A101481 -id:A101481 - cp's OEIS frontend

A101479 Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal.

1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 19, 9, 3, 1, 1, 191, 70, 18, 4, 1, 1, 2646, 795, 170, 30, 5, 1, 1, 46737, 11961, 2220, 335, 45, 6, 1, 1, 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1, 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1, 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1

Offset: 0

Paul D. Hanna, Jan 21 2005, Jul 26 2006, May 27 2007

Remarkably, T equals the product of these triangular matrices: T = A107867*A107862^-1 = A107870*A107867^-1 = A107873*A107870^-1; reversing the order of these products yields triangle A107876.

			Triangle begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1;
25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1;
735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1; ...
Row 4 starts with row 3 of T^3 which begins:
1;
3, 1;
6, 3, 1;
19, 9, 3, 1; ...
row 5 starts with row 4 of T^4 which begins:
1;
4, 1;
10, 4, 1;
34, 14, 4, 1;
191, 70, 18, 4, 1; ...
An ALTERNATE GENERATING METHOD is illustrated as follows.
For row 4:
Start with a '1' and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0;
1, 1, 1, 0;
1, 2, 3, 3;
1, 3, 6, 9;
1, 4,10,19.
Final nonzero terms form row 4: [19,9,3,1,1].
For row 5:
Start with a '1' and append 3 zeros,
take partial sums and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0, 0;
1, 1, 1, 1, 0,  0;
1, 2, 3, 4, 4,  4,  0;
1, 3, 6,10,14, 18, 18;
1, 4,10,20,34, 52, 70;
1, 5,15,35,69,121,191;
where the final nonzero terms form row 5: [191,70,18,4,1,1].
Likewise, for row 6:
1, 0, 0, 0,  0;
1, 1, 1, 1,  1,  0,  0,  0;
1, 2, 3, 4,  5,  5,  5,  5,   0,   0;
1, 3, 6,10, 15, 20, 25, 30,  30,  30,   0;
1, 4,10,20, 35, 55, 80,110, 140, 170, 170;
1, 5,15,35, 70,125,205,315, 455, 625, 795;
1, 6,21,56,126,251,456,771,1226,1851,2646;
where the final nonzero terms form row 6: [2646,795,170,30,5,1,1].
Continuing in this way generates all rows of this triangle.

Alois P. Heinz, Rows n = 0..140, flattened (first 31 rows from Paul D. Hanna)

Columns are A101481, A101482, A101483, row sums form A101484.
Cf. A107876 (dual triangle).
Cf. A304184, A304185, A304186, A304187.

b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(n-1)
    end:
T:= proc(n,k) option remember;
     `if`(n=k, 1, `if`(k>n, 0, b(n)[n,k+1]))
    end:
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 13 2020

b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], n-1];
T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 25 2020, after Alois P. Heinz *)

{T(n,k) = my(A=Mat(1),B); for(m=1,n+1, B=matrix(m,m); for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1, B[i,j] = (A^(i-2))[i-1,j]);)); A=B); return(A[n+1,k+1])}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))

{T(n,k) = my(A=vector(n+1),p); A[1]=1; for(j=1,n-k-1, p=(n-1)*(n-2)/2-(n-j-1)*(n-j-2)/2; A = Vec((Polrev(A)+x*O(x^p))/(1-x))); A = Vec((Polrev(A) +x*O(x^p)) / (1-x) ); A[p+1]}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))

A101483 Column 2 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

Original entry on oeis.org

1, 1, 3, 18, 170, 2220, 37149, 758814, 18301950, 508907970, 16023271660, 563336380167, 21870097514265, 929083211885920, 42864992943282825, 2134180938818369850, 114045356939250133338, 6510275427522198876426

Offset: 0

Paul D. Hanna, Jan 21 2005

nonn

			This sequence can also be generated in the following manner.
Start a table with the all 1's sequence in row 0; from then on, row n+1 can be formed from row n by dropping the initial n+1 terms of row n and taking partial sums of the remaining terms to obtain row n+1.
The table below illustrates this method:
[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, ...;
[3, 7, 12], 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, ...;
[18, 43, 76, 118], 170, 233, 308, 396, 498, 615, 748, 898, ...;
[170, 403, 711, 1107, 1605], 2220, 2968, 3866, 4932, 6185, ...;
[2220, 5188, 9054, 13986, 20171, 27816], 37149, 48420, ...;
[37149, 85569, 147471, 225363, 322075, 440785, 585046], 758814, ...; ...
In the above table, drop the initial n+1 terms in row n (enclosed in square brackets) and then take partial sums to obtain row n+1 for n>=0;
this sequence then forms the first column of the resultant table.
Note: column k of the above table equals column 2 of matrix power T^(k+1) where T=A101479, for k>=0.

