A110813 -id:A110813 - cp's OEIS frontend

A192575 Triangle T(n,0) = A040000(n), T(n,k)=0 (odd-numbered columns); T(n,k) = (-1)^(k/2)*A110813(n-k/2-1,k/2-1) (even-numbered columns, k>0).

1, 2, 0, 2, 0, -1, 2, 0, -3, 0, 2, 0, -5, 0, 1, 2, 0, -7, 0, 4, 0, 2, 0, -9, 0, 9, 0, -1, 2, 0, -11, 0, 16, 0, -5, 0, 2, 0, -13, 0, 25, 0, -14, 0, 1, 2, 0, -15, 0, 36, 0, -30, 0, 6, 0, 2, 0, -17, 0, 49, 0, -55, 0, 20, 0, -1

Offset: 0

Paul Curtz, Jul 04 2011

A zero-padded variant of A110813, which provides more information.

			1;
2 0;
2 0  -1;
2 0  -3 0;
2 0  -5 0  1;
2 0  -7 0  4 0;
2 0  -9 0  9 0  -1;
2 0 -11 0 16 0  -5 0;
2 0 -13 0 25 0 -14 0 1;
2 0 -15 0 36 0 -30 0 6 0;

Cf. A191662.

T(n,k) = T(n-1,k)-T(n-2,k-2), n>1.
T(n,2k+1)=0.
T(n,2k) = (-1)^k*binomial(n-k-1,k-1)*(2n-3k)/k , k>0. - R. J. Mathar, Aug 26 2011
T(n,0) = A040000(n).
sum_{k=0..n} T(n,k) = A057079(n).
sum_{k=0..n} |T(n,k)| = A000045(n+2). (See A129710).

A097207 Triangle read by rows: T(n,k) = binomial(n,k) + 2*binomial(n,k-1).

Original entry on oeis.org

1, 1, 3, 1, 4, 5, 1, 5, 9, 7, 1, 6, 14, 16, 9, 1, 7, 20, 30, 25, 11, 1, 8, 27, 50, 55, 36, 13, 1, 9, 35, 77, 105, 91, 49, 15, 1, 10, 44, 112, 182, 196, 140, 64, 17, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100, 21, 1, 13, 77, 275, 660

Offset: 0

N. J. A. Sloane, Sep 21 2004

			Triangle begins:
1
1 3
1 4 5
1 5 9 7
1 6 14 16 9

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
H. W. Gould, Power sum identities for arbitrary symmetric arrays, SIAM J. Appl. Math., 17 (1969), 307-316.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A029637, A110813 (row-reversed).

a097207 n k = a097207_tabl !! n !! k
a097207_row n = a097207_tabl !! n
a097207_tabl = map init $ tail a029635_tabl
-- Reinhard Zumkeller, Mar 12 2012

T[n_, k_] := Binomial[n, k] + 2Binomial[n, k - 1]; Flatten[ Table[ T[n, k], {n, 0, 10}, {k, 0, n}]] (* Robert G. Wilson v, Sep 21 2004 *)

T(n,k) = A029635(n+1,k), 0 <= k <= n. - Reinhard Zumkeller, Mar 12 2012

More terms from Robert G. Wilson v, Sep 21 2004

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 15, 11, 5, 1, 31, 26, 16, 6, 1, 63, 57, 42, 22, 7, 1, 127, 120, 99, 64, 29, 8, 1, 255, 247, 219, 163, 93, 37, 9, 1, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1, 2047, 2036, 1981, 1816, 1486, 1024, 562, 232, 67

