A107877 -id:A107877 - cp's OEIS frontend

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.

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.

A107878 Column 2 of triangle A107876.

Original entry on oeis.org

1, 1, 3, 15, 106, 975, 11100, 151148, 2401365, 43681578, 896371205, 20504034645, 517705752096, 14310162565395, 430020328711305, 13963933247986995, 487456219774434795, 18209055555140970945, 724952705958984299025, 30650849492427960893946, 1371796147488157950190065

Offset: 0

Paul D. Hanna, Jun 04 2005

nonn

			G.f. = 1 + x + 3*x^2 + 15*x^3 + 106*x^4 + 975*x^5 + 11100*x^6 + 151148*x^7 + ...
1 = 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 + 11100*x^6*(1-x)^21 +...

Alois P. Heinz, Table of n, a(n) for n = 0..390

Cf. A000217, A107876, A107877, A107879, A209440 (variant).

a:= proc(n) option remember; `if`(n=0, 1, add(a(j-1)*
      (-1)^(n-j)*binomial(j*(j+1)/2, n-j+1), j=1..n))
    end:
seq(a(n), n=0..22);  # Alois P. Heinz, Jul 10 2022

a[ n_, k_: 0, j_: 1] := 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+1)*(k+2)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^((k+1)*(k+2)/2).
From Benedict W. J. Irwin, Nov 29 2016: (Start)
Conjecture: a(n) is given by a series of nested sums,
a(1) = Sum_{i=1..1} 1,
a(2) = Sum_{i=1..1} Sum_{j=1..i+2} 1,
a(3) = Sum_{i=1..1} Sum_{j=1..i+2} Sum_{k=1..j+3} 1,
a(4) = Sum_{i=1..1} Sum_{j=1..i+2} Sum_{k=1..j+3} Sum_{l=1..k+4} 1. (End)

A107880 Matrix square of triangle A107876; equals matrix product of triangles: A107876^2 = A107862^-1A107870 = A107867^-1A107873.

Original entry on oeis.org

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, 9693, 7197, 2496, 550, 91, 13, 2, 1, 111522, 82910, 28612, 6195, 982, 127, 15, 2, 1, 1528112, 1136923, 391189, 83837, 12977, 1596, 169, 17, 2, 1, 24372513, 18141867, 6230646, 1326923, 202494, 24206, 2424, 217, 19, 2, 1

Offset: 0

Paul D. Hanna, Jun 04 2005

Column 0 is A107881. Column 1 is A107882. Column 3 equals A107883. Column 2 equals SHIFT_LEFT(A107877), where A107877 is column 1 of A107876.

			G.f. for column 0:
1 = T(0,0)*(1-x)^2 + T(1,0)*x*(1-x)^2 + T(2,0)*x^2*(1-x)^3 + T(3,0)*x^3*(1-x)^5 + T(4,0)*x^4*(1-x)^8 + T(5,0)*x^5*(1-x)^12 +...
= 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 +...
G.f. for column 1:
1 = T(1,1)*(1-x)^2 + T(2,1)*x*(1-x)^3 + T(3,1)*x^2*(1-x)^5 + T(4,1)*x^3*(1-x)^8 + T(5,1)*x^4*(1-x)^12 + T(6,1)*x^5*(1-x)^17 +...
= 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 +...
Triangle T 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;
  9693,7197,2496,550,91,13,2,1;
  111522,82910,28612,6195,982,127,15,2,1;
  ...

Cf. A107862, A107870, A107873, A107867, A107876, A107884, A107887.

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, 2];
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))^(2+(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)^(2+(k+j)*(k+j-1)/2-k*(k-1)/2).

A177450 G.f.: Sum_{n>=0} a(n)*x^n/(1+x)^(n^2+n) = 1+x.