Vaclav Kotesovec, Table of n, a(n) for n = 0..370

Cf. A101479, A101481, A101482.

{a(n)=local(A=Mat(1),B);for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=(A^i)[i-1,j]);));A=B);return(A[n+1,1])}
for(n=0,25,print1(a(n),", "))

{a(n)=local(A=[1]);for(i=1,n,A=Vec(Ser(A)/(1-x)^(#A));A=concat(A,A[#A]));A[#A]}
for(n=0,25,print1(a(n),", "))

a(n) ~ (2 + LambertW(-2*exp(-2))) * exp(n+2) * n^(n - 3/2) / (sqrt(Pi) * 2^(n + 1/2)). - Vaclav Kotesovec, Dec 03 2017

A101482 Column 1 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

Original entry on oeis.org

1, 1, 2, 9, 70, 795, 11961, 224504, 5051866, 132523155, 3969912160, 133678842902, 4997280555576, 205320100093953, 9195224163850830, 445775353262707365, 23255990676521697670, 1299028117862237432959, 77348967890083608924045

Offset: 0

Paul D. Hanna, Jan 21 2005

nonn

Number of Dyck paths whose ascent lengths are exactly {n, n-1, .. 1}, for example the a(2) = 2 paths are uududd and uuddud. - David Scambler, May 30 2012

			This sequence can also be generated in the following manner.
Start a table with the all 1's sequence in row 0; from then on, row n+1 can be formed from row n by dropping the initial n terms of row n and taking partial sums of the remaining terms to obtain row n+1.
The following table (A136730) illustrates this method:
  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, ...;
  [2, 5], 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, ...;
  [9, 23, 43], 70, 105, 149, 203, 268, 345, 435, 539, 658, ...;
  [70, 175, 324, 527], 795, 1140, 1575, 2114, 2772, 3565, ...;
  [795, 1935, 3510, 5624, 8396], 11961, 16471, 22096, 29025, ...;
  [11961, 28432, 50528, 79553, 117020, 164672], 224504, ...; ...
In the above table, drop the initial n terms in row n (enclosed in square brackets) and then take partial sums to obtain row n+1 for n>=0;
this sequence then forms the first column of the resultant table.
Note: column k of the above table equals column 1 of matrix power T^(k+1) where T=A101479, for k>=0.

Cf. A101479, A101481, A101483, A136730.

{a(n)=local(A=Mat(1),B);for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,B[i,j]=(A^(i-1))[i-1,j]);));A=B);return(A[n+1,1])}

Equals column 0 of array A136730.

A304184 G.f. A(x) satisfies: 0 = [x^n] (1+x)^(n*(n-1)/2) / A(x) for n>0.

Original entry on oeis.org

1, 0, 0, 1, 9, 117, 1851, 34923, 765933, 19155084, 538051164, 16771165230, 574424285076, 21443516818065, 866521903003641, 37683366660458208, 1754777541925339779, 87115221430910051901, 4592968693335470802627, 256294382115032521083411, 15090698035153332532531074

