A107880 -id:A107880 - cp's OEIS frontend

A107881 Column 0 of triangle A107880.

1, 2, 3, 7, 26, 141, 1034, 9693, 111522, 1528112, 24372513, 444461496, 9135985760, 209223309377, 5286985099830, 146225910921737, 4395973641599418, 142794782804641965, 4985911547339049132, 186284207595651780174

Offset: 0

Paul D. Hanna, Jun 04 2005

nonn

			1 = 1*(1-x)^2 + 2*x*(1-x)^2 + 3*x^2*(1-x)^3 +
7*x^3*(1-x)^5 + 26*x^4*(1-x)^8 + 141*x^5*(1-x)^12 +...

Cf. A107880, A107882, A107883.

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^(2+k*(k-1)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^(2 + k*(k-1)/2).

A107882 Column 1 of triangle A107880.

Original entry on oeis.org

1, 2, 5, 19, 104, 766, 7197, 82910, 1136923, 18141867, 330940109, 6803936050, 155839142185, 3938383850350, 108934529005948, 3275059508166297, 106388204134734785, 3714826559490125850, 138796913898027894261

Offset: 0

Paul D. Hanna, Jun 04 2005

nonn

			G.f. = 1 + 2*x + 5*x^2 + 19*x^3 + 104*x^4 + 766*x^5 + 7197*x^6 + 82910*x^7 + ...
1 = 1*(1-x)^2 + 2*x*(1-x)^3 + 5*x^2*(1-x)^5 +
19*x^3*(1-x)^8 + 104*x^4*(1-x)^12 + 766*x^5*(1-x)^17 +...

Cf. A107880, A107881, A107883.

a[ n_, k_: 2, j_: 0] := If[ n < 1, Boole[n >= 0], a[ n, k, j] = Sum[ a[ n - 1, i, j + 1], {i, k + j}]]; (* Michael Somos, Nov 26 2016 *)

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^(2+k*(k+1)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^(2 + k*(k+1)/2).
From Benedict W. J. Irwin, Nov 26 2016: (Start)
Conjecture: a(n) can be expressed with a series of nested sums,
a(2) = Sum_{i=1..2} i+1,
a(3) = Sum_{i=1..2}Sum_{j=1..i+1} j+2,
a(4) = Sum_{i=1..2}Sum_{j=1..i+1}Sum_{k=1..j+2} k+3,
a(5) = Sum_{i=1..2}Sum_{j=1..i+1}Sum_{k=1..j+2}Sum_{l=1..k+3} l+4. (End)

A107883 Column 3 of triangle A107880.

Original entry on oeis.org

1, 2, 9, 61, 550, 6195, 83837, 1326923, 24078588, 493309850, 11271757335, 284379843234, 7856320956198, 235986714918110, 7660827258318780, 267365373971139600, 9985779421324740445, 397508459931685273305

Offset: 0

Paul D. Hanna, Jun 04 2005

nonn

			1 = 1*(1-x)^2 + 2*x*(1-x)^5 + 9*x^2*(1-x)^9 +
61*x^3*(1-x)^14 + 550*x^4*(1-x)^20 + 6195*x^5*(1-x)^27 +...

Cf. A107880, A107881, A107882.

a[ n_, k_: 0, j_: 2] := If[n < 1, Boole[n >= 0], a[ n, k, j] = Sum[ a[ n - 1, i, j + 1], {i, k + j}]]; (* Michael Somos, Nov 26 2016 *)

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^((k+2)*(k+3)/2-1)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^((k+2)*(k+3)/2 - 1).

