A092683 -id:A092683 - cp's OEIS frontend

A092686 Triangle, read by rows, such that the convolution of each row with {1,2} produces a triangle which, when flattened, equals this flattened form of the original triangle.

1, 2, 2, 6, 4, 6, 16, 14, 12, 16, 46, 40, 40, 32, 46, 132, 120, 112, 110, 92, 132, 384, 352, 334, 312, 316, 264, 384, 1120, 1038, 980, 940, 896, 912, 768, 1120, 3278, 3056, 2900, 2776, 2704, 2592, 2656, 2240, 3278, 9612, 9012, 8576, 8256, 8000, 7840, 7552, 7758

Offset: 0

Paul D. Hanna, Mar 04 2004

First column and main diagonal forms A092687. Row sums form A092688.

This triangle is the cascadence of binomial (1+2x). More generally, the cascadence of polynomial F(x) of degree d, F(0)=1, is a triangle with d*n+1 terms in row n where the g.f. of the triangle, A(x,y), is given by: A(x,y) = ( x*H(x) - y*H(x*y^d) )/( x*F(y) - y ), where H(x) satisfies: H(x) = G*H(x*G^d)/x and G=G(x) satisfies: G(x) = x*F(G(x)) so that G = series_reversion(x/F(x)); also, H(x) is the g.f. of column 0. - Paul D. Hanna, Jul 17 2006

			Rows begin:
1;
2, 2;
6, 4, 6;
16, 14, 12, 16;
46, 40, 40, 32, 46;
132, 120, 112, 110, 92, 132;
384, 352, 334, 312, 316, 264, 384;
1120, 1038, 980, 940, 896, 912, 768, 1120;
3278, 3056, 2900, 2776, 2704, 2592, 2656, 2240, 3278;
9612, 9012, 8576, 8256, 8000, 7840, 7552, 7758, 6556, 9612;
28236, 26600, 25408, 24512, 23840, 23232, 22862, 22072, 22724, 19224, 28236; ...
Convolution of each row with {1,2} results in the triangle:
1, 2;
2, 6, 4;
6, 16, 14, 12;
16, 46, 40, 40, 32;
46, 132, 120, 112, 110, 92;
132, 384, 352, 334, 312, 316, 264;
384, 1120, 1038, 980, 940, 896, 912, 768; ...
which, when flattened, equals the original triangle in flattened form.

Paul D. Hanna, Table of n, a(n) for n = 0..495

Cf. A092683, A092687, A092688, A092689, A120894, A120898.

T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,2, if(k==n,T(n,0), 2*T(n-1,k)+T(n-1,k+1)))))
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))

/* Generate Triangle by the G.F.: */
{T(n,k)=local(A,F=1+2*x,d=1,G=x,H=1+2*x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Jul 17 2006

T(n, k) = 2*T(n-1, k) + T(n-1, k+1) for 0<=k

G.f.: A(x,y) = ( x*H(x) - y*H(x*y) )/( x*(1+2y) - y ), where H(x) satisfies: H(x) = H(x^2/(1-2x))/(1-2x) and H(x) is the g.f. of column 0 (A092687). - Paul D. Hanna, Jul 17 2006

A120898 Cascadence of 1+2x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,2,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.

Original entry on oeis.org

1, 2, 1, 2, 5, 6, 5, 2, 5, 16, 22, 18, 14, 12, 5, 16, 54, 78, 72, 58, 43, 38, 37, 16, 54, 186, 282, 280, 231, 182, 156, 128, 123, 124, 54, 186, 654, 1030, 1073, 924, 751, 622, 535, 498, 425, 418, 426, 186, 654, 2338, 3787, 4100, 3672, 3048, 2530, 2190, 1956, 1766

Offset: 0

Paul D. Hanna, Jul 14 2006

In this case, the g.f. of column 0, H(x), satisfies: H(x) = H(x*G^2)*G/x where G satisfies: G = x*(1+2G+G^2), so that 1+G = g.f. of Catalan numbers (A000108). More generally, the cascadence of polynomial F(x) of degree d, F(0)=1, is a triangle with d*n+1 terms in row n where the g.f. H(x) of column 0 satisfies: H(x) = H(x*G^d)*G/x where G = x*F(G); thus G = series_reversion(x/F(x)), or, equivalently, [x^n] G = [x^n] x*F(x)^n/n for n>=1.