Original entry on oeis.org

1, 1, 2, 9, 70, 805, 12480, 245847, 5909338, 168310515, 5556486450, 209003251240, 8835266400450, 415094928861530, 21473740362658640, 1213683089969940075, 74446121738526773490, 4927385997649620215895, 350145746700442604768346

Offset: 0

Paul D. Hanna, May 09 2010

nonn

			1+x = 1 + 1*x/(1+x)^2 + 2*x^2/(1+x)^6 + 9*x^3/(1+x)^12 + 70*x^4/(1+x)^20 + 805*x^5/(1+x)^30 +...
1/(1-x) = 1 + 1*x*(1-x) + 2*x^2*(1-x)^4 + 9*x^3*(1-x)^9 + 70*x^4*(1-x)^16 + 805*x^5*(1-x)^25 +...
Also forms the final terms in rows of the triangle where row n+1 equals the partial sums of row n with the final term repeated 2(n+1) times, starting with a '1' in row 0, as illustrated by:
  1;
  1, 1;
  1, 2,  2,  2,  2;
  1, 3,  5,  7,  9,  9,   9,   9,   9,   9;
  1, 4,  9, 16, 25, 34,  43,  52,  61,  70,  70,  70,  70,  70,  70,  70,  70;
  1, 5, 14, 30, 55, 89, 132, 184, 245, 315, 385, 455, 525, 595, 665, 735, 805, 805, 805, 805, 805, 805, 805, 805, 805, 805;
  ...

Alois P. Heinz, Table of n, a(n) for n = 0..356

Cf. A107877, A177447, A177448, A177449, A209440.

a:= proc(n) option remember; `if`(n=0, 1, -add(a(j)
      *(-1)^(n-j)*binomial(1+ j^2, n-j), j=0..n-1))
    end:
seq(a(n), n=0..19);  # Alois P. Heinz, Jul 08 2022

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

G.f.: Sum_{n>=0} a(n)*x^n*(1-x)^(n^2) = 1/(1-x).
G.f.: Sum_{n>=0} a(n)*x^n*C(-x)^(n^2+2n) = 1/C(-x) where C(x) is the Catalan function of A000108.
a(n) = number of subpartitions of partition consisting of the first n square numbers starting with zero for n>0; e.g., a(4) = subp([0,1,4,9]) = 70. See A115728 for the definition of subpartitions.

A127496 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the following rule: start with a single '1' in row n=0; subsequently, row n+1 equals the partial sums of row n with the final term repeated n+1 more times at the end.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 3, 5, 7, 7, 7, 7, 1, 4, 9, 16, 23, 30, 37, 37, 37, 37, 37, 1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268, 1, 6, 20, 50, 103, 186, 306, 463, 657, 888, 1156, 1424, 1692, 1960, 2228, 2496, 2496, 2496, 2496, 2496, 2496, 2496

Offset: 0

Paul D. Hanna, Jan 16 2007

Last term in each row forms A107877, the number of subpartitions of the partition consisting of the triangular numbers.

			To obtain row 4 from row 3:
  [1, 3, _5, _7, _7, _7, __7];
take partial sums with final term '37' repeated 4 more times:
  [1, 4, _9, 16, 23, 30, _37, _37, _37, _37, _37].
To obtain row 5, take partial sums of row 4 with the final term '268' repeated 5 more times at the end:
  [1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268,268,268,268,268].
Triangle begins:
  1;
  1, 1;
  1, 2, 2, 2;
  1, 3, 5, 7, 7, 7, 7;
  1, 4, 9, 16, 23, 30, 37, 37, 37, 37, 37;
  1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268;
  1, 6, 20, 50, 103, 186, 306, 463, 657, 888, 1156, 1424, 1692, 1960, 2228, 2496, 2496, 2496, 2496, 2496, 2496, 2496;
Final term in rows forms A107877 which satisfies the g.f. 1/(1-x) = 1 + 1*x*(1-x) + 2*x^2*(1-x)^3 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A107877 (leading edge); diagonals: A127497, A127498.

Cf. A305605, A305601.

nxt[h_] :=Module[{c = Accumulate[h]}, Join[c, PadRight[{}, c[[2]], c[[-1]]]]]; Join[{1},Flatten[NestList[nxt,{1,1},5]]] (* Harvey P. Dale, Mar 10 2020 *)

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

A107879 Column 3 of triangle A107876.

