A118433 -id:A118433 - cp's OEIS frontend

A118434 Row sums of self-inverse triangle A118433.

1, 0, 2, 4, -4, 0, -8, -16, 16, 0, 32, 64, -64, 0, -128, -256, 256, 0, 512, 1024, -1024, 0, -2048, -4096, 4096, 0, 8192, 16384, -16384, 0, -32768, -65536, 65536, 0, 131072, 262144, -262144, 0, -524288, -1048576, 1048576, 0, 2097152, 4194304, -4194304

Offset: 0

Paul D. Hanna, Apr 28 2006

Index entries for linear recurrences with constant coefficients, signature (0,0,0,-4).

Cf. A118433.

A118434 := proc(n) I*(1-I)^n-I*(1+I)^n+(-1-I)^n+(-1+I)^n ; expand(%/2) ;end proc:
# R. J. Mathar, Jan 18 2011

a[n_] := 2^(Floor[(n+1)/2]-3)*(-2*Mod[n, 8] + Mod[n+2 , 8] - Mod[n+3, 8] + 2*Mod[n+4, 8] - Mod[n+6, 8] + Mod[n+7, 8]); Table[a[n], {n, 0, 44}] (* Jean-François Alcover, May 23 2013 *)

{a(n)=polcoeff((1+2*x^2+4*x^3)/(1+4*x^4+x*O(x^n)),n)}

/* E.g.f.: */
{a(n)=local(x=X+X*O(X^n));n!*polcoeff(cos(x)*exp(-x)+sin(x)*exp(x),n,X)}

O.g.f.: A(x) = (1+2*x^2+4*x^3)/(1+4*x^4).
E.g.f.: A(x) = cos(x)*exp(-x) + sin(x)*exp(x).
2*a(n) = i*(1-i)^n-i*(1+i)^n + (-1-i)^n + (-1+i)^n with i=sqrt(-1). - R. J. Mathar, Jan 18 2011

A118435 Triangle T, read by rows, equal to the matrix product T = H[C^-1]H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

Original entry on oeis.org

1, 1, 1, -3, 2, 1, -11, 15, 3, 1, 25, -44, -18, 4, 1, 41, -115, -110, 50, 5, 1, -43, 246, 375, -220, -45, 6, 1, 29, 315, 861, -805, -385, 105, 7, 1, -335, 232, -1204, 2296, 1750, -616, -84, 8, 1, -1199, 3033, 1044, 3780, 5166, -2898, -924, 180, 9, 1

Offset: 0

Paul D. Hanna, Apr 28 2006

The matrix inverse of H*[C^-1]*H is H*C*H = A118438, where H^2 = I (identity). The matrix log, log(T) = A118441, is a matrix square root of a triangular matrix with a single diagonal (two rows down from the main diagonal).

			Triangle begins:
  1;
  1, 1;
  -3, 2, 1;
  -11, 15, 3, 1;
  25,-44,-18, 4, 1;
  41,-115,-110, 50, 5, 1;
  -43, 246, 375,-220,-45, 6, 1;
  29, 315, 861,-805,-385, 105, 7, 1;
  -335, 232,-1204, 2296, 1750,-616,-84, 8, 1;
  -1199, 3033, 1044, 3780, 5166,-2898,-924, 180, 9, 1;
  ...
The matrix log, log(T) = A118441, starts:
  0;
  1, 0;
  -4, 2, 0;
  -12, 12, 3, 0;
  32,-48,-24, 4, 0;
  80,-160,-120, 40, 5, 0;
  ...
where matrix square, log(T)^2, is a single diagonal:
  0;
  0,0;
  2,0,0;
  0,6,0,0;
  0,0,12,0,0;
  0,0,0,20,0,0;
  ...

Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (self-inverse H).

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

{T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2- (c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))));(M*C^-1*M)[n+1,k+1]}

Since T + T^-1 = C + C^-1, then [T^-1](n,k) = (1+(-1)^(n-k))*C(n,k) - T(n,k) is a formula for the matrix inverse T^-1 = A118438.

A118438 Triangle T, read by rows, equal to the matrix product T = HCH, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

Original entry on oeis.org

1, -1, 1, 5, -2, 1, 11, -9, -3, 1, -23, 44, 30, -4, 1, -41, 125, 110, -30, -5, 1, 45, -246, -345, 220, 75, -6, 1, -29, -301, -861, 875, 385, -63, -7, 1, 337, -232, 1260, -2296, -1610, 616, 140, -8, 1, 1199, -3015, -1044, -3612, -5166, 3150, 924, -108, -9, 1

Offset: 0

Paul D. Hanna, Apr 28 2006

The matrix inverse of H*C*H is H*[C^-1]*H = A118435, where H^2 = I (identity). The matrix log, log(T) = -A118441, is a matrix square root of a triangular matrix with a single diagonal (two rows down from the main diagonal).