Offset: 0

Gary W. Adamson, Mar 19 2005

This array (A104709) is the mirror of the fission, A054143, of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence (q(n,x): n >= 0) given by q(n,x) = x^n + x^(n-1) + ... + x + 1.  See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
The elements of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A110813(n,k) assuming the same offset in both triangles. - R. J. Mathar, Mar 15 2013
From Paul Curtz, Jun 12 2019: (Start)
Numerators of the triangle [Curtz, page 15, triangle (E)]:
     1/2;
     3/4,  1/4;
     7/8,  4/8,    1/8;
   15/16, 11/16,  5/16,  1/16;
   31/32, 26/31, 16/32,  6/32, 1/32;
   63/64, 57/64, 42/64, 22/64, 7/64, 1/64;
   ...
Denominators - Numerators: Triangle A054143.
  1;
  1, 3;
  1, 4, 7;
  1, 5, 11, 15;
  ...
(E) is a transform which accelerates the convergence of series.
For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have
  1*(1/2) = 1/2,
  1*(3/4) - (1/2)*(1/4) = 5/8,
  1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,
  1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,
  ...
This is A068566/A068565. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   3,  1;
   7,  4,  1;
  15, 11,  5,  1;
  31, 26, 16,  6,  1;
  63, 57, 42, 22,  7,  1;
  ...

Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.

A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

z = 10;
p[n_, x_] := (x + 1)^n;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A054143 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
(* Clark Kimberling, Aug 07 2011 *)

Begin with A055248 as a triangle, delete leftmost column.
The Riordan array factors as (1/(1-2*x), x)*(1/(1-x), x/(1-x)) - the sequence array for 2^n times Pascal's triangle. - Paul Barry, Aug 05 2005
T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - Paul Barry, Jan 12 2006
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
Working with an offset of 0, we have exp(x) * (e.g.f. for row n) = (e.g.f. for diagonal n). For example, for n = 3 we have exp(x)*(15 + 11*x + 5*x^2/2! + x^3/3!) = 15 + 26*x + 42*x^2/2! + 64*x^3/3! + 93*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), x/(1 - x)). - Peter Bala, Dec 21 2014
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: A(x,y) = Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - 3*x - x*y + 2*x^2 + 2*x^2*y) = 1/((1 - 2*x)*(1 - x*(y+1))).
The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).
Let B(x,y) be the bivariate o.g.f. of triangular array A054143. Because A054143 is the mirror image of the current array, we have A(x,y) = B(x*y, 1/y) and B(x,y) = A(x*y, 1/y). This makes it easy to identify lower diagonals of the array.
For example, if we want to identify the second lower diagonal of the array (i.e., 7, 11, 16, 22, ...), we take the 2nd derivative of B(x,y) with respect to y, set y = 0, and divide by 2!. (Note that columns in A054143 start at k = 0.) We get the g.f. x^2*(7 - 10*x + 4*x^2)/(1 - x)^3.
It is then easy to derive that T(n,n-2) = A000124(n+1) = (n+1)*(n+2)/2 + 1 for n >= 2 (by ignoring the first three terms of A000124). Of course, in the current case, it is much easier to use the formula for T(n,k) to find T(n,n-2). (End)
T(n,0) = 2^(n+1) - 1 for n >= 0; T(n,k) = T(n-1,k) + T(n-1,k-1) for 1 <= k <= n. - Peter Bala, Jan 30 2023
T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - Bernard Schott, Feb 22 2023

Name edited and offset changed by Petros Hadjicostas, Jun 04 2020

A110814 Inverse of a triangle of pyramidal numbers.

Original entry on oeis.org

1, -3, 1, 7, -4, 1, -15, 11, -5, 1, 31, -26, 16, -6, 1, -63, 57, -42, 22, -7, 1, 127, -120, 99, -64, 29, -8, 1, -255, 247, -219, 163, -93, 37, -9, 1, 511, -502, 466, -382, 256, -130, 46, -10, 1, -1023, 1013, -968, 848, -638, 386, -176, 56, -11, 1, 2047, -2036, 1981, -1816, 1486, -1024, 562, -232, 67, -12, 1, -4095, 4083

Offset: 0

Paul Barry, Aug 05 2005

Inverse of A110813. Array factors as (1/(1+2x),x)*(1/(1+x),x/(1+x)). Row sums are (-2)^n. Diagonal sums are (-1)^n*A008466(n+2). Signed version of A104709.