A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 7, 7, 3, 1, 1, 37, 37, 15, 4, 1, 1, 268, 268, 106, 26, 5, 1, 1, 2496, 2496, 975, 230, 40, 6, 1, 1, 28612, 28612, 11100, 2565, 425, 57, 7, 1, 1, 391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1, 6230646, 6230646, 2401365, 544423

Offset: 0

Paul D. Hanna, Jun 04 2005

Remarkably, T equals the product of these triangular matrices: T = A107862^-1*A107867 = A107867^-1*A107870 = A107870^-1*A107873; reversing the order of these products yields triangle A101479.
Column m of T^k is the number of subpartitions of the initial terms of the sequence (k-1)+n(m-1)+n(n-1)/2 (ignoring 0's above the diagonal). E.g., column 4 of T^3 is 1,3,15,106,975,.... The sequence above is 2,5,9,14,20,.... subp([]) = 1, subp([2]) = 3, subp([2,5]) = 15, subp([2,5,9]) = 106, etc. The matrix product of T^(k-1) * T computes the number of such subpartitions by looking at the first part index where the subpartition is maxed - for [2,5,9,14,20] the third term (9 maxed) has subp([1,4]) for the first two values (not maxed), times subp([5,11]) for the last two values (possibly maxed). - Franklin T. Adams-Watters, Jun 26 2006
T(n,k) is the number of Dyck paths whose sequence of ascent lengths is exactly k,k+1,...,n, for example the T(4,3) = 3 paths are UUUdUUUUd^6, UUUddUUUUd^5 and UUUdddUUUUd^4. - David Scambler, May 30 2012

			G.f. for column 1:
1 = T(1,1)*(1-x)^1 + T(2,1)*x*(1-x)^2 + T(3,1)*x^2*(1-x)^4 + T(4,1)*x^3*(1-x)^7 + T(5,1)*x^4*(1-x)^11 + T(6,1)*x^5*(1-x)^16 +...
= 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 +...
G.f. for column 2:
1 = T(2,2)*(1-x)^1 + T(3,2)*x*(1-x)^3 + T(4,2)*x^2*(1-x)^6 + T(5,2)*x^3*(1-x)^10 + T(6,2)*x^4*(1-x)^15 + T(7,2)*x^5*(1-x)^21 +...
= 1*(1-x)^1 + 1*x*(1-x)^3 + 3*x^2*(1-x)^6 + 15*x^3*(1-x)^10 + 106*x^4*(1-x)^15 + 975*x^5*(1-x)^21 +...
Triangle T begins:
       1;
       1,      1;
       1,      1,      1;
       2,      2,      1,     1;
       7,      7,      3,     1,    1;
      37,     37,     15,     4,    1,   1;
     268,    268,    106,    26,    5,   1,  1;
    2496,   2496,    975,   230,   40,   6,  1, 1;
   28612,  28612,  11100,  2565,  425,  57,  7, 1, 1;
  391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1; ...
where column 1 of T = SHIFT_LEFT(column 0 of T).
Matrix square T^2 begins:
     1;
     2,   1;
     3,   2,   1;
     7,   5,   2,  1;
    26,  19,   7,  2,  1;
   141, 104,  37,  9,  2, 1;
  1034, 766, 268, 61, 11, 2, 1; ...
Compare column 2 of T^2 with column 1 of T.
Matrix inverse begins:
   1;
  -1,    1;
   0,   -1,   1;
   0,   -1,  -1,   1;
   0,   -3,  -2,  -1,  1;
   0,  -15,  -9,  -3, -1,  1;
   0, -106, -61, -18, -4, -1, 1; ...
Compare column 1 of T^-1 with column 2 of T and
compare column 2 of T^-1 with column 3 of T^2.

Alois P. Heinz, Rows n = 0..50, flattened

Cf. A107862, A107865, A107867, A107870, A107877 (column 1), A107878 (column 2), A107879 (column 3), A107880 (matrix square), A107884 (matrix cube), A107889 (matrix inverse).
Cf. A115728, A115729, A101479 (dual triangle).
T(2n,n) gives A300954.

max = 10;
A107862 = Table[Binomial[If[nA107867 = Table[Binomial[If[nA107862].A107867;
Table[t[[n, k]], {n, 1, max+1}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012, after first comment, fixed by Vaclav Kotesovec, Jun 13 2018 *)

