A145661 -id:A145661 - cp's OEIS frontend

A119258 Triangle read by rows: T(n,0) = T(n,n) = 1 and for 0

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 17, 15, 1, 1, 9, 31, 49, 31, 1, 1, 11, 49, 111, 129, 63, 1, 1, 13, 71, 209, 351, 321, 127, 1, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023, 1

Offset: 0

Reinhard Zumkeller, May 11 2006

From Richard M. Green, Jul 26 2011: (Start)
T(n,n-k) is the (k-1)-st Betti number of the subcomplex of the n-dimensional half cube obtained by deleting the interiors of all half-cube shaped faces of dimension at least k.
T(n,n-k) is the (k-2)-nd Betti number of the complement of the k-equal real hyperplane arrangement in R^n.
T(n,n-k) gives a lower bound for the complexity of the problem of determining, given n real numbers, whether some k of them are equal.
T(n,n-k) is the number of nodes used by the Kronrod-Patterson-Smolyak cubature formula in numerical analysis. (End)

			Triangle begins as:
  1;
  1, 1;
  1, 3,  1;
  1, 5,  7,  1;
  1, 7, 17, 15,  1;
  1, 9, 31, 49, 31, 1;

Reinhard Zumkeller, Rows n=0..120 of triangle, flattened
A. Björner and V. Welker, The homology of "k-equal" manifolds and related partition lattices, Adv. Math., 110 (1995), 277-313.
Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO], 2023.
R. M. Green, Homology representations arising from the half cube, Adv. Math., 222 (2009), 216-239.
R. M. Green, Homology representations arising from the half cube, II, J. Combin. Theory Ser. A, 117 (2010), 1037-1048.
R. M. Green, Homology representations arising from the half cube, II, arXiv:0812.1208 [math.RT], 2008.
R. M. Green and Jacob T. Harper, Morse matchings on polytopes, arXiv preprint arXiv:1107.4993 [math.GT], 2011.
OEIS Wiki, Sequence of the Day for November 3.
K. Petras, On the Smolyak cubature error for analytic functions, Adv. Comput. Math., 12 (2000), 71-93.
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, arXiv:1107.1063 [math.CO], 2010.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A119259, A007318, A007051, A005408, A056220, A027608, A055580, A000337, A000225, A112857 (row reverse).

A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

a119258 n k = a119258_tabl !! n !! k
a119258_row n = a119258_tabl !! n
a119258_list = concat a119258_tabl
a119258_tabl = iterate (\row -> zipWith (+)
   ([0] ++ init row ++ [0]) $ zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Reinhard Zumkeller, Nov 15 2011

function T(n,k)
  if k eq 0 or k eq n then return 1;
  else return 2*T(n-1,k-1) + T(n-1,k);
  end if;
  return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 18 2019

# Case m = 2 of the more general:
A119258 := (n,k,m) -> (1-m)^(-n+k)-m^(k+1)*pochhammer(n-k, k+1) *hypergeom([1,n+1],[k+2],m)/(k+1)!;
seq(seq(round(evalf(A119258(n,k,2))),k=0..n), n=0..10); # Peter Luschny, Jul 25 2014

T[n_, k_] := Binomial[n, k] Hypergeometric2F1[-k, n-k, n-k+1, -1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 10 2017 *)

T(n,k) = if(k==0 || k==n, 1, 2*T(n-1, k-1) + T(n-1,k) ); \\ G. C. Greubel, Nov 18 2019

@CachedFunction
def T(n, k):
    if (k==0 or k==n): return 1
    else: return 2*T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 18 2019

T(2*n,n-1) = T(2*n-1,n) for n > 0;
central terms give A119259; row sums give A007051;
T(n,0) = T(n,n) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A056220(n-1) for n > 1;
T(n,n-4) = A027608(n-4) for n > 3;
T(n,n-3) = A055580(n-3) for n > 2;
T(n,n-2) = A000337(n-1) for n > 1;
T(n,n-1) = A000225(n) for n > 0.
T(n,k) = [k<=n]*(-1)^k*Sum_{i=0..k} (-1)^i*C(k-n,k-i)*C(n,i). - Paul Barry, Sep 28 2007
From Richard M. Green, Jul 26 2011: (Start)
T(n,k) = [k<=n] Sum_{i=n-k..n} (-1)^(n-k-i)*2^(n-i)*C(n,i).
T(n,k) = [k<=n] Sum_{i=n-k..n} C(n,i)*C(i-1,n-k-1).
G.f. for T(n,n-k): x^k/(((1-2x)^k)*(1-x)). (End)
T(n,k) = R(n,k,2) where R(n, k, m) = (1-m)^(-n+k)-m^(k+1)*Pochhammer(n-k,k+1)* hyper2F1([1,n+1], [k+2], m)/(k+1)!. - Peter Luschny, Jul 25 2014
From Peter Bala, Mar 05 2018: (Start)
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = 1 + 7*x + 17*x^2 + 15*x^3 + x^4 + O(x^5).
Row reverse of A112857. (End)

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Abdullahi Umar, Aug 25 2008

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A007051, A118801, A135233, columns: A000225, A000337, A055580, A027608.
Cf. also A038207, A250118, A250119.
Cf. A007318, A167374, A145661, A029635.