Further, the g.f. of the cascadence triangle for polynomial F(x) of degree d is given by: A(x,y) = ( x*H(x) - y*H(x*y^d) )/( x*F(y) - y ), where H(x) = G*H(x*G^d)/x and G = x*F(G). - Paul D. Hanna, Jul 17 2006

			Triangle begins:
1;
2, 1, 2;
5, 6, 5, 2, 5;
16, 22, 18, 14, 12, 5, 16;
54, 78, 72, 58, 43, 38, 37, 16, 54;
186, 282, 280, 231, 182, 156, 128, 123, 124, 54, 186;
654, 1030, 1073, 924, 751, 622, 535, 498, 425, 418, 426, 186, 654;
2338, 3787, 4100, 3672, 3048, 2530, 2190, 1956, 1766, 1687, 1456, 1452, 1494, 654, 2338; ...
Convolution of [1,2,1] with each row produces:
[1,2,1]*[1] = [1,2,1];
[1,2,1]*[2,1,2] = [2,5,6,5,2];
[1,2,1]*[5,6,5,2,5] = [5,16,22,18,14,12,5];
[1,2,1]*[16,22,18,14,12,5,16] = [16,54,78,72,58,43,38,37,16];
These convoluted rows, when concatenated, yield the sequence:
1,2,1, 2,5,6,5,2, 5,16,22,18,14,12,5, 16,54,78,72,58,43,38,37,16, ...
which equals the concatenated rows of this original triangle:
1, 2,1,2, 5,6,5,2,5, 16,22,18,14,12,5,16, 54,78,72,58,43,38,37,16,54,

Paul D. Hanna, Table of n, a(n) for n = 0..440; rows 0..20 of flattened triangle.

Cf. A120899 (column 0), A120901 (central terms), A120902 (row sums), A000108 (Catalan); variants: A092683, A092686, A120894.

PARI
```
T(n,k)=if(2*n
				
```

/* Generated by the G.F.: */
{T(n,k)=local(A,F=1+2*x+x^2,d=2,G=x,H=1+x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Jul 17 2006

G.f.: A(x,y) = ( x*H(x) - y*H(x*y^2) )/( x*F(y) - y ), where H(x) = G*H(x*G^2)/x, G = x*F(G), F(x)=1+2x+x^2. - Paul D. Hanna, Jul 17 2006

A120894 Cascadence of 1+x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,1,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 2, 1, 2, 5, 7, 6, 5, 3, 2, 5, 12, 18, 18, 14, 10, 10, 7, 5, 12, 30, 48, 50, 42, 34, 27, 22, 24, 17, 12, 30, 78, 128, 140, 126, 103, 83, 73, 63, 53, 59, 42, 30, 78, 206, 346, 394, 369, 312, 259, 219, 189, 175, 154, 131, 150, 108, 78, 206, 552, 946, 1109

Offset: 0

Paul D. Hanna, Jul 14 2006

In this case, the g.f. of column 0, H(x), satisfies: H(x) = H(x*G^2)*G/x where G satisfies: G = x*(1+G+G^2), so that G/x = g.f. of Motzkin numbers (A001006). More generally, the cascadence of polynomial F(x) of degree d, F(0)=1, is a triangle with d*n+1 terms in row n where the g.f. H(x) of column 0 satisfies: H(x) = H(x*G^d)*G/x where G = x*F(G); thus G = series_reversion(x/F(x)), or, equivalently, [x^n] G = [x^n] x*F(x)^n/n for n>=1.

			Triangle begins:
1;
1, 1, 1;
2, 3, 2, 1, 2;
5, 7, 6, 5, 3, 2, 5;
12, 18, 18, 14, 10, 10, 7, 5, 12;
30, 48, 50, 42, 34, 27, 22, 24, 17, 12, 30;
78, 128, 140, 126, 103, 83, 73, 63, 53, 59, 42, 30, 78;
206, 346, 394, 369, 312, 259, 219, 189, 175, 154, 131, 150, 108, 78, 206;
552, 946, 1109, 1075, 940, 790, 667, 583, 518, 460, 435, 389, 336, 392, 284, 206, 552;
1498, 2607, 3130, 3124, 2805, 2397, 2040, 1768, 1561, 1413, 1284, 1160, 1117, 1012, 882, 1042, 758, 552, 1498; ...
Convolution of [1,1,1] with each row produces:
[1,1,1]*[1] = [1,1,1];
[1,1,1]*[1,1,1] = [1,2,3,2,1];
[1,1,1]*[2,3,2,1,2] = [2,5,7,6,5,3,2];
[1,1,1]*[5,7,6,5,3,2,5] = [5,12,18,18,14,10,10,7,5];
[1,1,1]*[12,18,18,14,10,10,7,5,12] = [12,30,48,50,42,34,27,22,24,17,12]; ...
These convoluted rows, when concatenated, yield the sequence:
1,1,1, 1,2,3,2,1, 2,5,7,6,5,3,2, 5,12,18,18,14,10,10,7,5, ...
which equals the concatenated rows of this original triangle:
1, 1,1,1, 2,3,2,1,2, 5,7,6,5,3,2,5, 12,18,18,14,10,10,7,5,12, ...

