A111528 -id:A111528 - cp's OEIS frontend

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) = pT(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

Paul D. Hanna, Aug 06 2005

Column 0 equals A111529 (related to log of factorial series).

Column 2 (A111538) equals SHIFT_LEFT(column 0 of log(T)), where the matrix logarithm, log(T), equals the integer matrix A111541.

			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.

Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.

Cf. A111537 (column 1), A111538 (column 2), A111539 (row sums), A111540 (matrix inverse), A111541 (matrix log); related tables: A111528, A104980, A111544, A111553.

T[n_, k_] := T[n, k] = If[nJean-François Alcover, Jan 24 2017, adapted from PARI *)

PARI
```
T(n,k)=if(n
				
```

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.

A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2n-1)xR(n,x) = 1 + (2n-3)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)x^k.

Original entry on oeis.org

1, 1, 2, 1, 2, 6, 1, 2, 10, 26, 1, 2, 14, 74, 158, 1, 2, 18, 138, 706, 1282, 1, 2, 22, 218, 1686, 8162, 13158, 1, 2, 26, 314, 3194, 24162, 110410, 163354, 1, 2, 30, 426, 5326, 53890, 394254, 1708394, 2374078, 1, 2, 34, 554, 8178, 102722, 1019250, 7191018, 29752066, 39456386

Offset: 0

Peter Bala, Jul 15 2022

Compare with A111528, which has a similar definition.

			Square array begins
1, 2,  6,  26,   158,    1282,   13158,    163354,    2374078,     39456386, ...
1, 2, 10,  74,   706,    8162,  110410,   1708394,   29752066,    576037442, ...
1, 2, 14, 138,  1686,   24162,  394254,   7191018,  144786006,   3188449602, ...
1, 2, 18, 218,  3194,   53890, 1019250,  21256090,  483426010,  11895873410, ...
1, 2, 22, 314,  5326,  102722, 2197558,  51355514, 1297759918,  35208930050, ...
1, 2, 26, 426,  8178,  176802, 4206618, 108577674, 3011332338,  89141101506, ...
1, 2, 30, 554, 11846,  283042, 7396830, 208569034, 6288011206, 201404591042, ...
...

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A112934 (row 0), A000698 (row 1), A355722 (row 2), A355723 (row 3), A355724 (row 4), A355725 (row 5). Cf. A001147, A111528.

T := (n,k) -> coeff(series(hypergeom([n+1/2, 1], [], 2*x)/ hypergeom([n-1/2, 1], [], 2*x), x, 21), x, k):
# display as a sequence
seq(seq(T(n-k,k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n,k), k = 0..10)), n = 0..10);

Let d(n) = Product_{k = 1..n} 2*k-1 = A001147(n) denote the double factorial of odd numbers.
O.g.f. for row n: R(n,x) = ( Sum_{k >= 0} d(n+k)/d(n)*x^k )/( Sum_{k >= 0} d(n-1+k)/d(n-1)*x^k ).
R(n,x)/(1 - (2*n-1)*x*R(n,x)) = Sum_{k >= 0} d(n+k)/d(n)*x^k.
R(n,x) = 1/(1 + (2*n-1)*x - (2*n+1)*x/(1 + (2*n+1)*x - (2*n+3)*x/(1 + (2*n+3)*x - (2*n+5)*x/(1 + (2*n+5)*x - ... )))).
R(n,x) satisfies the Riccati differential equation 2*x^2*d/dx(R(n,x)) + (2*n-1)*x*R(n,x)^2 - (1 + (2*n-3)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 2*x/(1 - (2*n+1)*x/(1 - 4*x/(1 - (2*n+3)*x/(1 - 6*x/(1 - (2*n+5)*x/(1 - ... - 2*m*x/(1 - (2*n+2*m-1)*x/(1 - ... ))))))))), a continued fraction of Stieltjes type.
Row 0: A112934(n+1); Row 1; A000698(n+1).

A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3n-1)xR(n,x) = 1 + (3n-4)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)x^k.

Original entry on oeis.org

1, 1, 3, 1, 3, 15, 1, 3, 24, 111, 1, 3, 33, 282, 1131, 1, 3, 42, 507, 4236, 14943, 1, 3, 51, 786, 9609, 76548, 243915, 1, 3, 60, 1119, 17736, 212835, 1608864, 4742391, 1, 3, 69, 1506, 29103, 459768, 5350785, 38488152, 106912131, 1, 3, 78, 1947, 44196, 859143, 13333488

Offset: 0

Peter Bala, Jul 17 2022

Compare with A111528 and A355721, which have similar definitions and properties.