			Rows begin
    1;
   -3,   1;
    7,  -4,   1;
  -15,  11,  -5,   1;
   31, -26,  16,  -6,   1;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

A110814_row := proc(n) add((-1)^k*add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1); coeffs(sort(%)) end; seq(print(A110814_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

T[n_, k_] := Sum[(-2)^(n - j)*Binomial[j, k]*(-1)^(j - k), {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-2)^(n-j)*(-1)^(j-k)* binomial(j,k)), ", "))) \\ G. C. Greubel, Oct 19 2017

Number triangle T(n, k) = Sum_{j=0..n} (-2)^(n-j)*binomial(j, k)*(-1)^(j-k).
Riordan array (1/(1+3x+2x^2), x/(1+x)).
T(n,k) = -3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) + 2*T(n-2,k-1), T(0,0)=1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Original entry on oeis.org

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

A156319 Triangle by columns: (1, 2, 0, 0, 0, ...) in every column.

Original entry on oeis.org

1, 2, 1, 0, 2, 1, 0, 0, 2, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1

Offset: 1

Gary W. Adamson, Feb 07 2009

Binomial transform of the triangle = A110813.
Eigensequence of the triangle = A001045
Inverse = a triangle with (1, -2, 4, -8, 16, ...) in every column.
Triangle T(n,k), 0 <= k <= n, given by [2,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 08 2009

			First few rows of the triangle:
  1;
  2, 1;
  0, 2, 1;
  0, 0, 2, 1;
  0, 0, 0, 2, 1;
  0, 0, 0, 0, 2, 1;
  0, 0, 0, 0, 0, 2, 1;
  0, 0, 0, 0, 0, 0, 2, 1;
  0, 0, 0, 0, 0, 0, 0, 2, 1;
...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A001045, A097806, A110813.

T:= function(n,k)
    if k=n then return 1;
    elif k=n-1 then return 2;
    else return 0;
    fi;
  end;
Flat(List([1..15], n-> List([1..n], k-> T(n,k) ))); # G. C. Greubel, Sep 20 2019

T:= func< n,k | k eq n select 1 else k eq n-1 select 2 else 0 >;
[T(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Sep 20 2019

T:= proc (n) option remember;
if k=n then 1
elif k=n-1 then 2
else 0 fi;
end proc;
seq(seq(T(n,k), k=1..n), n = 1..15); # G. C. Greubel, Sep 20 2019

Table[If[k==n,1, If[k==n-1, 2, 0]], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Sep 20 2019 *)
Join[{1},Flatten[Table[PadRight[{2,1},n,0],{n,3,20}]]] (* Harvey P. Dale, Feb 28 2022 *)

T(n,k) = if(k==n, 1, if(k==n-1, 2, 0)); \\ G. C. Greubel, Sep 20 2019

def T(n,k):
    if (k==n): return 1
    elif (k==n-1): return 2
    else: return 0
[[T(n,k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Sep 20 2019

Triangle read by rows, T(n,k) = 1 if n=k, 2 if k = n-1, 0 otherwise.
By columns, (1, 2, 0, 0, 0, ...) in every column.
T(n,k) = A097806(n,k)*2^(n-k). - Philippe Deléham, Feb 08 2009
G.f.: (1+2*x)*x*y/(1-x*y). - R. J. Mathar, Aug 12 2015

More terms added by G. C. Greubel, Sep 20 2019

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1

Offset: 0

Philippe Deléham, Feb 22 2012

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Gheorghe Coserea, Rows n = 0..200, flattened

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013

seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016

P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

A209757 Triangle of coefficients of polynomials v(n,x) jointly generated with A013609; see the Formula section.

Original entry on oeis.org

1, 3, 2, 5, 8, 4, 7, 18, 20, 8, 9, 32, 56, 48, 16, 11, 50, 120, 160, 112, 32, 13, 72, 220, 400, 432, 256, 64, 15, 98, 364, 840, 1232, 1120, 576, 128, 17, 128, 560, 1568, 2912, 3584, 2816, 1280, 256, 19, 162, 816, 2688, 6048, 9408, 9984, 6912, 2816, 512

Offset: 1

Clark Kimberling, Mar 23 2012

For a discussion and guide to related arrays, see A208510.