			Triangle begins:
1;
-1, 1;
5,-2, 1;
11,-9,-3, 1;
-23, 44, 30,-4, 1;
-41, 125, 110,-30,-5, 1;
45,-246,-345, 220, 75,-6, 1;
-29,-301,-861, 875, 385,-63,-7, 1;
337,-232, 1260,-2296,-1610, 616, 140,-8, 1;
1199,-3015,-1044,-3612,-5166, 3150, 924,-108,-9, 1; ...

Cf. A118439 (column 0), A118440 (row sums), A118435 (matrix inverse), A118441 (-matrix log); A118433 (self-inverse H).

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

{T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2- (c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))));(M*C*M)[n+1,k+1]}

Since T + T^-1 = C + C^-1, then [T^-1](n,k) = (1+(-1)^(n-k))*C(n,k) - T(n,k) is a formula for the matrix inverse T^-1 = A118435.

A144220 Triangle read by rows, A118433 (the self-inverse triangle) * A000012^(-1).

Original entry on oeis.org

1, 2, -1, -3, 1, 1, -4, 0, 4, -1, 5, 2, -10, 3, 1, 6, 5, -20, 5, 6, -1, -7, -9, 35, 5, -21, 5, 1, -8, -14, 56, 0, -56, 14, 8, -1, 9, 20, -84, -14, 126, -28, -36, 7, 1, 10, 27, -120, -42, 252, -42, -120, 27, 10, -1, -11, -35, 165, 90, -462, 42, 330, -75, -55, 9, 1

Offset: 0

Gary W. Adamson, Sep 14 2008

Row sums = (1, 1, -1, -1, 1, 1, ...).

			First few rows of the triangle:
   1;
   2,  -1;
   3,   1,    1;
  -4,   0,    4,  -1
   5,   2,  -10,   3,   1;
   6,   5,  -20,   5,   6,  -1;
  -7,  -9,   35,  -5, -21,   5,    1;
  -8, -14,   56,   0, -56,  14,    8, -1;
   9,  20,  -84, -14, 126, -28,  -36,  7,  1;
  10,  27, -120, -42, 252, -42, -120, 27, 10, -1;
  ...

Cf. A118433.

A118433 * A000012^(-1), where A000012^(-1) is an infinite lower triangular matrix with 1's in the main diagonal and (-1)'s in the subdiagonal, with the rest zeros.

A144221 Triangle read by rows, A118433 * A007318^(-1) * A000012.

Original entry on oeis.org

1, -1, -1, 1, 1, -1, -1, 5, -1, 1, 5, -11, 1, 1, 1, 9, -11, -11, 9, -1, -1, -9, -9, 51, -29, 1, 1, -1, -1, -57, 111, -29, -29, 13, -1, 1, -15, 113, -111, -111, 169, -55, 1, 1, 1, -31, 113, 113, -559, 449, -55, -55, 17, -1

Offset: 0

Gary W. Adamson, Sep 14 2008

Row sums = (1, 0, 1, 2, -3, -4, 5, 6, -7, -8,...).

			First few rows of the triangle =
1;
1, -1;
-1, 1, 1;
-1, -1, 5, -1;
1, 5, -11, 1, 1;
1, 9, -11, -11, 9, -1;
-1, -9, -9, 51, -29, 1, 1
-1, -1, -57, 111, -29, -29, 13, -1;
1, -15, 113, -111, -111, 169, -55, 1, 1;
...

A118433

Triangle read by rows, A118433 * A007318^(-1) * A000012.

A118433 = the self-inverse triangle. A007318^(-1) = the inverse of Pascal's triangle and A000012 = an infinite lower triangular matrix with all 1's.

A118441 Triangle L, read by rows, equal to the matrix log of A118435, with the property that L^2 consists of a single diagonal (two rows down from the main diagonal).

Original entry on oeis.org

0, 1, 0, -4, 2, 0, -12, 12, 3, 0, 32, -48, -24, 4, 0, 80, -160, -120, 40, 5, 0, -192, 480, 480, -240, -60, 6, 0, -448, 1344, 1680, -1120, -420, 84, 7, 0, 1024, -3584, -5376, 4480, 2240, -672, -112, 8, 0, 2304, -9216, -16128, 16128, 10080, -4032, -1008, 144, 9, 0

Offset: 0

Paul D. Hanna, Apr 28 2006

L = log(A118435) = log(H*[C^-1]*H], where C=Pascal's triangle and H=A118433 where H^2 = I (identity matrix).