			Square array begins
1, 3, 15,  111,  1131,   14943,    243915,    4742391,    106912131, ...
1, 3, 24,  282,  4236,   76548,   1608864,   38488152,   1032125136, ...
1, 3, 33,  507,  9609,  212835,   5350785,  149961675,   4628365305, ...
1, 3, 42,  786, 17736,  459768,  13333488,  425600976,  14791250688, ...
1, 3, 51, 1119, 29103,  859143,  28091463, 1002057591,  38606468343, ...
1, 3, 60, 1506, 44196, 1458588,  52917360, 2080630776,  87823112496, ...
1, 3, 69, 1947, 63501, 2311563,  91949469, 3943276347, 180679742061, ...
1, 3, 78, 2442, 87504, 3477360, 150259200, 6970190160, 344116224960, ...

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A112936 (row 0), A355794 (row 1), A355795 (row 2), A355796 (row 3), A355797 (row 4). Cf. A008544, A111528, A355721.

T := (n,k) -> coeff(series(hypergeom([n+2/3, 1], [], 3*x)/ hypergeom([n-1/3, 1], [], 3*x), x, 21), x, k):
# display as a sequence
seq(seq(T(n-k,k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n,k), k = 0..10)), n = 0..10);

Let t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
O.g.f. for row n >= 0: R(n,x) = ( Sum_{k >= 0} t(n+k)/t(n)*x^k )/( Sum_{k >= 0} t(n-1+k)/t(n-1)*x^k ).
R(n,x)/(1 - (3*n-1)*x*R(n,x)) = Sum_{k >= 0} t(n+k)/t(n)*x^k.
R(n,x) = 1/(1 + (3*n-1)*x - (3*n+2)*x/(1 + (3*n+2)*x - (3*n+5)*x/(1 + (3*n+5)*x - (3*n+8)*x/(1 + (3*n+8)*x - ... )))) (continued fraction).
R(n,x) satisfies the Riccati differential equation 3*x^2*d/dx(R(n,x)) + (3*n-1)*x*R(n,x)^2 - (1 + (3*n-4)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Applying Stokes 1982 gives R(n,x) = 1/(1 - 3*x/(1 - (3*n+2)*x/(1 - 6*x/(1 - (3*n+5)*x/(1 - 9*x/(1 - (3*n+8)*x/(1 - 12*x/(1 - ...)))))))), a continued fraction of Stieltjes type.

A355794 Row 1 of A355793.

Original entry on oeis.org

1, 3, 24, 282, 4236, 76548, 1608864, 38488152, 1032125136, 30670171248, 1000637672064, 35571839009952, 1368990872569536, 56720594992438848, 2517761078627172864, 119222916630934484352, 5999613754698100628736, 319763269764299852744448, 17994913747767982690289664

Offset: 0

Peter Bala, Jul 19 2022

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A355793 (table).
Cf. A112936 (row 0), A355795 (row 2), A355796 (row 3), A355797 (row 4).
Cf. A008544, A111528, A355721.

n := 1: seq(coeff(series( hypergeom([n+2/3, 1], [], 3*x)/hypergeom([n-1/3, 1], [], 3*x ), x, 21), x, k), k = 0..20);

O.g.f.: A(x) = ( Sum_{k >= 0} t(k+1)/t(1)*x^k )/( Sum_{k >= 0} t(k)/t(0)*x^k ), where t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
A(x)/(1 - 2*x*A(x)) = Sum_{k >= 0} t(k+1)/t(1)*x^k.
A(x) = 1/(1 + 2*x - 5*x/(1 + 5*x - 8*x/(1 + 8*x - 11*x/(1 + 11*x - ... )))) (continued fraction).
A(x) satisfies the Riccati differential equation 3*x^2*A(x)' + 2*x*A(x)^2 - (1 - x)*A(x) + 1 = 0 with A(0) = 1.
Hence by Stokes, A(x) = 1/(1 - 3*x/(1 - 5*x/(1 - 6*x/(1 - 8*x/(1 - 9*x/(1 - 11*x/(1 - 12*x/(1 - ... )))))))), a continued fraction of Stieltjes type.

A355795 Row 2 of A355793.