Subtriangle of the triangle given by (1, 2, -2, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 24 2012

			First five rows:
  1;
  3,  2;
  5,  8,  4;
  7, 18, 20,  8;
  9, 32, 56, 48, 16;
First three polynomials v(n,x):
  1
  3 + 2x
  5 + 8x + 4x^2.
From _Philippe Deléham_, Mar 24 2012: (Start)
(1, 2, -2, 1, 0, 0, ...) DELTA (0, 2, 0, 0, 0, ...) begins:
  1;
  1,  0;
  3,  2,  0;
  5,  8,  4,  0;
  7, 18, 20,  8,  0;
  9, 32, 56, 48, 16,  0; (End)

Cf. A013609, A208510, A110813.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A013609 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209757 *)

u(n,x) = x*u(n-1,x) + x*v(n-1,x) + 1,
v(n,x) = (x+1)*u(n-1,x) + (x+1)*v(n-1,x) + 1,
where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Mar 24 2012: (Start)
As DELTA-triangle T(n,k) with 0 <= k <= n:
G.f.: (1 - x - 2*y*x + 2*x^2 + 2*x^2*y)/(1 - 2*x - 2*y*x + x^2 + 2*y*x^2).
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-2), T(0,0) = T(1,0) = 1, T(1,1) = T(2,2) = 0, T(2,0) = 3, T(2,1) = 2 and T(n,k) = 0 if k < 0 or if k > n.
T(n,k) = 2^k*binomial(n-1,k)*(2*n-k-1)/(k+1). (End)
From Peter Bala, Dec 21 2014: (Start)
Following remarks assume an offset of 0.
T(n,k) = 2^k * A110813(n,k).
Riordan array ((1+x)/(1-x)^2, 2*x/(1-x)).
exp(2*x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(2*x)*(7 + 18*x + 20*x^2/2! + 8*x^3/3!) = 7 + 32*x + 120*x^2/2! + 400*x^3/3! + 1232*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), 2*x/(1-x)). (End)

A029655 Numbers in the (2,1)-Pascal triangle A029653 that are different from 2.

Original entry on oeis.org

1, 1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385

Offset: 0

Mohammad K. Azarian

			1;
1;
3,1;
5,4,1;
7,9,5,1;
9,16,14,6,1;

Cf. A110813.

More terms from James Sellers

A111526 Number triangle T(n,k)=C((n+k)/2,k)(n+1)(1+(-1)^(n-k))/(2(k+1)); T(n,k)=(-1)^((n-k)/2)*A053120(n+1,k+1)/2^k; Riordan array ((1+x^2)/(1-x^2)^2,x/(1-x^2)).

Original entry on oeis.org

1, 0, 1, 3, 0, 1, 0, 4, 0, 1, 5, 0, 5, 0, 1, 0, 9, 0, 6, 0, 1, 7, 0, 14, 0, 7, 0, 1, 0, 16, 0, 20, 0, 8, 0, 1, 9, 0, 30, 0, 27, 0, 9, 0, 1, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1, 13, 0, 91, 0, 182, 0, 156, 0, 65, 0, 13, 0, 1, 0

Offset: 0

Paul Barry, Aug 05 2005

A scaled Chebyshev triangle.

Row sums are A001350(n+1). Diagonal sums are A033484, with interpolated zeros. Inverse is A111527.

			Triangle starts
1;
0,1;
3,0,1;
0,4,0,1;
5,0,5,0,1;
0,9,0,6,0,1;
7,0,14,0,7,0,1;

Cf. A110813.

cp's OEIS Frontend

A192575 Triangle T(n,0) = A040000(n), T(n,k)=0 (odd-numbered columns); T(n,k) = (-1)^(k/2)*A110813(n-k/2-1,k/2-1) (even-numbered columns, k>0).

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A097207 Triangle read by rows: T(n,k) = binomial(n,k) + 2*binomial(n,k-1).

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A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)*binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)*(1-2*x)), x/(1-x)).

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Maple

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A110814 Inverse of a triangle of pyramidal numbers.

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A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

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A156319 Triangle by columns: (1, 2, 0, 0, 0, ...) in every column.

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GAP

Magma

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Sage

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A207974 Triangle related to A152198.

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A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).