 1; ...
where column k of P^3 = column 0 of P^(3k+3) such that column 0 of P^3 = column 0 of P shift one place left.
MATHEMATICA
nmax = 9;
row[0] = Table[1, {nmax^2}];
row[n_] := row[n] = Accumulate[Delete[row[n-1], Table[{Floor[m((m+7)/6)+1] }, {m, 0, (1/2)(-7 + Sqrt[1 + 24 Length[row[n-1]]]) // Floor}]]];
R = row /@ Range[0, nmax];
T[n_, k_] := R[[n+1, k+1]];
PROG
(PARI) {T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==(m*(m+7))\6, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}
1, 1, 3, 15, 108, 1036, 12569, 185704, 3247546, 65762269, 1515642725, 39211570981, 1125987938801, 35554753133312, 1224882431140838, 45731901253649898, 1839804317195739634, 79355626796692509253, 3653687500034925338348
COMMENTS
P = A136220 is a triangular matrix 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 place left. Tables A136217 and A136218 are defined by recurrences seemingly unrelated to the matrix product recurrence of A136220 and yet they all generate this same sequence in column 0.
EXAMPLE
Equals column 0 of triangle P= A136220, which begins: 1;
1, 1;
3, 2, 1;
15, 10, 3, 1;
108, 75, 21, 4, 1;
1036, 753, 208, 36, 5, 1;
12569, 9534, 2637, 442, 55, 6, 1;
185704, 146353, 40731, 6742, 805, 78, 7, 1; ...
where column k of P^3 = column 0 of P^(3k+3) such that
column 0 of P^3 = column 0 of P shift one place left.
Surprisingly, column 0 of P is also found in square A136218: (1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),...;
(3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),...;
(15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),...;
(108),(414),1036,(2116),3493,(5555),8040,(11477),15483,...;
(1036),(4529),12569,(28052),48800,(82328),124335,(186261),...;
(12569),(61369),185704,(446560),811111,(1438447),2250731,...;
...
and has a recurrence similar to that of square array A136212which generates the triple factorials.
PROG
(PARI) /* Generate using matrix product recurrences of triangle A136220: */ {a(n)=local(P=Mat(1), U, PShR); if(n>0, for(i=0, n, PShR=matrix(#P, #P, r, c, if(r>=c, if(r==c, 1, if(c==1, 0, P[r-1, c-1])))); U=P*PShR^2; U=matrix(#P+1, #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, 1])))); P=matrix(#U, #U, r, c, if(r>=c, if(r<#R, P[r, c], (U^c)[r-c+1, 1]))))); P[n+1, 1]} (PARI) /* Generated as column 0 in triangle A136218 (faster): */ {a(n)=local(A=[1], B); if(n>0, for(i=1, n, m=1; B=[0]; for(j=1, #A, if(j+m-1==(m*(m+7))\6, m+=1; B=concat(B, 0)); B=concat(B, A[j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B))))))); A[1]}
Triangle U, read by rows, where U(n,k) = Product_{j=k..n-1} (3*j+1) for n > k with U(n,n) = 1.
+10
7
1, 1, 1, 4, 4, 1, 28, 28, 7, 1, 280, 280, 70, 10, 1, 3640, 3640, 910, 130, 13, 1, 58240, 58240, 14560, 2080, 208, 16, 1, 1106560, 1106560, 276640, 39520, 3952, 304, 19, 1, 24344320, 24344320, 6086080, 869440, 86944, 6688, 418, 22, 1
COMMENTS
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n, k, p) = G(n-1, n-k, p) then T(n, k, 1) = A094587(n, k), T(n, k, 2) = A112292(n, k) and T(n, k, 3) is this sequence. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
FORMULA
Matrix powers: column 0 of U^(k+1) = column k of A136216 for k >= 0; simultaneously, column k = column 0 of A136216^(3k+1) for k >= 0. Element in column 0, row n, of matrix power U^(k+1) = A007559(n)*C(n+k,k), where A007559 are triple factorials found in column 0 of this triangle.