{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(1+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))

/* Print the Triangular Matrix to the Power p: */
{T(n,k,p)=polcoeff(1- sum(j=0,n-k-1,T(j+k,k,p)*x^j*(1-x+x*O(x^n))^(j*(j-1)/2+j*k+p)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k,1),", ")); print(""))

G.f. for column k of T^m, the m-th matrix power of this triangle T:
(1) 1 = Sum_{j>=0} T(k+j, k) * x^j * (1-x)^(1+(k+j)*(k+j-1)/2-k*(k-1)/2) for m=1.
(2) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j*(1-x)^(m+(k+j)*(k+j-1)/2-k*(k-1)/2) for all m and k>=0.
(3) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j / C(x)^(m-j+(k+j)*(k+j-1)/2-k*(k-1)/2) where C(x)=2/(1+sqrt(1-4*x)) is g.f. for A000108 (Catalan numbers).
Matrix inverse of this triangle T satisfies:
(4) [T^-1](n,k) = -[T^k](n,k+1) for n>k>=0.

A107884 Matrix cube of triangle A107876; equals the product of triangular matrices: A107876^3 = A107862^-1*A107873.

Original entry on oeis.org

1, 3, 1, 6, 3, 1, 16, 9, 3, 1, 63, 37, 12, 3, 1, 351, 210, 67, 15, 3, 1, 2609, 1575, 498, 106, 18, 3, 1, 24636, 14943, 4701, 975, 154, 21, 3, 1, 284631, 173109, 54298, 11100, 1689, 211, 24, 3, 1, 3909926, 2381814, 745734, 151148, 22518, 2688, 277, 27, 3, 1

Offset: 0

Paul D. Hanna, Jun 04 2005

Column 0 is A107885.
Column 1 is A107886.
Column 2 equals A107887.
Column 3 equals SHIFT_LEFT(A107878), where A107878 is column 2 of A107876.
Column 4 equals A107888.

			G.f. for column 0:
1 = T(0,0)*(1-x)^3 + T(1,0)*x*(1-x)^3 + T(2,0)*x^2*(1-x)^4 + T(3,0)*x^3*(1-x)^6 + T(4,0)*x^4*(1-x)^9 + T(5,0)*x^5*(1-x)^13 + ...
  = 1*(1-x)^3 + 3*x*(1-x)^3 + 6*x^2*(1-x)^4 + 16*x^3*(1-x)^6 + 63*x^4*(1-x)^9 + 351*x^5*(1-x)^13 + ...
G.f. for column 1:
1 = T(1,1)*(1-x)^3 + T(2,1)*x*(1-x)^4 + T(3,1)*x^2*(1-x)^6 + T(4,1)*x^3*(1-x)^9 + T(5,1)*x^4*(1-x)^13 + T(6,1)*x^5*(1-x)^18 + ...
  = 1*(1-x)^3 + 3*x*(1-x)^4 + 9*x^2*(1-x)^6 + 37*x^3*(1-x)^9 + 210*x^4*(1-x)^13 + 1575*x^5*(1-x)^18 + ...
Triangle begins:
       1;
       3,      1;
       6,      3,     1;
      16,      9,     3,     1;
      63,     37,    12,     3,    1;
     351,    210,    67,    15,    3,   1;
    2609,   1575,   498,   106,   18,   3,  1;
   24636,  14943,  4701,   975,  154,  21,  3, 1;
  284631, 173109, 54298, 11100, 1689, 211, 24, 3, 1;
  ...

Cf. A107862, A107870, A107873, A107867, A107876, A107880, A107884, A107885, A107886, A107887, A107888.

max = 10;
A107862 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n - k], n - k], {n, 0, max}, {k, 0, max}];
A107867 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n-k+1], n - k], {n, 0, max}, {k, 0, max}];
T = MatrixPower[Inverse[A107862].A107867, 3];
Table[T[[n+1, k+1]], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 31 2024 *)

{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(3+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}

G.f. for column k: 1 = Sum_{j>=0} T(k+j, k)*x^j*(1-x)^(3 + (k+j)*(k+j-1)/2 - k*(k-1)/2).

