A101482 -id:A101482 - 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

A101481 Column 0 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, 1, 3, 19, 191, 2646, 46737, 1003150, 25330125, 735180292, 24103027865, 880627477269, 35469795883964, 1561107221731851, 74528004538789830, 3835467005270307663, 211648845813188932595, 12465477924609075602136

Offset: 0

Paul D. Hanna, Jan 21 2005

nonn

From Gerhard Kirchner, Apr 20 2017: (Start)
Also: Let U(n,i,k), k<=i<=n, be a triangular matrix with elements selected as "0" or "1" such that the partial sum of the first m rows is s(m)<=m for 1<=mA101481(n+1) is the number of possible selections.
U(n,i,k)  has r(n) = n*(n+1)/2 elements. There are c(n) = binomial(r(n), n) ways to select n elements, some of which, however, are forbidden, see example. This yields the estimation a(n+1) < c(n).
Derivation of the recurrence:
s(m-1) <= m-1, say s(m-1)= j with 0 <= j <= m-1. Let f(m-1, j) be the number of associated submatrices. In order to determine f(m, k), k>=j, we have to distribute "1" k-j times among the m elements of row number m. There are binomial(m, k-j) ways to do that. The distribution must be repeated for each j. The recurrence describes this procedure. (End)

			G.f. = 1 + x + x^2 + 3*x^3 + 19*x^4 + 191*x^5 + 2646*x^6 + 46737*x^7 + ...
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, ...;
[1], 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, ...;
[3, 9], 19, 34, 55, 83, 119, 164, 219, 285, 363, 454, ...;
[19, 53, 108], 191, 310, 474, 693, 978, 1341, 1795, 2354, ...;
[191, 501, 975, 1668], 2646, 3987, 5782, 8136, 11169, 15017, ...;
[2646, 6633, 12415, 20551, 31720], 46737, 66570, 92358, ...; ...;
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>=1;
this sequence then forms the first column of the resultant table.
Note: column k of the above table equals column 0 of matrix power T^(k+1) where T=A101479, for k>=0.
From _Gerhard Kirchner_, Apr 20 2017: (Start)
n=3: 0         0        1       forbidden: 1
     0 0       1 0      0 1                1 1
     1 1 1     0 1 1    1 0 1              0 0 0
     s(2)=0    s(2)=1   s(2)=2             s(2)=3>2
c(3) = binomial(6,3) = 20. There is only one forbidden matrix.
Thus: a(n+1) = a(4) = 19
Using the recurrence:
f(2,0) = f(1,0) = 1
f(2,1) = 2*f(1,0) + f(1,1) = 3
a(3) = f(2,2) = f(1,0) + 2*f(1,1) = 3
a(4) = f(3,3) = f(2,0) + 3*f(2,1) + 3*f(2,2) = 19
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 0..300

Cf. A101479, A101482, A101483, A226775.

a[ n_, k_: 1] := a[n, k] = If[ n < 2, Boole[n >= 0], Sum[a[n - 1, i], {i, n + k - 2}]]; (* Michael Somos, Nov 29 2016 *)
f[m_, k_] := f[m, k] = If[(m == 0 && k == 0) || (m == 0 && k == 1) || (m == 1 && k == 0) || (m == 1 && k == 1), 1, Sum[f[m-1, j]*Binomial[m, k-j], {j, 0, m-1}]]; Flatten[{1, Table[f[n-1, n-1], {n, 1, 20}]}] (* Vaclav Kotesovec, Apr 20 2017 after Gerhard Kirchner *)

{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-2))[i-1,j]);));A=B);return(A[n+1,1])}

{a(n, k=1) = if( n<2, n>=0, sum(i=1, n+k-2, a(n-1, i)))}; /* Michael Somos, Nov 29 2016 */

From Benedict W. J. Irwin, Nov 29 2016: (Start)
Conjecture: a(n) is described by a series of nested sums,
a(2) = Sum_{i=1..1} 1,
a(3) = Sum_{i=1..1+1}Sum_{j=1..i} 1,
a(4) = Sum_{i=1..1+2}Sum_{j=1..i+1}Sum_{k=1..j} 1,
a(5) = Sum_{i=1..1+3}Sum_{j=1..i+2}Sum_{k=1..j+1}Sum_{l=1..k} 1,
(End)
From Gerhard Kirchner, Apr 20 2017: (Start)
Recurrence: f(m,k) = Sum_{j=0..m-1} f(m-1,j)*binomial(m,k-j) with f(1,0) = f(1,1)= 1. a(n+1) = f(n,n). (End)
a(n) ~ c * exp(n) * n^(n-3/2) / 2^n, where c = (2 + LambertW(-2*exp(-2))) / (exp(2) * sqrt(2*Pi)) = 0.08604131405842589281435... - Vaclav Kotesovec, Apr 20 2017, updated Dec 03 2017

A136730 Square array, read by antidiagonals, where T(n,k) = T(n,k-1) + T(n-1,k+n-1) for n>0, k>0, such that T(n,0) = T(n-1,n-1) for n>0 with T(0,k)=1 for k>=0.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 9, 5, 3, 1, 70, 23, 9, 4, 1, 795, 175, 43, 14, 5, 1, 11961, 1935, 324, 70, 20, 6, 1, 224504, 28432, 3510, 527, 105, 27, 7, 1, 5051866, 523290, 50528, 5624, 795, 149, 35, 8, 1, 132523155, 11587072, 913377, 79553, 8396, 1140, 203, 44, 9, 1, 3969912160

