A118434 -id:A118434 - cp's OEIS frontend

A144181 INVERT transform of A118434, = row sums of triangle A144182.

1, 1, 3, 9, 11, 17, 35, 57, 91, 161, 275, 457, 779, 1329, 2243, 3801, 6459, 10945, 18547, 31465, 53355, 90449, 153379, 260089, 440987, 747745, 1267923, 2149897, 3645387, 6181233, 10481027, 17771801, 30134267, 51096321, 86639923, 146908457, 249101099

Offset: 0

Gary W. Adamson, Sep 13 2008

A118434 = row sums of the self-inverse triangle A118433 (a generator for the Rao Uppuluri-Carpenter numbers, A000587).

A144181 = row sums of triangle A144182.

			a(3) = 9 = sum of row 3 terms, triangle A144182: (4 + 2 + 0 + 3).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,2).

Cf. A118434.

Cf. A144182, A000587.

Vec((1+2*x^2+4*x^3)/(1-x-2*x^3) + O(x^40)) \\ Colin Barker, Aug 21 2016

Equals row sums of triangle A144182 and INVERT transform of A118434: (1, 0, 2, 4, -4, 0, -8, -16, 16, 0, 32,...).
From Colin Barker, Aug 21 2016: (Start)
a(n) = a(n-1)+2*a(n-3) for n>3.
G.f.: (1+2*x^2+4*x^3) / (1-x-2*x^3).
(End)

More terms from Alois P. Heinz, May 23 2015

A118433 Self-inverse triangle H, read by rows; a nontrivial matrix square-root of identity: H^2 = I, where H(n,k) = C(n,k)*(-1)^(floor((n+1)/2) - floor(k/2) + n - k) for n >= k >= 0.

Original entry on oeis.org

1, 1, -1, -1, 2, 1, -1, 3, 3, -1, 1, -4, -6, 4, 1, 1, -5, -10, 10, 5, -1, -1, 6, 15, -20, -15, 6, 1, -1, 7, 21, -35, -35, 21, 7, -1, 1, -8, -28, 56, 70, -56, -28, 8, 1, 1, -9, -36, 84, 126, -126, -84, 36, 9, -1, -1, 10, 45, -120, -210, 252, 210, -120, -45, 10, 1

Offset: 0

Paul D. Hanna, Apr 28 2006

There are an infinite number of integer square-roots of the identity matrix.

			Triangle H begins:
   1;
   1, -1;
  -1,  2,   1;
  -1,  3,   3,   -1;
   1, -4,  -6,    4,    1;
   1, -5, -10,   10,    5,   -1;
  -1,  6,  15,  -20,  -15,    6,   1;
  -1,  7,  21,  -35,  -35,   21,   7,   -1;
   1, -8, -28,   56,   70,  -56, -28,    8,   1;
   1, -9, -36,   84,  126, -126, -84,   36,   9, -1;
  -1, 10,  45, -120, -210,  252, 210, -120, -45, 10, 1; ...
G.f.s for columns:
k=0: (x + 1)/(1+x^2);
k=1: (x^2 + 2*x - 1)/(1+x^2)^2;
k=2: (-x^3 - 3*x^2 + 3*x + 1)/(1+x^2)^3;
k=3: (-x^4 - 4*x^3 + 6*x^2 + 4*x - 1)/(1+x^2)^4;
k=4: (x^5 + 5*x^4 - 10*x^3 - 10*x^2 + 5*x + 1)/(1+x^2)^5;
k=5: (x^6 + 6*x^5 - 15*x^4 - 20*x^3 + 15*x^2 + 6*x - 1)/(1+x^2)^6.
The g.f. of column k is thus:
G_k(x) = (Sum_{j=0..k+1} -H(k+1,j)*(-x)^(k+1-j))/(1+x^2)^(k+1).
The triangle formed from above polynomial numerators of column g.f.s is described by the e.g.f.: cos(x*y)*exp(-x) - sin(x*y)*exp(x).

Cf. A118434 (row sums), A118435 (H*[C^-1]*H).