Paul D. Hanna, Table of n, a(n) for n = 0..440

Cf. A120895 (column 0), A120896 (central terms), A120897 (row sums), A001006 (Motzkin numbers); variants: A092683, A092686, A120898.

PARI
```
T(n,k)=if(2*n
				
```

/* Generated by the G.F.: */
{T(n,k)=local(A,F=1+x+x^2,d=2,G=x,H=1+x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Jul 17 2006

G.f.: A(x,y) = ( x*H(x) - y*H(x*y^2) )/( x*F(y) - y ), where H(x) = G*H(x*G^2)/x, G = x*F(G), F(x)=1+x+x^2. - Paul D. Hanna, Jul 17 2006

A092687 First column and main diagonal of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.

Original entry on oeis.org

1, 2, 6, 16, 46, 132, 384, 1120, 3278, 9612, 28236, 83072, 244752, 722048, 2132704, 6306304, 18666190, 55300732, 163968612, 486528288, 1444571068, 4291629384, 12756459936, 37934818112, 112855778768, 335867740704, 999895548736

Offset: 0

Paul D. Hanna, Mar 04 2004

nonn

Conjecture: Limit n->infinity a(n)^(1/n) = 3. - Vaclav Kotesovec, Jun 29 2015

Vaclav Kotesovec, Table of n, a(n) for n = 0..850

Cf. A092683, A092686, A092688, A092689.

m = 27; A[] = 1; Do[A[x] = A[x^2/(1-2x)]/(1-2x) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)

T(n,k)=if(n<0||k>n,0, if(n==0&k==0,1, if(n==1&k<=1,2, if(k==n,T(n,0), 2*T(n-1,k)+T(n-1,k+1)))))
a(n)=T(n,0)
for(n=0,30,print1(a(n),", "))

a(n)=local(A=1+x);for(i=0,n\2,A=subst(A,x,x^2/(1-2*x+x*O(x^n)))/(1-2*x));polcoeff(A,n) \\ Paul D. Hanna, Jul 10 2006

/* Using Recurrence: */
a(n)=if(n==0, 1, sum(k=0, n\2, binomial(n-k, k)*2^(n-2*k)*a(k)))
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Jul 10 2006

G.f. satisfies: A(x) = A( x^2/(1-2x) )/(1-2x). Recurrence: a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*2^(n-2k)*a(k). - Paul D. Hanna, Jul 10 2006

A092689 Triangle, read by rows, such that the convolution of each row with {1,2} produces a triangle which, after the main diagonal is divided by 2 and the triangle is flattened, equals this flattened form of the original triangle.

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 7, 5, 3, 7, 19, 13, 13, 7, 19, 51, 39, 33, 33, 19, 51, 141, 111, 99, 85, 89, 51, 141, 393, 321, 283, 259, 229, 243, 141, 393, 1107, 925, 825, 747, 701, 627, 675, 393, 1107, 3139, 2675, 2397, 2195, 2029, 1929, 1743, 1893, 1107, 3139, 8953, 7747

Offset: 0

Paul D. Hanna, Mar 04 2004

First column and main diagonal forms the central trinomial coefficients (A002426). Row sums form A092690.

			Rows begin:
{1},
{1,1},
{3,1,3},
{7,5,3,7},
{19,13,13,7,19},
{51,39,33,33,19,51},
{141,111,99,85,89,51,141},
{393,321,283,259,229,243,141,393},
{1107,925,825,747,701,627,675,393,1107},
{3139,2675,2397,2195,2029,1929,1743,1893,1107,3139},
{8953,7747,6989,6419,5987,5601,5379,4893,5353,3139,8953},...
Convolution of each row with {1,2} forms the triangle:
{1,2},
{1,3,2},
{3,7,5,6},
{7,19,13,13,14},
{19,51,39,33,33,38},
{51,141,111,99,85,89,102},
{141,393,321,283,259,229,243,282},...
which, after the main diagonal is divided by 2 and the triangle is flattened, equals the original triangle in flattened form: {1,1,1,3,1,3,7,5,3,7,19,...}.