Original entry on oeis.org

1, 3, 33, 507, 9609, 212835, 5350785, 149961675, 4628365305, 155913036915, 5692874399025, 224034935130075, 9456933847187625, 426402330032719875, 20460268520575152225, 1041301103429870128875, 56040353252589013121625, 3180443637298592493577875, 189863589771186976073108625

Offset: 0

Peter Bala, Jul 21 2022

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A355793 (table).
Cf. A112936 (row 0), A355794 (row 1), A355796 (row 3), A355797 (row 4).
Cf. A008544, A111528, A355721.

n := 2: seq(coeff(series( hypergeom([n+2/3, 1], [], 3*x)/hypergeom([n-1/3, 1], [], 3*x ), x, 21), x, k), k = 0..20);

O.g.f.: A(x) = ( Sum_{k >= 0} t(k+2)/t(2)*x^k )/( Sum_{k >= 0} t(k+1)/t(1)*x^k ), where t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
A(x)/(1 - 5*x*A(x)) = Sum_{k >= 0} t(k+2)/t(2)*x^k.
A(x) = 1/(1 + 5*x - 8*x/(1 + 8*x - 11*x/(1 + 11*x - 14*x/(1 + 14*x - ... )))) (continued fraction).
A(x) satisfies the Riccati differential equation 3*x^2*A(x)' + 5*x*A(x)^2 - (1 + 2*x)*A(x) + 1 = 0 with A(0) = 1.
Hence by Stokes, A(x) = 1/(1 - 3*x/(1 - 8*x/(1 - 6*x/(1 - 11*x/(1 - 9*x/(1 - 14*x/(1 - 12*x/(1 - ... )))))))), a continued fraction of Stieltjes type.

A355796 Row 3 of A355793.

Original entry on oeis.org

1, 3, 42, 786, 17736, 459768, 13333488, 425600976, 14791250688, 555381292800, 22398626084352, 965768866650624, 44347055502428160, 2161455366606034944, 111489317304231616512, 6069676735484389779456, 347921629212782938472448, 20950823605616500202323968, 1322561808699778749456678912

Offset: 0

Peter Bala, Jul 21 2022

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A355793 (table).
Cf. A112936 (row 0), A355794 (row 1), A355795 (row 2), A355797 (row 4).
Cf. A008544, A111528, A355721.

n := 3: seq(coeff(series( hypergeom([n+2/3, 1], [], 3*x)/hypergeom([n-1/3, 1], [], 3*x ), x, 21), x, k), k = 0..20);

Let t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
O.g.f.: A(x) = ( Sum_{k >= 0} t(k+3)/t(3)*x^k )/( Sum_{k >= 0} t(k+2)/t(2)*x^k ).
A(x)/(1 - 8*x*A(x)) = Sum_{k >= 0} t(k+3)/t(3)*x^k.
A(x) = 1/(1 + 8*x - 11*x/(1 + 11*x - 14*x/(1 + 14*x - 17*x/(1 + 17*x - ... )))) (continued fraction).
A(x) satisfies the Riccati differential equation 3*x^2*d/dx(A(x)) + 8*x*R(n,x)^2 - (1 + 5*x)*R(n,x) + 1 = 0 with A(0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 3*x/(1 - 11*x/(1 - 6*x/(1 - 14*x/(1 - 9*x/(1 - 17*x/(1 - 12*x/(1 - ...)))))))), a continued fraction of Stieltjes type.

A355797 Row 4 of A355793.

Original entry on oeis.org

1, 3, 51, 1119, 29103, 859143, 28091463, 1002057591, 38606468343, 1595167432599, 70315835952471, 3293268346004439, 163337193581191575, 8554718468806548951, 471976737725208306327, 27369722655919760451159, 1664858070989667129693975, 106029602841882346657155543

Offset: 0

Peter Bala, Jul 21 2022

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A355793 (table).
Cf. A112936 (row 0), A355794 (row 1), A355795 (row 2), A355796 (row 3).
Cf. A008544, A111528, A355721.

n := 4: seq(coeff(series( hypergeom([n+2/3, 1], [], 3*x)/hypergeom([n-1/3, 1], [], 3*x ), x, 21), x, k), k = 0..20);

Let t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
O.g.f.: A(x) = ( Sum_{k >= 0} t(k+4)/t(4)*x^k )/( Sum_{k >= 0} t(k+3)/t(3)*x^k ).
A(x)/(1 - 11*x*A(x)) = Sum_{k >= 0} t(k+4)/t(4)*x^k.
A(x) = 1/(1 + 11*x - 14*x/(1 + 14*x -17*x/(1 + 17*x - 20*x/(1 + 20*x - ... )))) (continued fraction).
A(x) satisfies the Riccati differential equation 3*x^2*d/dx(A(x)) + 11*x*A(x)^2 - (1 + 8*x)*A(x) + 1 = 0 with A(0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 3*x/(1 - 14*x/(1 - 6*x/(1 - 17*x/(1 - 9*x/(1 - 20*x/(1 - 12*x/(1 - 23*x/(1 - ...))))))))), a continued fraction of Stieltjes type.