Offset: 0

Paul D. Hanna, May 08 2018

nonn

			G.f.: A(x) = 1 + x^3 + 9*x^4 + 117*x^5 + 1851*x^6 + 34923*x^7 + 765933*x^8 + 19155084*x^9 + 538051164*x^10 + 16771165230*x^11 + 574424285076*x^12 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n-1)/2) / A(x) begins:
n=0: [1, 0, 0, -1, -9, -117, -1850, -34905, -765618, ...];
n=1: [1, 0, 0, -1, -9, -117, -1850, -34905, -765618, ...];
n=2: [1, 1, 0, -1, -10, -126, -1967, -36755, -800523, ...];
n=3: [1, 3, 3, 0, -12, -147, -2229, -40815, -876000, ...];
n=4: [1, 6, 15, 19, 0, -180, -2706, -47955, -1005279, ...];
n=5: [1, 10, 45, 119, 191, 0, -3335, -59840, -1214055, ...];
n=6: [1, 15, 105, 454, 1341, 2646, 0, -73965, -1545531, ...];
n=7: [1, 21, 210, 1329, 5955, 19833, 46737, 0, -1913457, ...];
n=8: [1, 28, 378, 3275, 20438, 97533, 364936, 1003150, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^(n*(n-1)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 1, 3, 19, 191, 2646, 46737, 1003150, 25330125,  ...]
yields A101481, column 0 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Cf. A101481, A304185, A304186, A304187, A304189, A101479.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m-1)*(m-2)/2)/Ser(A) )[m] );A[n+1]}
for(n=0,30, print1(a(n),", "))

[x^n] (1+x)^(n*(n+1)/2) / A(x) = A101481(n+1) = A101479(n+1,0) for n>=0.

[x^n] (1+x)^((n+1)*(n+2)/2) / A(x) = Sum_{k=0..n} A101479(n+2,k+1) * A101479(k+1,0) for n>=0.

A304185 G.f. A(x) satisfies: 0 = [x^n] (1+x)^(n*(n+1)/2) / A(x) for n>0.

Original entry on oeis.org

1, 1, 1, 6, 51, 609, 9284, 171779, 3729929, 92828134, 2602268335, 81082146565, 2778537934965, 103826098384590, 4200631499008965, 182912327481135887, 8528448938007615767, 423908532350099458532, 22375155669520993773517, 1249871928128324730985042, 73662719257076005954584046

Offset: 0

Paul D. Hanna, May 08 2018

nonn

			G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 51*x^4 + 609*x^5 + 9284*x^6 + 171779*x^7 + 3729929*x^8 + 92828134*x^9 + 2602268335*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n+1)/2) / A(x) begins:
n=0: [1, -1, 0, -5, -40, -513, -8092, -153395, -3388500, ...];
n=1: [1, 0, -1, -5, -45, -553, -8605, -161487, -3541895, ...];
n=2: [1, 2, 0, -7, -56, -648, -9756, -179250, -3873474, ...];
n=3: [1, 5, 9, 0, -75, -837, -11875, -210518, -4441140, ...];
n=4: [1, 9, 35, 70, 0, -1096, -15664, -263340, -5357885, ...];
n=5: [1, 14, 90, 345, 795, 0, -20260, -352235, -6842115, ...];
n=6: [1, 20, 189, 1115, 4510, 11961, 0, -452166, -9245340, ...];
n=7: [1, 27, 350, 2893, 17019, 74282, 224504, 0, -11809259, ...];
n=8: [1, 35, 594, 6505, 51545, 312984, 1483340, 5051866, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^(n*(n+1)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 2, 9, 70, 795, 11961, 224504, 5051866, 132523155, ...]
yields A101482, column 1 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Cf. A101481, A304184, A304186, A304187, A101479.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^(m*(m-1)/2)/Ser(A) )[m] );A[n+1]}
for(n=0,30, print1(a(n),", "))

[x^n] (1+x)^((n+1)*(n+2)/2) / A(x) = A101482(n+1) = A101479(n+2,1) for n>=0.

[x^n] (1+x)^((n+2)*(n+3)/2) / A(x) = Sum_{k=0..n} A101479(n+3,k+2) * A101479(k+2,1) for n>=0.

A304186 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+1)*(n+2)/2) / A(x) for n>0.