H[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2]-Quotient[k, 2]+n-k);
Table[H[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)

{H(n,k)=binomial(n,k)*(-1)^((n+1)\2-k\2+n-k)}
for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))

/* Using E.G.F.: */
{H(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k));n!*polcoeff(polcoeff( cos(x)*exp(-x*y)+sin(x)*exp(x*y),n,X),k,Y)}
for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))

/* Using O.G.F.: */
{H(n,k)=polcoeff(polcoeff((1+x*(1-y)+x^2*(1+2*y-y^2)+x^3*(1+y+y^2+y^3))/(1+2*x^2*(1-y^2)+x^4*(1+y^2)^2+x*O(x^n)+y*O(y^k)),n,x),k,y)}
for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))

E.g.f.: A(x,y) = cos(x)*exp(-x*y) + sin(x)*exp(x*y).

O.g.f.: A(x,y) = (1 + x*(1-y) + x^2*(1+2*y-y^2) + x^3*(1+y+y^2+y^3)) / (1 + 2*x^2*(1-y^2) + x^4*(1+y^2)^2).

A118436 Column 0 of triangle A118435.

Original entry on oeis.org

1, 1, -3, -11, 25, 41, -43, 29, -335, -1199, 3117, 6469, -10295, -8839, -16123, -108691, 354145, 873121, -1721763, -2521451, 1476985, -6699319, 34182197, 103232189, -242017775, -451910159, 597551757, 130656229, 2465133865, 10513816601, -29729597083, -66349305331

Offset: 0

Paul D. Hanna, Apr 28 2006

sign

Binomial transform of A118434 = (1, 1, 3, 11, 25, 41, 43, -29, -335, -1199, ...). - Gary W. Adamson, Sep 19 2008

Index entries for linear recurrences with constant coefficients, signature (0, -5, 0, -19, 0, 25).

Cf. A118434, A118435 (triangle), A118437 (row sums).

LinearRecurrence[{0, -5, 0, -19, 0, 25}, {1, 1, -3, -11, 25, 41}, 32] (* Jean-François Alcover, Apr 08 2024 *)
CoefficientList[Series[(1+x+2x^2-6x^3+29x^4+5x^5)/((1-x^2)(1+6x^2+25x^4)),{x,0,40}],x] (* Harvey P. Dale, Oct 17 2024 *)

{a(n)=polcoeff((1+x+2*x^2-6*x^3+29*x^4+5*x^5)/(1-x^2)/(1+6*x^2+25*x^4+x*O(x^n)),n)}

G.f.: (1 + x + 2*x^2 - 6*x^3 + 29*x^4 + 5*x^5)/((1-x^2)*(1 + 6*x^2 + 25*x^4)).

A144182 Eigentriangle, row sums = A144181.

Original entry on oeis.org

1, 0, 1, 2, 0, 2, 4, 2, 0, 3, -4, 4, 2, 0, 9, 0, -4, 4, 6, 0, 11, -8, 0, -4, 12, 18, 0, 17, -16, -8, 0, -12, 36, 22, 0, 35, 16, -16, -8, 0, -36, 44, 34, 0, 57, 0, 16, -16, -24, 0, -44, 68, 70, 0, 91, 32, 0, 16, -48, -72, 0, -68, 140, 114, 0, 161

Offset: 0

Gary W. Adamson, Sep 13 2008

Row sums = A144181: (1, 1, 3, 9, 11, 17, 35,...).
Left border = A118434: (1, 0, 2, 4, -4, 0, -8,...); (i.e. row sums of the self-inverse triangle A118433).
Triangle A144183 = partial sums starting from the right of A144182.
Sum of n-th row terms = rightmost term of next row.

			First few rows of the triangle are:
1;
0, 1;
2, 0, 1;
4, 2, 0, 3;
-4, 4, 2, 0, 9;
0, -4, 4, 6, 0, 11;
-8, 0, -4, 12, 18, 0, 17;
-16, -8, 0, -12, 36, 22, 0, 35;
...
row 3 = (4, 2, 0, 3) = termwise products of (4, 2, 0, 1) and (1, 1, 1, 3) = (4*1, 2*1, 0*1, 1*3).