Original entry on oeis.org

1, 1, 4, 26, 230, 2565, 34516, 544423, 9857583, 201664780, 4603336725, 116059191472, 3204682702923, 96226940232235, 3122975927539860, 108970956192622980, 4069312064491308140, 161969813446983961395, 6846708764857861662741, 306381118887919045527510

Offset: 0

Paul D. Hanna, Jun 04 2005

nonn

			1 = 1*(1-x)^1 + 1*x*(1-x)^4 + 4*x^2*(1-x)^8 + 26*x^3*(1-x)^13 + 230*x^4*(1-x)^19 + 2565*x^5*(1-x)^26 + 34516*x^6*(1-x)^34 +...

Alois P. Heinz, Table of n, a(n) for n = 0..389

Cf. A107876, A107877, A107878.

a:= proc(n) option remember; `if`(n=0, 1, add(a(j-1)*
      (-1)^(n-j)*binomial((j+1)*(j+2)/2-2, n-j+1), j=1..n))
    end:
seq(a(n), n=0..22);  # Alois P. Heinz, Jul 10 2022

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

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

A305601 G.f. A(x) satisfies: [x^k] (1+x)^(n(n+1)/2) A(x) = 0 for k = n(n-1)/2 + 1 through k = n(n+1)/2 for n >= 1.

Original entry on oeis.org

1, -1, 0, 2, -7, 21, -56, 125, -209, 154, 572, -3404, 11930, -35394, 99144, -274550, 757813, -2057867, 5392764, -13383194, 30829315, -63995085, 112304664, -133488537, -70063101, 1177164984, -5212740024, 17744267816, -53365305260, 149452146536, -401301906464, 1053638673004, -2741495192679, 7122821187935, -18514807074104, 48019064944442, -123571120120435, 313403811733896, -778001059367703, 1877334690759250, -4370190271978998

Offset: 0

Paul D. Hanna, Jun 14 2018

sign

The diagonals in the table of coefficients of x^k in (1+x)^n * A(x) forms the rows of irregular triangle A127496.

Equals the inverse binomial transform of A305605.

			G.f. A(x) = 1 - x + 2*x^3 - 7*x^4 + 21*x^5 - 56*x^6 + 125*x^7 - 209*x^8 + 154*x^9 + 572*x^10 - 3404*x^11 + 11930*x^12 - 35394*x^13 + 99144*x^14 - 274550*x^15 + 757813*x^16 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^n * A(x) begins:
n= 0: [1, -1,  0,   2,  -7,   21,  -56,  125, -209,  154, 572, -3404, ...];
n= 1: [1,  0, -1,   2,  -5,   14,  -35,   69,  -84,  -55, 726, -2832, ...];
n= 2: [1,  1, -1,   1,  -3,    9,  -21,   34,  -15, -139, 671, -2106, ...];
n= 3: [1,  2,  0,   0,  -2,    6,  -12,   13,   19, -154, 532, -1435, ...];
n= 4: [1,  3,  2,   0,  -2,    4,   -6,    1,   32, -135, 378,  -903, ...];
n= 5: [1,  4,  5,   2,  -2,    2,   -2,   -5,   33, -103, 243,  -525, ...];
n= 6: [1,  5,  9,   7,   0,    0,    0,   -7,   28,  -70, 140,  -282, ...];
n= 7: [1,  6, 14,  16,   7,    0,    0,   -7,   21,  -42,  70,  -142, ...];
n= 8: [1,  7, 20,  30,  23,    7,    0,   -7,   14,  -21,  28,   -72, ...];
n= 9: [1,  8, 27,  50,  53,   30,    7,   -7,    7,   -7,   7,   -44, ...];
n=10: [1,  9, 35,  77, 103,   83,   37,    0,    0,    0,   0,   -37, ...];
n=11: [1, 10, 44, 112, 180,  186,  120,   37,    0,    0,   0,   -37, ...];
n=12: [1, 11, 54, 156, 292,  366,  306,  157,   37,    0,   0,   -37, ...];
n=13: [1, 12, 65, 210, 448,  658,  672,  463,  194,   37,   0,   -37, ...];
n=14: [1, 13, 77, 275, 658, 1106, 1330, 1135,  657,  231,  37,   -37, ...];
n=15: [1, 14, 90, 352, 933, 1764, 2436, 2465, 1792,  888, 268, 0, 0, 0, 0, 0, -268, ...]; ...
which illustrates the occurrences of zeros in the table.
RELATED SEQUENCES.
Notice that [x^(n*(n-1)/2)] (1+x)^(n*(n+1)/2) * A(x) = A107877(n), which begins:
[1, 1, 2, 7, 37, 268, 2496, 28612, 391189, 6230646, 113521387, ...].
Also, note that the coefficient of x^(n*(n-1)/2) in (1+x)^(n*(n+1)/2) * A(x) yields -A107877(n).
Also, observe that [x^n] (1+x)^(2*n) * A(x) = A127497(n), which begins:
[1, 1, 2, 7, 23, 83, 306, 1135, 4257, 16095, 61222, 233956, ...].
The initial terms of the diagonals in the above table forms the rows of irregular triangle A127496:
1;
1, 1;
1, 2, 2, 2;
1, 3, 5, 7, 7, 7, 7;
1, 4, 9, 16, 23, 30, 37, 37, 37, 37, 37;
1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268;
1, 6, 20, 50, 103, 186, 306, 463, 657, 888, 1156, 1424, 1692, 1960, 2228, 2496, 2496, 2496, 2496, 2496, 2496, 2496; ...
in which row n equals the partial sums of the prior row with the final term repeated n more times at the end.