A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011

Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -7, 5, -1, 1, -15, 17, -7, 1, 1, -31, 49, -31, 9, -1, 1, -63, 129, -111, 49, -11, 1, 1, -127, 321, -351, 209, -71, 13, -1, 1, -255, 769, -1023, 769, -351, 97, -15, 1, 1, -511, 1793, -2815, 2561, -1471, 545, -127, 17, -1, 1, -1023, 4097, -7423, 7937, -5503, 2561, -799, 161, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).
T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

			Formulas for initial columns are, for n>=0:
T(n+1,1) = 1 - 2^(n+1);
T(n+2,2) = 1 + 2^(n+1)*n;
T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
Triangle begins:
1;
1,-1;
1,-3,1;
1,-7,5,-1;
1,-15,17,-7,1;
1,-31,49,-31,9,-1;
1,-63,129,-111,49,-11,1;
1,-127,321,-351,209,-71,13,-1;
1,-255,769,-1023,769,-351,97,-15,1;
1,-511,1793,-2815,2561,-1471,545,-127,17,-1;
1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...
The matrix square, T^2, starts:
1;
0,1;
-1,0,1;
-2,-1,0,1;
-3,-2,-1,0,1;
-4,-3,-2,-1,0,1; ...
where all columns are the same.
The matrix product C^-1*T*C = T^-1*C^2 is:
1;
-1,-1;
0, 1, 1;
0, 0,-1,-1;
0, 0, 0, 1, 1; ...
where C(n,k) = n!/(n-k)!/k!.

M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011
Cf. A007318, A032807, A097805, A112857, A130595, A156644, A167374, A200139, A239473.

Table[(1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]) - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

{T(n,k)=if(n==0&k==0,1,1+(-1)^k*2^(n-k+1)*sum(j=0,k\2,binomial(n-2*j-2,k-2*j-1)))}

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.
For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
Signed P^2 = signed A032807 = T*A167374. (End)

A193844 Triangular array: the fission of ((x+1)^n) by ((x+1)^n); i.e., the self-fission of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 17, 15, 1, 9, 31, 49, 31, 1, 11, 49, 111, 129, 63, 1, 13, 71, 209, 351, 321, 127, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023

Offset: 0

Clark Kimberling, Aug 07 2011

See A193842 for the definition of fission of two sequences of polynomials or triangular arrays.
A193844 is also the fission of (p1(n,x)) by (q1(n,x)), where p1(n,x)=x^n+x^(n-1)+...+x+1 and q1(n,x)=(x+2)^n.
Essentially A119258 but without the main diagonal. - Peter Bala, Jul 16 2013
From Robert Coquereaux, Oct 02 2014: (Start)
This is also a rectangular array A(n,p) read down the antidiagonals:
1 1 1 1 1 1 1 1 1
3 5 7 9 11 13 15 17 19
7 17 31 49 71 97 127 161 199
15 49 111 209 351 545 799 1121 1519
31 129 351 769 1471 2561 4159 6401 9439
...
Calling Gr(n) the Grassmann algebra with n generators, A(n,p) is the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence. If p is odd A(n,p) is the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n). If p is even, the dimension of this cohomology group is A(n,p)+1. A(n,p) = 2^n*A059260(p,n-1)-(-1)^p.
(End)
The n-th row are also the coefficients of the polynomial P=sum_{k=0..n} (X+2)^k (in falling order, i.e., that of X^n first). - M. F. Hasler, Oct 15 2014

			First six rows:
1
1....3
1....5....7
1....7....17....15
1....9....31....49....31
1....11...49....111...129...63

Jean-François Chamayou, A Random Difference Equation with Dufresne Variables revisited, arXiv:1410.1708 [math.PR], 2014.
R. Coquereaux and E. Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
R. Coquereaux and E. Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
C. Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.

Cf. A193842, A193845, A119258.
Columns, diagonals: A000225, A000337, A055580, A027608, A211386, A211388, A000012, A005408, A056220, A199899.
A145661 is an essentially identical triangle.

A193844 := (n,k) -> 2^k*binomial(n+1,k)*hypergeom([1,-k],[-k+n+2],1/2);
for n from 0 to 5 do seq(round(evalf(A193844(n,k))),k=0..n) od; # Peter Luschny, Jul 23 2014
# Alternatively
p := (n,x) -> add(x^k*(1+2*x)^(n-k), k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od; # Peter Luschny, Jun 18 2017

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
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}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)

# uses[fission from A193842]
p = lambda n,x: (x+1)^n
A193844_row = lambda n: fission(p, p, n)
for n in range(7): print(A193844_row(n)) # Peter Luschny, Jul 23 2014

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = sum {i = 0..k} (-1)^i*binomial(n+1,k-i)*2^(k-i).
O.g.f.: 1/( (1 - x*t)*(1 - (2*x + 1)*t) ) = 1 + (1 + 3*x)*t + (1 + 5*x + 7*x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x+1)*( (2*x+1)^(n+1) - x^(n+1) ). (End)
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 3, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014
T(n,k) = 2^k*binomial(n+1,k)*hyper2F1(1,-k,-k+n+2, 1/2). - Peter Luschny, Jul 23 2014

cp's OEIS Frontend

A119258 Triangle read by rows: T(n,0) = T(n,n) = 1 and for 0

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Sage

Formula

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A118801 Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A193844 Triangular array: the fission of ((x+1)^n) by ((x+1)^n); i.e., the self-fission of Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.