A107889 Triangular matrix T, read by rows, that satisfies: [T^-k](n,k) = -T(n,k-1) for n >= k > 0, or, equivalently, (column k of T^-k) = -SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored. Also, matrix inverse of triangle A107876.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 0, -1, -1, 1, 0, -3, -2, -1, 1, 0, -15, -9, -3, -1, 1, 0, -106, -61, -18, -4, -1, 1, 0, -975, -550, -154, -30, -5, -1, 1, 0, -11100, -6195, -1689, -310, -45, -6, -1, 1, 0, -151148, -83837, -22518, -4005, -545, -63, -7, -1, 1, 0, -2401365, -1326923, -353211, -61686, -8105, -875, -84, -8, -1, 1

Offset: 0

Paul D. Hanna, Jun 05 2005

SHIFT_LEFT(column 1) = -A107878.
SHIFT_LEFT(column 2) = -A107883.
SHIFT_LEFT(column 3) = -A107888.

			G.f. for column 1:
1 = T(1,1)*(1-x)^-1 + T(2,1)*x*(1-x)^0 + T(3,1)*x^2*(1-x)^2 + T(4,1)*x^3*(1-x)^5 + T(5,1)*x^4*(1-x)^9 + T(6,1)*x^5*(1-x)^14 + ...
  = 1*(1-x)^-1 - 1*x*(1-x)^0 - 1*x^2*(1-x)^2 - 3*x^3*(1-x)^5 - 15*x^4*(1-x)^9 - 106*x^5*(1-x)^14 - 975*x^6*(1-x)^20 + ...
G.f. for column 2:
1 = T(2,2)*(1-x)^-1 + T(3,2)*x*(1-x)^1 + T(4,2)*x^2*(1-x)^4 + T(5,2)*x^3*(1-x)^8 + T(6,2)*x^4*(1-x)^13 + T(7,2)*x^5*(1-x)^19 + ...
  = 1*(1-x)^-1 - 1*x*(1-x)^1 - 2*x^2*(1-x)^4 - 9*x^3*(1-x)^8 - 61*x^4*(1-x)^13 - 550*x^5*(1-x)^19 - 6195*x^6*(1-x)^26 + ...
Triangle begins:
   1;
  -1,      1;
   0,     -1,     1;
   0,     -1,    -1,     1;
   0,     -3,    -2,    -1,    1;
   0,    -15,    -9,    -3,   -1,   1;
   0,   -106,   -61,   -18,   -4,  -1,  1;
   0,   -975,  -550,  -154,  -30,  -5, -1,  1;
   0, -11100, -6195, -1689, -310, -45, -6, -1, 1;
  ...

Cf. A107876, A107880, A107884.

max = 10;
A107862 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n - k], n - k], {n, 0, max}, {k, 0, max}];
A107867 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n-k+1], n - k], {n, 0, max}, {k, 0, max}];
T = Inverse[Inverse[A107862].A107867];
Table[T[[n + 1, k + 1]], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 31 2024 *)

{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(-1+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}

G.f. for column k: 1 = Sum_{j>=0} T(k+j, k)*x^j*(1-x)^(-1 + (k+j)*(k+j-1)/2 - k*(k-1)/2).

cp's OEIS Frontend

A107881 Column 0 of triangle A107880.

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A107882 Column 1 of triangle A107880.

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A107883 Column 3 of triangle A107880.

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A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

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A107884 Matrix cube of triangle A107876; equals the product of triangular matrices: A107876^3 = A107862^-1*A107873.

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A107889 Triangular matrix T, read by rows, that satisfies: [T^-k](n,k) = -T(n,k-1) for n >= k > 0, or, equivalently, (column k of T^-k) = -SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored. Also, matrix inverse of triangle A107876.

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