A144181, Cf. A118434, A144183

Triangle read by rows, T(n,k) = A118434(n-k)*A144181(k-1); where A144181(k-1) = A144181 shifted to (1, 1, 1, 3, 9, 11, 17, 35, 57, 91, 161,...).

A144183 Triangle read by rows, A144182 * A000012.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 9, 5, 3, 3, 11, 15, 11, 9, 9, 17, 17, 21, 17, 11, 11, 35, 43, 43, 47, 35, 17, 17, 57, 73, 81, 81, 93, 57, 35, 35, 91, 75, 91, 99, 99, 135, 91, 57, 57, 161, 161, 145, 161, 185, 185, 229, 161, 91, 91, 275, 243, 243, 227, 275, 347, 347, 415, 275, 161, 161

Offset: 0

Gary W. Adamson, Sep 13 2008

Left border = A144181: (1, 1, 3, 9, 11, 17, 35,...) = INVERT transform of A118434. Right border = A144181 shifted.

			First few rows of the triangle =
1;
1, 1;
3, 1, 1;
9, 5, 3, 3;
11, 15, 11, 9, 9;
17, 17, 21, 17, 11, 11;
35, 43, 43, 47, 35, 17, 17;
57, 73, 81, 81, 93, 57, 35, 35;
91, 75, 91, 99, 99, 135, 91, 57, 57;
...
Row 3 = (9, 5, 3, 3) = partial sums from the right of row 3, triangle A144182: (4, 2, 0, 3).

A144182, Cf. A144181, A118434

Triangle read by rows, A144182 * A000012; (equivalent to taking partial row sums

of A144182 starting from the right). A000012 = an infinite lower triangular matrix with all 1's and the rest zeros.

A144185 Triangle, row sums = a signed, shifted version of A000587, the Rao Uppuluri-Carpenter numbers.

Original entry on oeis.org

1, 1, -1, -1, 2, 0, -1, 3, 0, -1, 1, -4, 0, 4, 1, 1, -5, 0, 10, 5, -2, -1, 6, 0, -20, -15, 12, 9, -1, 7, 0, -35, -35, 42, 63, 9, 1, -8, 0, 56, 70, -112, -252, -72, 50, 1, -9, 0, 84, 126, -252, -756, -324, 450, 267, -1, 10, 0, -120, -210, 504, 1890, 1080, -2250, -2670

Offset: 0

Gary W. Adamson, Sep 13 2008

Right border = A000587, the Rao Uppuluri-Carpenter numbers, with different signs: (1, 1, 0, 1, 1, 2, 9, -9, 50, -267, -413, -2180,...).
Row sums = the same sequence shifted: (1, 0, 1, 1, 2, 9,...).
Let A = the self-inverse triangle, A118433. Shift the triangle down one place placing "1" at (0,0). Lim_{n->oo} A^n, = a signed version B of the Rao Uppuluri-Carpenter numbers (A000587), as follows: (1, 1, 0, 1, 1, 2, 9, -9, 50, -267, -413, -2180,...). This triangle = (A * (an infinite lower triangular matrix with B as the main diagonal and the rest zeros)). These operations are equivalent to (by rows), taking termwise products of A118433 row terms and B, the signed version of the Rao Uppuluri-Carpenter numbers.

			First few rows of the triangle =
  1;
  1, -1;
  -1, 2, 0;
  -1, 3, 0, -1,
  1, -4, 0, 4, 1;
  1, -5, 0, 10, 5, -2;
  -1, 6, 0, -20, -15, 12, 9;
  -1, 7, 0, -35, -35, 42, 63, 9;
  1, -8, 0, 56, 70, -112, -252, -72, 50;
  1, -9, 0, 84, 126, -252, -756, -324, 450, 267;
  -1, 10, 0, -120, -210, 504, 1890, 1080, -2250, -2670, -413;
...
Example: row 5 = (1, -5, 0, 10, 5, -2) = termwise products of row 5 of the self-inverse triangle A118433: (1, -5, -10, 10, 5, -1) and the first 6 terms of the "B" signed version of A000587 (the Rao Uppuluri-Carpenter numbers): (1, 1, 0, 1, 1, 2) = (1*1, -5*1, 0*0, 10*1, 5*1, -5*2).