Paul D. Hanna, Table of n, a(n) for n = 0..2080

Cf. A107877, A127497, A127496, A305605, A305600.

/* Informal code to generate terms */
{A=[1, -1]; for(i=1, 465, A=concat(A, 0); m=floor(sqrt(2*#A-2) + 1/2); A[#A] = -polcoeff( (1+x +x*O(x^#A))^(m*(m+1)/2)*Ser(A), #A-1) ; print1(#A, ", ")); A}
/* Show that the definition is satisfied: */
for(n=1, floor(sqrt(2*#A) + 1/2), print1(n": "); for(k=n*(n-1)/2+1, n*(n+1)/2, print1(polcoeff( (1+x +x*O(x^#A))^(n*(n+1)/2)*Ser(A) , k), ", ")); print(""))

G.f. A(x) also satisfies:
(1) A(x) = 1/(1+x) - Sum_{n>=1} A107877(n) * (x/(1+x))^(n*(n+1)/2+1) / (1+x)^n.
(2) [x^(n*(n-1)/2)] (1+x)^(n*(n+1)/2) * A(x) = A107877(n) for n >= 0.
(3) [x^(n*(n+1)/2 + 1)] (1+x)^(n*(n+1)/2) * A(x) = -A107877(n) for n >= 0.
(4) [x^n] (1+x)^(2*n) * A(x) = A127497(n) = A127496(n,n) for n>=0.
(5) [x^k] (1+x)^(n+k) * A(x) = A127496(n,k) for k = 0..n*(n+1)/2, for n>=0.
(6) [x^n] (1+x)^n * A(x) = A305605(n) for n >= 0.

A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n(n+1)/2 + 1 through k = (n+1)(n+2)/2 for n >= 0.

Original entry on oeis.org

1, 0, -1, 0, -2, 2, 0, -7, 14, -7, 0, -37, 111, -111, 37, 0, -268, 1072, -1608, 1072, -268, 0, -2496, 12480, -24960, 24960, -12480, 2496, 0, -28612, 171672, -429180, 572240, -429180, 171672, -28612, 0, -391189, 2738323, -8214969, 13691615, -13691615, 8214969, -2738323, 391189, 0, -6230646, 49845168, -174458088, 348916176, -436145220, 348916176, -174458088, 49845168, -6230646, 0