EXAMPLE
Triangle begins:
1;
1, 1;
4, 4, 1;
28, 28, 7, 1;
280, 280, 70, 10, 1;
3640, 3640, 910, 130, 13, 1;
58240, 58240, 14560, 2080, 208, 16, 1;
1106560, 1106560, 276640, 39520, 3952, 304, 19, 1; ...
Matrix inverse begins:
1;
-1, 1;
0, -4, 1;
0, 0, -7, 1;
0, 0, 0, -10, 1;
0, 0, 0, 0, -13, 1; ...
MAPLE
nmax:=8; for n from 0 to nmax do U(n, n):=1 od: for n from 0 to nmax do for k from 0 to n do if n > k then U(n, k) := mul((3*j+1), j = k..n-1) fi: od: od: for n from 0 to nmax do seq(U(n, k), k=0..n) od: seq(seq(U(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jul 04 2011, revised Nov 23 2012
MATHEMATICA
Table[Product[3*j+1, {j, k, n-1}], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 14 2019 *)
PROG
(PARI) T(n, k)=if(n==k, 1, prod(j=k, n-1, 3*j+1))
(Magma) [[n eq 0 select 1 else k eq n select 1 else (&*[3*j+1: j in [k..n-1]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Jun 14 2019 (Sage)
def T(n, k):
if (k==n): return 1
else: return product(3*j+1 for j in (k..n-1))
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 14 2019
Triple factorial triangle, read by rows of 3n(n+1)/2+1 terms, where row n+1 is generated from row n by first inserting zeros in row n at positions {[m*(m+5)/6], m=0..3n-1} and then taking partial sums, starting with a '1' in row 0.
+10
6
1, 1, 1, 1, 1, 4, 4, 4, 4, 3, 3, 2, 2, 1, 1, 28, 28, 28, 28, 24, 24, 20, 20, 16, 16, 12, 9, 9, 6, 4, 4, 2, 1, 1, 280, 280, 280, 280, 252, 252, 224, 224, 196, 196, 168, 144, 144, 120, 100, 100, 80, 64, 64, 48, 36, 27, 27, 18, 12, 8, 8, 4, 2, 1, 1, 3640, 3640, 3640, 3640, 3360
COMMENTS
Square array A136212 is generated by a complementary process. This is the triple factorial variant of triangles A135877 (double factorials) and A127452 (factorials).
FORMULA
Column 0 forms the triple factorials A007559.
EXAMPLE
Triangle begins:
1;
1,1,1,1;
4,4,4,4,3,3,2,2,1,1;
28,28,28,28,24,24,20,20,16,16,12,9,9,6,4,4,2,1,1;
280,280,280,280,252,252,224,224,196,196,168,144,144,120,100,100,80,64,64,48,36,27,27,18,12,8,8,4,2,1,1;
3640,3640,3640,3640,3360,3360,3080,3080,2800,2800,2520,2268,2268,2016,1792,1792,1568,1372,1372,1176,1008,864,864,720,600,500,500,400,320,256,256,192,144,108,81,81,54,36,24,16,16,8,4,2,1,1;
...
To generate row 3, start with row 2:
[4,4,4,4,3,3,2,2,1,1];
insert zeros at positions [0,1,2,4,6,8,11,14,17] to get:
[0,0,0,4,0,4,0,4,0,4,3,0,3,2,0,2,1,0,1],
then take reverse partial sums (from right to left) to obtain row 3:
[28,28,28,28,24,24,20,20,16,16,12,9,9,6,4,4,2,1,1].
Continuing in this way will generate all the rows of this triangle.
PROG
(PARI) {T(n, k)=local(A=[1], B); if(n>0, for(i=1, n, m=1; B=[0, 0]; for(j=1, #A, if(j+m-1==(m*(m+7))\6, m+=1; B=concat(B, 0)); B=concat(B, A[j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B))))))); if(k+1>#A, 0, A[k+1])}
Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {m*(m+1)/2, m>=0} and then taking partial sums, starting with all 1's in row 0.