Cf. A002426, A092683, A092686, A092690.

T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,1, if(k==n-1,T(n-1,0), if(k==n,T(n,0), 2*T(n-1,k)+T(n-1,k+1))))))

T(n, k) = 2*T(n-1, k) + T(n-1, k+1) for 0A002426(n), T(0, 0)=1, T(0, 1)=T(1, 0)=1.

A120919 Cascadence of (1+x)^3; a triangle, read by rows of 3n+1 terms, that retains its original form upon convolving each row with [1,3,3,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 3n+1 terms remain in row n for n>=0.

Original entry on oeis.org

1, 3, 3, 1, 3, 12, 19, 18, 15, 10, 3, 12, 55, 111, 138, 128, 96, 66, 55, 39, 12, 55, 276, 636, 930, 1005, 876, 669, 498, 360, 263, 240, 177, 55, 276, 1464, 3666, 5979, 7317, 7242, 6138, 4737, 3506, 2607, 2046, 1569, 1212, 1170, 883, 276, 1464, 8058, 21369, 37716

Offset: 0

Paul D. Hanna, Jul 17 2006

			Triangle begins:
1;
3, 3, 1, 3;
12, 19, 18, 15, 10, 3, 12;
55, 111, 138, 128, 96, 66, 55, 39, 12, 55;
276, 636, 930, 1005, 876, 669, 498, 360, 263, 240, 177, 55, 276;
Convolution of [1,3,3,1] with each row produces:
[1,3,3,1]*[1] = [1,3,3,1];
[1,3,3,1]*[3,3,1,3] = [3,12,19,18,15,10,3];
[1,3,3,1]*[12,19,18,15,10,3,12] = [12,55,111,138,128,96,66,55,39,12];
[1,3,3,1]*[55,111,138,128,96,66,55,39,12,55] =
[55,276,636,930,1005,876,669,498,360,263,240,177,55];
These convoluted rows, when concatenated, yield the sequence:
1,3,3,1, 3,12,19,18,15,10,3, 12,55,111,138,128,96,66,55,39,12, 55,...
which equals the concatenated rows of this original triangle:
1, 3,3,1,3, 12,19,18,15,10,3,12, 55,111,138,128,96,66,55,39,12,55, ...

Cf. A120920 (column 0), A120922 (central terms), A120923 (row sums), A001764 (ternary trees); variants: A092683, A092686, A120894, A120898, A120919.

T[n_, k_] := T[n, k] = If[3*n < k || k < 0, 0, If[n == 0 && k == 0, 1, If[k == 3*n, T[n, 0], T[n - 1, k + 1] + 3*T[n - 1, k] + 3*T[n - 1, k - 1] + T[n - 1, k - 2]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, 3 n}] // Flatten (* Jean-François Alcover, Jan 24 2018, translated from PARI *)

/* Generate Triangle by the Recurrence: */
{T(n,k)=if(3*n

/* Generate Triangle by the G.F.: */
{T(n,k)=local(A,F=(1+x)^3,d=3,G=x,H=1+x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0, 10, for(k=0, 3*n, print1(T(n, k), ", ")); print(""))

G.f.: A(x,y) = ( x*H(x) - y*H(x*y^3) )/( x*(1+y)^3 - y ), where H(x) satisfies: H(x) = G*H(x^4*G^3) and G(x) is g.f. of A001764: G(x) = 1 + x*G(x)^3.

A120914 Cascadence of (1+2x)^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,4,4] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.

Original entry on oeis.org

1, 4, 4, 4, 20, 36, 32, 16, 20, 116, 256, 288, 212, 144, 80, 116, 720, 1776, 2388, 2144, 1504, 1012, 784, 464, 720, 4656, 12372, 18800, 19632, 15604, 10848, 7648, 5712, 4736, 2880, 4656, 30996, 86912, 144320, 169332, 151792, 113456, 79696, 58176

Offset: 0

Paul D. Hanna, Jul 17 2006

More generally, the cascadence of polynomial F(x) of degree d, F(0)=1, is a triangle with d*n+1 terms in row n where the g.f. of the triangle, A(x,y), is given by: A(x,y) = ( x*H(x) - y*H(x*y^d) )/( x*F(y) - y ), where H(x) satisfies: H(x) = G*H(x*G^d)/x and G satisfies: G = x*F(G) so that G = series_reversion(x/F(x)); also, H(x) is the g.f. of column 0.