Offset: 0

Paul D. Hanna, Jan 19 2008

			The square array begins:
1, 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...];
(1), 2, 3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,...;
(2,5), 9, 14,20,27,35,44,54,65,77,90,104,119,135,152,170,189,209,230,..;
(9,23,43), 70, 105,149,203,268,345,435,539,658,793,945,1115,1304,1513,.;
(70,175,324,527), 795, 1140,1575,2114,2772,3565,4510,5625,6929,8442,...;
(795,1935,3510,5624,8396), 11961, 16471,22096,29025,37467,47652,59832,..;
(11961,28432,50528,79553,117020,164672), 224504, 298786,390087,501300,..;
(224504,523290,913377,1414677,2050345,2847156,3835910), 5051866, 6535206,.;
(5051866,11587072,19918602,30410985,43486800,59633775,79412515,103464895),.;
where the rows are generated as follows.
Start row 0 with all 1's; from then on,
remove the first n terms (shown in parenthesis) from row n
and then take partial sums to yield row n+1.
Note that the main diagonal forms column 0 and equals A101482:
[1,1,2,9,70,795,11961,224504,5051866,132523155,3969912160,...]
which equals column 1 of triangle A101479:
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; ...
where row n equals row (n-1) of T^(n-1) with appended '1'.

Cf. A101479; columns: A101482, A136731, A136732; variants: A136733, A136737.

{T(n,k)=if(k<0,0,if(n==0,1,T(n,k-1) + T(n-1,k+n-1)))}

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.

A136731 Column 1 of square array A136730.

Original entry on oeis.org

1, 2, 5, 23, 175, 1935, 28432, 523290, 11587072, 299942890, 8886126540, 296438370794, 10993731095695, 448604373236731, 19971257117211555, 963142501803505255, 50015707804752012825, 2782336529985704607295

Offset: 0

Paul D. Hanna, Jan 19 2008

nonn

Cf. A136730; A101482, A136732.

A136732 Column 2 of square array A136730.

Original entry on oeis.org

1, 3, 9, 43, 324, 3510, 50528, 913377, 19918602, 509040779, 14918466255, 493115508126, 18143982947900, 735340631600946, 32542320101428755, 1561227609244084205, 80700623119099359600, 4470904603875492038790

Offset: 0

Paul D. Hanna, Jan 19 2008

nonn

Cf. A136730; A101482, A136731.

A165999 Triangle read by rows: T(0,0) = 1, T(n,k) = T(n-1,k-1) + T(n-1,k) for n > 0, 0 < k <= trinv(n), where trinv(n) = floor((1+sqrt(1+8*n))/2), and entries outside triangle are 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 2, 1, 4, 5, 1, 5, 9, 1, 6, 14, 9, 1, 7, 20, 23, 1, 8, 27, 43, 1, 9, 35, 70, 1, 10, 44, 105, 70, 1, 11, 54, 149, 175, 1, 12, 65, 203, 324, 1, 13, 77, 268, 527, 1, 14, 90, 345, 795, 1, 15, 104, 435, 1140, 795, 1, 16, 119, 539, 1575, 1935, 1, 17, 135, 658, 2114

Offset: 0

Gerald McGarvey, Oct 03 2009

There are trinv(n) terms in row n (see A002024). Related to A136730.

			Triangle begins: [1] [1, 1] [1, 2] [1, 3, 2] [1, 4, 5] [1, 5, 9] [1, 6, 14, 9] [1, 7, 20, 23] [1, 8, 27, 43] [1, 9, 35, 70] [1, 10, 44, 105, 70] [1, 11, 54, 149, 175] [1, 12, 65, 203, 324] [1, 13, 77, 268, 527] [1, 14, 90, 345, 795] [1, 15, 104, 435, 1140, 795]

A101482 (diagonal T(A000217(n), n))

trinv(n) = floor((1+sqrt(1+8*n))/2); f(n) = trinv(n-1); s=19;M=matrix(s,s);for(n=1,s,M[n,1]=1); for(n=2,s,for(k=2,f(n),M[n,k]=M[n-1,k-1]+M[n-1,k])); for(n=1,s,for(k=1,f(n),print1(M[n,k],", ")))

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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A101481 Column 0 of triangular matrix T=A101479, in which row n equals row (n-1) of T^(n-1) followed by '1'.

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A136730 Square array, read by antidiagonals, where T(n,k) = T(n,k-1) + T(n-1,k+n-1) for n>0, k>0, such that T(n,0) = T(n-1,n-1) for n>0 with T(0,k)=1 for k>=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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A136731 Column 1 of square array A136730.

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A136732 Column 2 of square array A136730.

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A165999 Triangle read by rows: T(0,0) = 1, T(n,k) = T(n-1,k-1) + T(n-1,k) for n > 0, 0 < k <= trinv(n), where trinv(n) = floor((1+sqrt(1+8*n))/2), and entries outside triangle are 0.

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