Offset: 0

Paul D. Hanna, Jun 14 2018

sign

			G.f.: A(x) = 1 - x^2 - 2*x^4 + 2*x^5 - 7*x^7 + 14*x^8 - 7*x^9 - 37*x^11 + 111*x^12 - 111*x^13 + 37*x^14 - 268*x^16 + 1072*x^17 - 1608*x^18 + 1072*x^19 - 268*x^20 - 2496*x^22 + 12480*x^23 - 24960*x^24 + 24960*x^25 - 12480*x^26 + 2496*x^27 - 28612*x^29 + 171672*x^30 + ...
The table of coefficients of x^k in A(x) / (1-x)^n, for n >= 0, begins:
[1, 0, -1,  0,  -2,   2,   0,   -7,  14,    -7,    0, -37, 111,-111,   37, ...];
[1, 1,  0,  0,  -2,   0,   0,   -7,   7,     0,    0, -37,  74, -37,    0, ...];
[1, 2,  2,  2,   0,   0,   0,   -7,   0,     0,    0, -37,  37,   0,    0, ...];
[1, 3,  5,  7,   7,   7,   7,    0,   0,     0,    0, -37,   0,   0,    0, ...];
[1, 4,  9, 16,  23,  30,  37,   37,  37,    37,   37,   0,   0,   0,    0, ...];
[1, 5, 14, 30,  53,  83, 120,  157,  194,  231,  268, 268, 268, 268,  268, ...];
[1, 6, 20, 50, 103, 186, 306,  463,  657,  888, 1156,1424,1692,1960, 2228, ...];
[1, 7, 27, 77, 180, 366, 672, 1135, 1792, 2680, 3836,5260,6952,8912,11140, ...]; ...
illustrating the occurrence of zeros.
Note that the initial terms of the rows in the above table forms the rows of irregular triangle A127496.
TRIANGULAR FORM.
This sequence may be arranged into a triangle like so:
1,
0, -1,
0, -2, 2,
0, -7, 14, -7,
0, -37, 111, -111, 37,
0, -268, 1072, -1608, 1072, -268,
0, -2496, 12480, -24960, 24960, -12480, 2496,
0, -28612, 171672, -429180, 572240, -429180, 171672, -28612,
...
in which the g.f. of the rows equal -x * A107877(n) * (1-x)^(n-1) for n > 0.

Paul D. Hanna, Table of n, a(n) for n = 0..2556

Cf. A305601, A107877, A127496.

/* Informal code to generate terms */
{A=[1, 0]; for(i=1, 465, A=concat(A, 0); m=floor(sqrt(2*#A-2) + 1/2); A[#A] = -polcoeff( Ser(A)/(1-x +x*O(x^#A))^(m-1), #A-1) ; print1(#A, ", ")); A}
/* Show that the definition is satisfied: */
for(n=0, sqrtint(2*#A)-1, print1(n": "); for(k=n*(n+1)/2+1, (n+1)*(n+2)/2, print1(polcoeff( Ser(A)/(1-x +x*O(x^#A))^n , k), ", ")); print(""))

G.f. A(x) = Sum_{n>=0} a(n) * x^n satisfies:
(1) A(x) = 1 - x*Sum_{n>=1} A107877(n) * x^(n*(n+1)/2) * (1-x)^(n-1).
(2) [x^k] A(x) / (1-x)^n = 0 for k = n*(n+1)/2 + 1 through (n+1)*(n+2)/2, n >= 0.
(3) [x^k] A(x) / (1-x)^n = A107877(n) for k = n*(n-1)/2 through n*(n+1)/2, n >= 0.
(4) [x^k] A(x) / (1-x)^n = A127496(n,k) for k = 0..n*(n+1)/2 for n >= 0.
(5) [x^n] A(x) / (1-x)^n = A127497(n) for n >= 0.
FORMULAS INVOLVING TERMS.
a(n*(n+1)/2) = 0 for n >= 1.
a(n*(n-1)/2) = (-1)^n * A107877(n) for n >= 0.
a(n*(n+1)/2 + 1) = -A107877(n) for n >= 0.
a(n) = [x^n] (1+x)^n * G(x) where G(x) is the g.f. of A305601, which is the inverse binomial transform of this sequence.

A121434 Matrix inverse of triangle A098568, where A098568(n, k) = C( (k+1)*(k+2)/2 + n-k-1, n-k) for n>=k>=0.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 2, -3, 1, 0, -7, 12, -6, 1, 0, 37, -67, 39, -10, 1, 0, -268, 498, -311, 95, -15, 1, 0, 2496, -4701, 3045, -1015, 195, -21, 1, 0, -28612, 54298, -35901, 12560, -2675, 357, -28, 1, 0, 391189, -745734, 499157, -179717, 40635, -6097, 602, -36, 1, 0, -6230646, 11911221, -8034267, 2945010

Offset: 0

Paul D. Hanna, Aug 27 2006

			Triangle begins:
  1;
  0, 1;
  0, -1, 1;
  0, 2, -3, 1;
  0, -7, 12, -6, 1;
  0, 37, -67, 39, -10, 1;
  0, -268, 498, -311, 95, -15, 1;
  0, 2496, -4701, 3045, -1015, 195, -21, 1;
  0, -28612, 54298, -35901, 12560, -2675, 357, -28, 1;
  0, 391189, -745734, 499157, -179717, 40635, -6097, 602, -36, 1; ...