			Triangle begins:
1;
4, 4, 4;
20, 36, 32, 16, 20;
116, 256, 288, 212, 144, 80, 116;
720, 1776, 2388, 2144, 1504, 1012, 784, 464, 720;
4656, 12372, 18800, 19632, 15604, 10848, 7648, 5712, 4736, 2880, 4656;
Convolution of [1,4,4] with each row produces:
[1,4,4]*[1] = [1,4,4];
[1,4,4]*[4,4,4] = [4,20,36,32,16];
[1,4,4]*[20,36,32,16,20] = [20,116,256,288,212,144,80];
[1,4,4]*[116,256,288,212,144,80,116] =
[116,720,1776,2388,2144,1504,1012,784,464];
These convoluted rows, when concatenated, yield the sequence:
1,4,4, 4,20,36,32,16, 20,116,256,288,212,144,80, 116,720,1776,2388,...
which equals the concatenated rows of this original triangle:
1, 4,4,4, 20,36,32,16,20, 116,256,288,212,144,80,116, 720,1776,2388,...

Cf. A120915 (column 0), A120917 (central terms), A120918 (row sums), A000108 (Catalan); variants: A092683, A092686, A120894, A120898, A120919.

/* Generate Triangle by the Recurrence: */
{T(n,k)=if(2*n

/* Generate Triangle by the G.F.: */
{T(n,k)=local(A,F=1+4*x+4*x^2,d=2,G=x,H=1+x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

G.f.: A(x,y) = ( x*H(x) - y*H(x*y^2) )/( x*(1+2y)^2 - y ), where H(x) satisfies: H(x) = G*H(x*G^2)/x and G satisfies: G = x*(1 + 2G)^2 ; also, H(x) is the g.f. of column 0.

A092688 Row sums of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.

Original entry on oeis.org

1, 4, 16, 58, 204, 698, 2346, 7774, 25480, 82774, 266946, 855674, 2728702, 8663402, 27400862, 86376186, 271488444, 851099874, 2661967502, 8308462182, 25883429326, 80497346294, 249956869434, 775048966478, 2400067860090

Offset: 0

Paul D. Hanna, Mar 04 2004

nonn

Cf. A092683, A092686, A092687, A092689.

{T(n,k)=if(n<0 || k>n,0, if(n==0 && k==0,1, if(n==1 && k<=1,2, if(k==n,T(n,0), 2*T(n-1,k)+T(n-1,k+1)))))}
a(n)=sum(k=0,n,T(n,k))

{a(n)=local(A,F=1+2*x,d=1,G=x,H=1+2*x,S=ceil(log(n+1)/log(d+1))); for(i=0,n,G=x*subst(F,x,G+x*O(x^n)));for(i=0,S,H=subst(H,x,x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H,x,x*y^d +x*O(x^n)))/(x*subst(F,x,y)-y); sum(k=0,d*n,polcoeff(polcoeff(A,n,x),k,y))} \\ Paul D. Hanna, Jul 17 2006

G.f.: A(x) = H(x)*(1-x)/(1-3*x), where H(x) satisfies: H(x) = H(x^2/(1-2x))/(1-2x) and H(x) is the g.f. of A092687. - Paul D. Hanna, Jul 17 2006

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A092684 First column and main diagonal of triangle A092683, in which the convolution of each row with {1,1} produces a triangle that, when flattened, equals the flattened form of A092683.

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A092685 Row sums of triangle A092683, in which the convolution of each row with {1,1} produces a triangle that, when flattened, equals the flattened form of A092683.

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A092686 Triangle, read by rows, such that the convolution of each row with {1,2} produces a triangle which, when flattened, equals this flattened form of the original triangle.

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A120898 Cascadence of 1+2x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,2,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.

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A120894 Cascadence of 1+x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,1,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.

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A092687 First column and main diagonal of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.

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Mathematica

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A092689 Triangle, read by rows, such that the convolution of each row with {1,2} produces a triangle which, after the main diagonal is divided by 2 and the triangle is flattened, equals this flattened form of the original triangle.

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