Original entry on oeis.org

1, 3, 6, 24, 189, 2199, 33495, 625743, 13778469, 348258723, 9916111584, 313642702743, 10901017499472, 412730651160567, 16902257604014685, 744247858158013245, 35058132248539742325, 1758994171367292095805, 93646661943861501833100, 5272500660870261306912750

Offset: 0

Paul D. Hanna, May 08 2018

nonn

			G.f.: A(x) = 1 + 3*x + 6*x^2 + 24*x^3 + 189*x^4 + 2199*x^5 + 33495*x^6 + 625743*x^7 + 13778469*x^8 + 348258723*x^9 + 9916111584*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)*(n+2)/2) / A(x) begins:
n=0: [1, -2, 0, -12, -105, -1434, -23877, -473730, -10881882, ...];
n=1: [1, 0, -3, -14, -129, -1656, -26850, -522918, -11853219, ...];
n=2: [1, 3, 0, -22, -180, -2088, -32219, -608565, -13504179, ...];
n=3: [1, 7, 18, 0, -255, -2937, -41739, -750711, -16140285, ...];
n=4: [1, 12, 63, 170, 0, -3996, -58877, -991308, -20341875, ...];
n=5: [1, 18, 150, 748, 2220, 0, -78435, -1401570, -27251715, ...];
n=6: [1, 25, 297, 2211, 11271, 37149, 0, -1843458, -38615364, ...];
n=7: [1, 33, 525, 5343, 38745, 207663, 758814, 0, -50361381, ...];
n=8: [1, 42, 858, 11340, 108630, 797100, 4541805, 18301950, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^((n+1)*(n+2)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 3, 18, 170, 2220, 37149, 758814, 18301950, 508907970, ...]
yields A101483, column 2 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Cf. A101481, A304184, A304185, A304187, A101479.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^(m*(m+1)/2)/Ser(A) )[m] );A[n+1]}
for(n=0,30, print1(a(n),", "))

[x^n] (1+x)^((n+2)*(n+3)/2) / A(x) = A101483(n+1) = A101479(n+3,2) for n>=0.

[x^n] (1+x)^((n+3)*(n+4)/2) / A(x) = Sum_{k=0..n} A101479(n+4,k+3) * A101479(k+3,2) for n>=0.

A304187 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+2)*(n+3)/2) / A(x) for n>0.

Original entry on oeis.org

1, 6, 21, 86, 606, 6756, 102316, 1931046, 43250376, 1114876536, 32394654066, 1045240099026, 37027935179016, 1427410628324856, 59449956448178106, 2659215814433980056, 127108810653344820456, 6464722863550156435146, 348541208165221134718986, 19854709880058367829287716, 1191556960000156185148449636

Offset: 0

Paul D. Hanna, May 08 2018

nonn

			G.f.: A(x) = 1 + 6*x + 21*x^2 + 86*x^3 + 606*x^4 + 6756*x^5 + 102316*x^6 + 1931046*x^7 + 43250376*x^8 + 1114876536*x^9 + 32394654066*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)*(n+2)/2) / A(x) begins:
n=0: [1, -3, 0, -22, -216, -3180, -56540, -1186170, -28599870, ...];
n=1: [1, 0, -6, -30, -285, -3894, -66750, -1365546, -32331180, ...];
n=2: [1, 4, 0, -50, -440, -5238, -84162, -1657080, -38209725, ...];
n=3: [1, 9, 30, 0, -645, -7917, -115248, -2134920, -47391375, ...];
n=4: [1, 15, 99, 335, 0, -11046, -171920, -2957874, -62097600, ...];
n=5: [1, 22, 225, 1378, 4984, 0, -233730, -4379370, -86791905, ...];
n=6: [1, 30, 429, 3850, 23610, 92652, 0, -5860422, -127938780, ...];
n=7: [1, 39, 735, 8875, 76350, 483684, 2065146, 0, -169402725, ...];
n=8: [1, 49, 1170, 18100, 203065, 1743201, 11567124, 53636520, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^((n+2)*(n+3)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 4, 30, 335, 4984, 92652, 2065146, 53636520, 1589752230, ...]
yields column 3 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Cf. A101481, A304184, A304185, A304186, A101479.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m+1)*(m+2)/2)/Ser(A) )[m] );A[n+1]}
for(n=0,30, print1(a(n),", "))

A101479(n+4,3) = [x^n] (1+x)^((n+3)*(n+4)/2) / A(x) for n>=0.

A295928 Number of triangular matrices T(n,i,k), k <= i <= n, with entries "0" or "1" with the property that each triple {T(n,i,k), T(n,i,k+1), T(n,i-1,k)} containing a single "0" can be successively replaced by {1, 1, 1} until finally no "0" entry remains.