A111184 Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 6, 1, 0, 24, 34, 12, 1, 0, 120, 210, 110, 20, 1, 0, 720, 1452, 974, 270, 30, 1, 0, 5040, 11256, 8946, 3248, 560, 42, 1, 0, 40320, 97296, 87504, 38338, 8792, 1036, 56, 1

Offset: 0

Philippe Deléham and Paul D. Hanna, Oct 16 2005

			Rows begin:
  1;
  0,     1;
  0,     2,     1;
  0,     6,     6,     1;
  0,    24,    34,    12,     1;
  0,   120,   210,   110,    20,    1;
  0,   720,  1452,   974,   270,   30,    1;
  0,  5040, 11256,  8946,  3248,  560,   42,  1;
  0, 40320, 97296, 87504, 38338, 8792, 1036, 56, 1;
  ...

Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
R. Cori, Indecomposable permutations, hypermaps and labeled Dyck paths, J. Comb. Theory A 116 (2009) 1326-1343, end of Section 1.2.2.

Cf. A003319, A004208, A111528.

DELTA[r_, s_, m_] := Module[{p, q, t, x, y}, q[k_] := x r[[k+1]] + y s[[k+1]]; p[0, ] = 1; p[, -1] = 0; p[n_ /; n >= 1, k_ /; k >= 0] := p[n, k] = p[n, k-1] + q[k] p[n-1, k+1] // Expand; t[n_, k_] := Coefficient[p[n, 0], x^(n-k) y^k]; t[0, 0] = p[0, 0]; Table[t[n, k], {n, 0, m}, {k, 0, n}]];
DELTA[LinearRecurrence[{1, 1, -1}, {0, 2, 1}, 10], Mod[Range[10], 2], 10] // Flatten (* Jean-François Alcover, Jul 27 2018 *)

{T(n,k)=local(A=1+x*y);for(i=1,n,A=1-x*deriv(log(1+x-x*y-x*A +x*O(x^n))));polcoeff(polcoeff(A,n,x),k,y)} /* Paul D. Hanna */

{T(n, k)=local(A=1+x*y); for(i=1, n, A=(1 + x^2*A')/(1 + x - x*y - x*A +x*O(x^n))); polcoeff(polcoeff(A, n, x), k, y)} /* Paul D. Hanna */
/* Print 10 Rows of the triangle: */
for(n=0,10,for(k=0,n,print1(T(n,k),","));print(""))

O.g.f. satisfies: A(x,y) = (1 + x^2*A'(x,y)) / (1+x - x*y - x*A(x,y)), where A'(x,y) = d/dx A(x,y). - Paul D. Hanna, Jul 31 2011
O.g.f. satisfies: A(x,y) = 1 - x * d/dx log(1+x - x*y - x*A(x,y)). - Paul D. Hanna, Jul 30 2011
Sum_{k=0..n} T(n, k) = A003319(n+1).
Sum_{k=0..n} T(n, k)*2^(n-k) = A004208(n).
From Mikhail Kurkov, Jul 15 2025: (Start)
Conjecture 1: Sum_{k=1..n} T(n,k)*q^(k-1) = A111528(q,n) for n > 0, q >= 0.
Conjecture 2: Sum_{k=1..n} T(n,k)*(-1/q)^(k-1) = R(n,q)/q^(n-1) for n > 0, q > 0 where log(1 + x + q*x*[Sum_{k>=1} R(k,q)*x^k]) = Sum_{k>=1} R(k,q)/k*x^k.
Conjecture 3: n-th row polynomial is x*v_n for n > 0 where we start with vector v of fixed length m with elements v_i = 1, and for i=1..m-1, j=i+1..m apply A := v_i (at the beginning of each cycle for i) and also apply A := A + v_j, v_j := (j-i+x-1)*v_j + A.
Here last conjecture provides fast and simple algorithm, that allow to compute sums in the previous conjectures by substituting x = q and x = -1/q respectively. (End)

cp's OEIS Frontend

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.

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A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2*n-1)*x*R(n,x) = 1 + (2*n-3)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)*x^k.

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A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3*n-1)*x*R(n,x) = 1 + (3*n-4)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)*x^k.

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A355794 Row 1 of A355793.

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A355795 Row 2 of A355793.

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A355796 Row 3 of A355793.

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A355797 Row 4 of A355793.

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Formula

A111184 Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938.

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Mathematica

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) = pT(p+m,p+2) for all m>=1 and p>=-2.

A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2n-1)xR(n,x) = 1 + (2n-3)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)x^k.

A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3n-1)xR(n,x) = 1 + (3n-4)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)x^k.