Cf. A000587, A118433, A118434.

A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2x)^k)(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 7, 1, 1, 26, 30, 10, 1, 1, 57, 102, 58, 13, 1, 1, 120, 303, 256, 95, 16, 1, 1, 247, 825, 955, 515, 141, 19, 1, 1, 502, 2116, 3178, 2310, 906, 196, 22, 1, 1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1

Offset: 0

Peter Luschny, Feb 03 2020

The triangle is the matrix inverse of the Riordan square (see A321620) generated by (1 + x - sqrt(1 - 6*x + x^2))/(4*x) (see A172094), where we take the absolute value of the terms.

T(n,k) is the number of evil-avoiding (2413, 3214, 4132, and 4213 avoiding) permutations of length (n+2) that start with 1 and whose inverse has k descents. - Donghyun Kim, Aug 16 2021

			Triangle starts:
[0] [1]
[1] [1,    1]
[2] [1,    4,    1]
[3] [1,   11,    7,    1]
[4] [1,   26,   30,   10,    1]
[5] [1,   57,  102,   58,   13,    1]
[6] [1,  120,  303,  256,   95,   16,    1]
[7] [1,  247,  825,  955,  515,  141,   19,   1]
[8] [1,  502, 2116, 3178, 2310,  906,  196,  22,  1]
[9] [1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1]
...
Seen as a square array (the triangle is formed by descending antidiagonals):
1,  1,   1,    1,    1,     1,      1,      1,       1, ... [A000012]
1,  4,  11,   26,   57,   120,    247,    502,    1013, ... [A000295]
1,  7,  30,  102,  303,   825,   2116,   5200,   12381, ... [A045889]
1, 10,  58,  256,  955,  3178,   9740,  28064,   77093, ... [A055583]
1, 13,  95,  515, 2310,  9078,  32354, 106970,  333295, ...
1, 16, 141,  906, 4746, 21504,  87374, 326084, 1136799, ...
1, 19, 196, 1456, 8722, 44758, 204204, 849180, 3275931, ...

Donghyun Kim and Lauren Williams, Schubert polynomials and the inhomogeneous TASEP on a ring, arXiv:2102.00560 [math.CO], 2021.

Row sums A006012, alternating row sums A118434 with different signs, central column A091527.
T(n, 1) = A000295(n+1) for n >= 1, T(n, 2) = A045889(n-2) for n >= 2, T(n, 3) = A055583(n-3) for n >= 3.
Cf. A172094 (inverse up to sign).

gf := k -> 1/(((1-2*x)^k)*(1-x)^(k+1)): ser := k -> series(gf(k), x, 32):
# Prints the triangle:
seq(lprint(seq(coeff(ser(k), x, n-k), k=0..n)), n=0..6);
# Prints the square array:
seq(lprint(seq(coeff(ser(k), x, n), n=0..8)), k=0..6);

(* The function RiordanSquare is defined in A321620; returns the triangle as a lower triangular matrix. *)
M := RiordanSquare[(1 + x - Sqrt[1 - 6 x + x^2])/(4 x), 9];
Abs[#] & /@ Inverse[PadRight[M]]

cp's OEIS Frontend

A144181 INVERT transform of A118434, = row sums of triangle A144182.

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A118433 Self-inverse triangle H, read by rows; a nontrivial matrix square-root of identity: H^2 = I, where H(n,k) = C(n,k)*(-1)^(floor((n+1)/2) - floor(k/2) + n - k) for n >= k >= 0.

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A118436 Column 0 of triangle A118435.

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A144182 Eigentriangle, row sums = A144181.

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A144183 Triangle read by rows, A144182 * A000012.

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A144185 Triangle, row sums = a signed, shifted version of A000587, the Rao Uppuluri-Carpenter numbers.

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A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2*x)^k)*(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

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A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2x)^k)(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.