Original entry on oeis.org

1, 3, 16, 122, 1188, 13844, 185448, 2781348, 45868268, 821096828, 15804092592, 324709899276, 7081361097108, 163179784397820, 3958519452109912, 100778473796398524

Offset: 1

Gerhard Kirchner, Nov 30 2017

A triple {T(n,i,k), T(n,i,k+1), T(n,i-1,k)} will be called a primitive triangle. It is easy to see that b(n) = n(n-1)/2 is the number of such triangles. At each step, exactly one primitive triangle is completed (replaced by {1, 1, 1}). So there are b(n) "0"- and n "1"-terms. Thus the starting matrix has no complete primitive triangle. Furthermore, any triangular submatrix T(m,i,k), k <= i <= m < n cannot have more than m "1"-terms because otherwise it would have less "0"-terms than primitive triangles. The replacement of at least one "0"-term would complete more than one primitive triangle. This has been excluded.
So T(n, i, k) is a special case of U(n, i, k), described in A101481: a(n) < A101481(n+1).
A start matrix may serve as a pattern for a number wall used on worksheets for elementary mathematics, see link "Number walls". That is why I prefer the more descriptive name "fill matrix".
The algorithm for the sequence is rather slow because each start matrix is constructed separately. There exists a faster recursive algorithm which produces the same terms and therefore is likely to be correct, but it is based on a conjecture. For the theory of the recurrence, see "Recursive aspects of fill matrices". Probable extension a(10)-a(14): 821096828, 15804092592, 324709899276, 7081361097108, 163179784397820.
The number of fill matrices with n rows and all "1"- terms concentrated on the last two rows, is A001960(n).
See link "Reconstruction of a sequence".
Number of 3-permutations of size n avoiding the patterns (12,12) and (312, 231) (explicit bijection with fill matrices). - Juliette Schabanel, Apr 14 2025

			Example (n=2):    0     1    1
    a(2)=3       1 1   0 1  1 0
Example for completing a 3-matrix (3 bottom terms):
    1        1       1       1
   0 0  ->  1 0 ->  1 1 ->  1 1
  1 1 0    1 1 0   1 1 0   1 1 1
Example for a 3-matrix which cannot be completed:
    1        1
   1 1  or  0 0
  0 0 0    1 0 1

Nicolas Bonichon and Pierre-Jean Morel Baxter d-permutations and other pattern avoiding classes, arXiv:2202.12677 [math.CO], 2022.
Gerhard Kirchner, Recursive aspects of fill matrices
Gerhard Kirchner, Number walls
Gerhard Kirchner, VB-program
Gerhard Kirchner, Reconstruction of a sequence
Ville Salo, Cutting Corners, arXiv:2002.08730 [math.DS], 2020.
Juliette Schabanel, 3D permutations and triangle solitaire, 2025
Yuan Yao and Fedir Yudin, Fine Mixed Subdivisions of a Dilated Triangle, arXiv:2402.13342 [math.CO], 2024.

Cf. A101481, A001960.

From Juliette Schabanel, Apr 14 2025: (Start)
G.f. satisfies T(x)-1-x=I^3(x)∆^3T(x)-3I^2(x)∆^2T(x)+ 3I(x)(∆T(x)-1) with ∆T(x) = (xT(x))' and T(x)= 1 +I(x)∆T(x) (proved).
a(n) ~ cn!e^(sqrt(12n))n^(5/12) (conjectured). (End)

a(10)-a(16) from Juliette Schabanel, Apr 14 2025

cp's OEIS Frontend

A101479 Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal.

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A101483 Column 2 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

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A101482 Column 1 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

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A304184 G.f. A(x) satisfies: 0 = [x^n] (1+x)^(n*(n-1)/2) / A(x) for n>0.

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A304185 G.f. A(x) satisfies: 0 = [x^n] (1+x)^(n*(n+1)/2) / A(x) for n>0.

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A304186 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+1)*(n+2)/2) / A(x) for n>0.

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A304187 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+2)*(n+3)/2) / A(x) for n>0.

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A295928 Number of triangular matrices T(n,i,k), k <= i <= n, with entries "0" or "1" with the property that each triple {T(n,i,k), T(n,i,k+1), T(n,i-1,k)} containing a single "0" can be successively replaced by {1, 1, 1} until finally no "0" entry remains.

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