			The matrix log, L = log(H*[C^-1]*H], begins:
     0;
     1,     0;
    -4,     2,      0;
   -12,    12,      3,     0;
    32,   -48,    -24,     4,     0;
    80,  -160,   -120,    40,     5,     0;
  -192,   480,    480,  -240,   -60,     6,     0;
  -448,  1344,   1680, -1120,  -420,    84,     7,   0;
  1024, -3584,  -5376,  4480,  2240,  -672,  -112,   8,  0;
  2304, -9216, -16128, 16128, 10080, -4032, -1008, 144,  9,  0;
  ...
The matrix square, L^2, is a single diagonal:
  0;
  0, 0;
  2, 0,  0;
  0, 6,  0,  0;
  0, 0, 12,  0,  0;
  0, 0,  0, 20,  0,  0;
  0, 0,  0,  0, 30,  0,  0;
  ...
From _Peter Luschny_, Apr 23 2020: (Start)
In unsigned form and without the main diagonal, as computed by the Maple script:
  [0], [0]
  [1], [1]
  [2], [4,   2]
  [3], [12,  12,   3]
  [4], [32,  48,   24,   4]
  [5], [80,  160,  120,  40,   5]
  [6], [192, 480,  480,  240,  60,  6]
  [7], [448, 1344, 1680, 1120, 420, 84, 7] (End)

Cf. A118435 (exp(L)), A118442 (column 0), A118443 (row sums), A027471 (unsigned row sums); A118433 (self-inverse triangle), A001815 (column 1?), A001789 (third of column 2?).

# Generalized Worpitzky transform of the harmonic numbers.
CL := p -> PolynomialTools:-CoefficientList(expand(p), x):
H := n -> add(1/k, k=1..n):
Trow := proc(n) local k,v; if n=0 then return [0] fi;
add(add((-1)^(n-v)*binomial(k,v)*H(k)*(-x+v-1)^n, v=0..k), k=0..n); CL(%) end:
for n from 0 to 7 do Trow(n) od; # Peter Luschny, Apr 23 2020

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

/* From definition of L as matrix log of H*C^-1*H: */
{L(n,k)=local(H=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2-(c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))),N=(H*C^-1*H)); Log=sum(p=1,n+1,-(N^0-N)^p/p);Log[n+1,k+1]}
for(n=0, 10, for(k=0, n, print1(L(n, k), ", ")); print(""))

/* The matrix power L^m is given by: */
{L(n,k,m)=if(m%2==0,if(n==k+m,n!/k!*2^(n-k-m)/(n-k-m)!), if(n>=k+m,n!/k!*2^(n-k-m)/(n-k-m)!*(-1)^(m\2+(n+1)\2-k\2+n-k)))}
for(n=0, 10, for(k=0, n, print1(L(n, k,1), ", ")); print(""))

For even exponents of L, L^(2m) is a single diagonal:
if n == k+2m, then [L^(2m)](n,k) = n!/k!*2^(n-k-2m)/(n-k-2m)!; else if n != k+2m: [L^(2m)](n,k) = 0.
For odd exponents of L:
if n >= k+2m+1, then [L^(2m+1)](n,k) = n!/k!*2^(n-k-2m-1)/(n-k-2m-1)!*(-1)^(m+[(n+1)/2]-[k/2]+n-k); else if n < k+2m+1: [L^(2m)](n,k) = 0.
Unsigned row sums equals A027471(n+1) = n*3^(n-1).

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

Original entry on oeis.org

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

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,...).

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.

cp's OEIS Frontend

A118434 Row sums of self-inverse triangle A118433.

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A118435 Triangle T, read by rows, equal to the matrix product T = H*[C^-1]*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

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A118438 Triangle T, read by rows, equal to the matrix product T = H*C*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

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A144220 Triangle read by rows, A118433 (the self-inverse triangle) * A000012^(-1).

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A144221 Triangle read by rows, A118433 * A007318^(-1) * A000012.

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A118441 Triangle L, read by rows, equal to the matrix log of A118435, with the property that L^2 consists of a single diagonal (two rows down from the main diagonal).

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

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A144182 Eigentriangle, row sums = A144181.

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A118435 Triangle T, read by rows, equal to the matrix product T = H[C^-1]H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

A118438 Triangle T, read by rows, equal to the matrix product T = HCH, where H is the self-inverse triangle A118433 and C is Pascal's triangle.