Cf. A098568, A107876; unsigned columns: A107877, A107887.

/* Matrix Inverse of A098568 */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((c-1)*(c-2)/2+r-2,r-c)))); return((M^-1)[n+1,k+1])

/* Obtain by G.F. */ T(n,k)=polcoeff(1-sum(j=0, n-k-1, T(j+k,k)*x^j/(1-x+x*O(x^n))^(j*(j+1)/2+j*k+k*(k+1)/2)), n-k)

(1) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2)](n,k); i.e., column k equals signed column k of A107876^(k*(k+1)/2).
G.f.s for column k:
(2) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2);
(3) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2).

A121435 Matrix inverse of triangle A122175, where A122175(n,k) = C( k*(k+1)/2 + n-k, n-k) for n>=k>=0.

Original entry on oeis.org

1, -1, 1, 1, -2, 1, -2, 5, -4, 1, 7, -19, 18, -7, 1, -37, 104, -106, 49, -11, 1, 268, -766, 809, -406, 110, -16, 1, -2496, 7197, -7746, 4060, -1210, 216, -22, 1, 28612, -82910, 90199, -48461, 15235, -3032, 385, -29, 1, -391189, 1136923, -1244891, 678874, -220352, 46732, -6699, 638, -37, 1

Offset: 0

Paul D. Hanna, Aug 27 2006

			Triangle begins:
1;
-1, 1;
1, -2, 1;
-2, 5, -4, 1;
7, -19, 18, -7, 1;
-37, 104, -106, 49, -11, 1;
268, -766, 809, -406, 110, -16, 1;
-2496, 7197, -7746, 4060, -1210, 216, -22, 1;
28612, -82910, 90199, -48461, 15235, -3032, 385, -29, 1;
-391189, 1136923, -1244891, 678874, -220352, 46732, -6699, 638, -37, 1; ...

Cf. A098568, A107876; unsigned columns: A107877, A107882.

/* Matrix Inverse of A122175 */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((c-1)*(c-2)/2+r-1,r-c)))); return((M^-1)[n+1,k+1])

/* Obtain by G.F. */ T(n,k)=polcoeff(1-sum(j=0, n-k-1, T(j+k,k)*x^j/(1-x+x*O(x^n))^(j*(j+1)/2+j*k+k*(k+1)/2+1)), n-k)

(1) T(n,k) = A121434(n-1,k) - A121434(n-1,k+1).
(2) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2 + 1)](n,k); i.e., column k equals signed column k of matrix power A107876^(k*(k+1)/2 + 1).
G.f.s for column k:
(3) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2 + 1);
(4) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2 + 1).

cp's OEIS Frontend

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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A107878 Column 2 of triangle A107876.

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A107880 Matrix square of triangle A107876; equals matrix product of triangles: A107876^2 = A107862^-1*A107870 = A107867^-1*A107873.

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A177450 G.f.: Sum_{n>=0} a(n)*x^n/(1+x)^(n^2+n) = 1+x.

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A127496 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the following rule: start with a single '1' in row n=0; subsequently, row n+1 equals the partial sums of row n with the final term repeated n+1 more times at the end.

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A107879 Column 3 of triangle A107876.

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A305601 G.f. A(x) satisfies: [x^k] (1+x)^(n*(n+1)/2) * A(x) = 0 for k = n*(n-1)/2 + 1 through k = n*(n+1)/2 for n >= 1.

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A107880 Matrix square of triangle A107876; equals matrix product of triangles: A107876^2 = A107862^-1A107870 = A107867^-1A107873.

A305601 G.f. A(x) satisfies: [x^k] (1+x)^(n(n+1)/2) A(x) = 0 for k = n(n-1)/2 + 1 through k = n(n+1)/2 for n >= 1.

A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n(n+1)/2 + 1 through k = (n+1)(n+2)/2 for n >= 0.