+10
5
1, 1, 1, 3, 2, 1, 14, 8, 3, 1, 86, 45, 14, 4, 1, 645, 318, 86, 22, 5, 1, 5662, 2671, 645, 152, 31, 6, 1, 56632, 25805, 5662, 1251, 232, 41, 7, 1, 633545, 280609, 56632, 11869, 2026, 327, 53, 8, 1, 7820115, 3381993, 633545, 126987, 20143, 2991, 457, 66, 9, 1
EXAMPLE
Table begins:
(1),(1),1,(1),1,1,(1),1,1,1,(1),1,1,1,1,(1),1,...;
(1),(2),3,(4),5,6,(7),8,9,10,(11),12,13,14,15,(16),...;
(3),(8),14,(22),31,41,(53),66,80,95,(112),130,149,169,190,...;
(14),(45),86,(152),232,327,(457),606,775,965,(1202),1464,1752,2067,...;
(86),(318),645,(1251),2026,2991,(4455),6207,8274,10684,(13934),17653,...;
(645),(2671),5662,(11869),20143,30827,(48480),70355,96990,128959,...;
(5662),(25805),56632,(126987),223977,352936,(582183),874664,1240239,...;
(56632),(280609),633545,(1508209),2748448,4438122,(7641111),11831184,...;
(633545),(3381993),7820115,(19651299),36837937,60743909,...; ...
where row n equals the partial sums of row n-1 after removing terms
at positions {m*(m+1)/2, m>=0} (marked by parenthesis in above table).
For example, to generate row 3 from row 2:
[3,8, 14, 22, 31,41, 53, 66,80,95, 112, 130,...]
remove terms at positions {0,1,3,6,10,...}, yielding:
[14, 31,41, 66,80,95, 130,149,169,190, ...]
then take partial sums to obtain row 3:
[14, 45,86, 152,232,327, 457,606,775,965, ...].
Continuing in this way generates all rows of this table.
RELATION TO POWERS OF A SPECIAL TRIANGULAR MATRIX.
Columns 0 and 1 are found in triangle T= A152400, which begins: 1;
1, 1;
3, 2, 1;
14, 8, 3, 1;
86, 45, 15, 4, 1;
645, 318, 99, 24, 5, 1;
5662, 2671, 794, 182, 35, 6, 1;
56632, 25805, 7414, 1636, 300, 48, 7, 1; ...
where column k of T = column 0 of matrix power T^(k+1) for k>=0.
Furthermore, matrix powers of triangle T= A152400 satisfy: column k of T^(j+1) = column j of T^(k+1) for all j>=0, k>=0.
Column 3 of this square array = column 1 of T^2:
1;
2, 1;
8, 4, 1;
45, 22, 6, 1;
318, 152, 42, 8, 1;
2671, 1251, 345, 68, 10, 1;
25805, 11869, 3253, 648, 100, 12, 1; ...
1;
1, 1, 1;
1, 2, 2, 3, 3, 3;
1, 3, 5, 5, 8, 11, 11, 14, 14, 14;
1, 4, 9, 14, 14, 22, 33, 44, 44, 58, 72, 72, 86, 86, 86;...
where right border = column 0 of this square array.
PROG
(PARI) {T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==m*(m+1)/2, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}
Triangle T, read by rows, where T(n,k) = A007559(n-k)*C(n,k) where A007559 equals the triple factorials in column 0.
+10
4
1, 1, 1, 4, 2, 1, 28, 12, 3, 1, 280, 112, 24, 4, 1, 3640, 1400, 280, 40, 5, 1, 58240, 21840, 4200, 560, 60, 6, 1, 1106560, 407680, 76440, 9800, 980, 84, 7, 1, 24344320, 8852480, 1630720, 203840, 19600, 1568, 112, 8, 1, 608608000, 219098880, 39836160
COMMENTS
Comments from Peter Bala, Jul 10 2008: (Start) This array is the particular case P(1,3) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below n\k|0....................1...............2.........3.....4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
See A094587 for some general properties of these arrays. Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159 and P(2,3) = A136216. (End) The generalized Pascal matrix that Bala refers to is itself a special case of application of the formalism of A133314 to fundamental matrices derived from infinitesimal generators described in A133314, of which the fundamental Pascal ( A007318), unsigned Lah ( A105278) and associated Laguerre ( A135278) matrices are special examples. The formalism gives, among other relations, the inverse of T as TI(n,k) = b(n-k)*C(n,k) where the sequence b is given by the list partition transform ( A133314) of A007559; i.e., b = LPT( A007559) = (1,- A008544)= (1,-1,-2,-10,-80,...). The formalism of A132382 may also be applied with the double factorial A001147 replaced by the triple factorial A007559 (see also A133480). - Tom Copeland, Aug 18 2008 Exponential Riordan array [1/(1 - 3*y)^(1/3), y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = sum {k = 0..n} binomial(n,k)*y^(n-k)*R(k,x).
Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = product {k = 0..n-1} (x + 3*k) with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x).
For example, row 3 is (28, 12, 3, 1) so P(3,x + 1) = (x + 1)*(x + 4)*(x + 7) = 28 + 12*x + 3*x*(x + 3) + x*(x + 3)*(x + 6). (End)
FORMULA
Column k of T = column 0 of U^(k+1) (matrix power) for k>=0 where U = A136214. Matrix square equals A136216, where A136216(n,k) = A008544(n-k)*C(n,k) where A008544 are also triple factorials. T(n,k) = (3*n-3*k-2)*T(n-1,k) + T(n-1,k-1).
E.g.f. exp(x*y)/(1-3*y)^(1/3) = 1 + (1+x)*y + (4+2*x+x^2)*y^2/2! + ... . (End)
EXAMPLE
Column k of T = column 0 of U^(k+1), while
column k of U = column 0 of T^(3k+1) where U = A136214 and column k of V = column 0 of T^(3k+2) where V = A112333. This triangle T begins:
1;
1, 1;
4, 2, 1;
28, 12, 3, 1;
280, 112, 24, 4, 1;
3640, 1400, 280, 40, 5, 1;
58240, 21840, 4200, 560, 60, 6, 1;
1106560, 407680, 76440, 9800, 980, 84, 7, 1; ...
1;
1, 1;
4, 4, 1;
28, 28, 7, 1;
280, 280, 70, 10, 1;
3640, 3640, 910, 130, 13, 1; ...
with triple factorials A007559 in column 0. 1;
2, 1;
10, 5, 1;
80, 40, 8, 1;
880, 440, 88, 11, 1;
12320, 6160, 1232, 154, 14, 1; ...
with triple factorials A008544 in column 0.
MATHEMATICA
T[n_, k_]:= Binomial[n, k]*If[n - k == 0, 1, Product[3*j + 1, {j, 0, n - k - 1}]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 10 2018 *)
PROG
(PARI) T(n, k)=binomial(n, k)*if(n-k==0, 1, prod(j=0, n-k-1, 3*j+1))
Triangle T, read by rows, where T(n,k) = A008544(n-k)*C(n,k) where A008544 equals the triple factorials in column 0.
+10
4
1, 2, 1, 10, 4, 1, 80, 30, 6, 1, 880, 320, 60, 8, 1, 12320, 4400, 800, 100, 10, 1, 209440, 73920, 13200, 1600, 150, 12, 1, 4188800, 1466080, 258720, 30800, 2800, 210, 14, 1, 96342400, 33510400, 5864320, 689920, 61600, 4480, 280, 16, 1
COMMENTS
This array is the particular case P(2,3) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown in the comments to A094587. - Peter Bala, Jul 10 2008 The row polynomials form an Appell sequence. - Tom Copeland, Dec 03 2013
FORMULA
Column k of T = column 0 of V^(k+1) for k>=0 where V = A112333. Equals the matrix square of triangle A136215. T(n,k) = (3*n-3*k-1)*T(n-1,k) + T(n-1,k-1). - Peter Bala, Jul 10 2008 Using the formalism of A132382 modified for the triple rather than the double factorial (replace 2 by 3 in basic formulas), the e.g.f. for the row polynomials is exp(x*t)*(1-3x)^(-2/3). - Tom Copeland, Aug 18 2008 Exponential Riordan array [1/(1 - 3*y)^(2/3), y].
The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = sum {k = 0..n} binomial(n,k)*y^(n-k)*R(k,x).
Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = product {k = 0..n-1} (2*x + 3*k) with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (80, 30, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 5)*(2*x + 8) = 80 + 20*(2*x) + 6*(2*x*(2*x + 3)) + (2*x)*(2*x + 3)*(2*x + 6). (End)
EXAMPLE
Triangle begins:
1;
2, 1;
10, 4, 1;
80, 30, 6, 1;
880, 320, 60, 8, 1;
12320, 4400, 800, 100, 10, 1;
209440, 73920, 13200, 1600, 150, 12, 1;
4188800, 1466080, 258720, 30800, 2800, 210, 14, 1; ...
MATHEMATICA
(* The function RiordanArray is defined in A256893. *)
PROG
(PARI) {T(n, k) = binomial(n, k)*if(n-k==0, 1, prod(j=0, n-k-1, 3*j+2))}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions 0 and {(m+1)*(m+2)/2-2, m>0} and then taking partial sums, starting with all 1's in row 0.
+10
2
1, 1, 1, 3, 2, 1, 13, 7, 3, 1, 71, 33, 13, 4, 1, 461, 191, 71, 20, 5, 1, 3447, 1297, 461, 120, 28, 6, 1, 29093, 10063, 3447, 836, 181, 38, 7, 1, 273343, 87669, 29093, 6616, 1333, 270, 49, 8, 1, 2829325, 847015, 273343, 58576, 11029, 2150, 375, 61, 9, 1
EXAMPLE
To generate the array, start with all 1's in row 0; from then on,
obtain row n+1 from row n by first removing terms in row n at
positions 0 and {(m+1)*(m+2)/2-2,m>0} and then taking partial sums.
This square array A begins:
(1), (1), 1, 1, (1), 1, 1, 1, (1), 1, 1, 1, 1, (1), 1, 1, 1, 1, 1, ...;
(1), (2), 3, 4, (5), 6, 7, 8, (9), 10, 11, 12, 13, (14), 15, 16, ...;
(3), (7), 13, 20, (28), 38, 49, 61, (74), 89, 105, 122, 140, (159),...;
(13), (33), 71, 120, (181), 270, 375, 497, (637), 817, 1019, 1244, ...;
(71), (191), 461, 836, (1333), 2150, 3169, 4413, (5906), 8001, ...;
(461), (1297), 3447, 6616, (11029), 19030, 29483, 42775, (59324),...;
(3447), (10063), 29093, 58576, (101351), 185674, 300329, 451277, ...;
(29093), (87669), 273343, 573672, (1024949), 1982310, 3330651, ...;
(273343), (847015), 2829325, 6159976, (11320359), 23009602, 39998897, ...;
where terms in parenthesis at positions {0,1,4,8,13,..} in a row
are removed before taking partial sums to obtain the next row.
...
RELATION TO SPECIAL TRIANGLE.
1;
1, 1;
3, 2, 1;
13, 7, 3, 1;
71, 33, 13, 4, 1;
461, 191, 71, 21, 5, 1;
3447, 1297, 461, 133, 31, 6, 1;
29093, 10063, 3447, 977, 225, 43, 7, 1; ...
in which column 0 and column 1 are found in square array A.
...
1;
2, 1;
8, 4, 1;
42, 20, 6, 1;
266, 120, 38, 8, 1;
1954, 836, 270, 62, 10, 1;
16270, 6616, 2150, 516, 92, 12, 1;
151218, 58576, 19030, 4688, 882, 128, 14, 1; ...
where column 1 and column 2 are also found in square array A.
PROG
(PARI) {T (n, k)=local (A=0, b=2, c=1, d=0); if (n==0, A=1, until (d>k, if (c==b* (b+1)/2-2, b+=1, A+=T (n-1, c); d+=1); c+=1)); A}
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