cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A135278 Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 12, 66, 220, 495, 792, 924, 792
Offset: 0

Views

Author

Zerinvary Lajos, Dec 02 2007

Keywords

Comments

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a simplex is the convex hull of a set of (n + 1) affinely independent points in some Euclidean space of dimension n or higher, i.e., a set of points such that no m-plane contains more than (m + 1) of them. Such points are said to be in general position.
Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
From Tom Copeland, Dec 07 2007: (Start)
T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows that T is related to combinations of letters defined by connectivity of regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1, then T is a fundamental type of matrix that is discussed in A133314 and the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with index n for the Laguerre polynomials of order 1. Other formulas for the action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) = exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 }. [ O.g.f. for row sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. [Sign correction noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] = A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292. (End)
Given matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. The e.g.f. for the row polynomials of A is {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. - Tom Copeland, Aug 21 2008
A007318*A097806 as infinite lower triangular matrices. - Philippe Deléham, Feb 08 2009
Riordan array (1/(1-x)^2, x/(1-x)). - Philippe Deléham, Feb 22 2012
The elements of the matrix inverse are T^(-1)(n,k)=(-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Relation to K-theory: T acting on the column vector (-0,d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168 and also A104712, A111492, and A238363. - Tom Copeland, Apr 11 2014
Number of walks of length p>0 between any two distinct vertices of the complete graph K_(n+2) is W(n+2,p)=(-1)^(p-1)*Sum_{k=0..p-1} T(p-1,k)*(-n-2)^k = ((n+1)^p - (-1)^p)/(n+2) = (-1)^(p-1)*Sum_{k=0..p-1} (-n-1)^k. This is equal to (-1)^(p-1)*Phi(p,-n-1), where Phi is the cyclotomic polynomial when p is an odd prime. For K_3, see A001045; for K_4, A015518; for K_5, A015521; for K_6, A015531; for K_7, A015540. - Tom Copeland, Apr 14 2014
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
See A074909 for associations among this array, the Bernoulli polynomials and their umbral compositional inverses, and the face polynomials of permutahedra and their duals (cf. A019538). - Tom Copeland, Nov 14 2014
From Wolfdieter Lang, Dec 10 2015: (Start)
A(r, n) = T(n+r-2, r-1) = risefac(n,r)/r! = binomial(n+r-1, r), for n >= 1 and r >= 1, gives the array with the number of independent components of a symmetric tensors of rank r (number of indices) and dimension n (indices run from 1 to n). Here risefac(n, k) is the rising factorial.
As(r, n) = T(n+1, r+1) = fallfac(n, r)/r! = binomial(n, r), r >= 1 and n >= 1 (with the triangle entries T(n, k) = 0 for n < k) gives the array with the number of independent components of an antisymmetric tensor of rank r and dimension n. Here fallfac is the falling factorial. (End)
The h-vectors associated to these f-vectors are given by A000012 regarded as a lower triangular matrix. Read as bivariate polynomials, the h-polynomials are the complete homogeneous symmetric polynomials in two variables, found in the compositional inverse of an e.g.f. for A008292, the h-vectors of the permutahedra. - Tom Copeland, Jan 10 2017
For a correlation between the states of a quantum system and the combinatorics of the n-simplex, see Boya and Dixit. - Tom Copeland, Jul 24 2017

Examples

			The triangle T(n, k) begins:
   n\k  0  1   2   3   4   5   6   7   8  9 10 11 ...
   0:   1
   1:   2  1
   2:   3  3   1
   3:   4  6   4   1
   4:   5 10  10   5   1
   5:   6 15  20  15   6   1
   6:   7 21  35  35  21   7   1
   7:   8 28  56  70  56  28   8   1
   8:   9 36  84 126 126  84  36   9   1
   9:  10 45 120 210 252 210 120  45  10  1
  10:  11 55 165 330 462 462 330 165  55 11  1
  11:  12 66 220 495 792 924 792 495 220 66 12  1
  ... reformatted by _Wolfdieter Lang_, Mar 23 2015
Production matrix begins
   2   1
  -1   1   1
   1   0   1   1
  -1   0   0   1   1
   1   0   0   0   1   1
  -1   0   0   0   0   1   1
   1   0   0   0   0   0   1   1
  -1   0   0   0   0   0   0   1   1
   1   0   0   0   0   0   0   0   1   1
- _Philippe Deléham_, Jan 29 2014
From _Wolfdieter Lang_, Nov 08 2018: (Start)
Recurrence [_Philippe Deléham_]: T(7, 3) = 2*35 + 35 - 15 - 20 = 70.
Recurrence from Riordan A- and Z-sequences: [1,1,repeat(0)] and [2, repeat(-1, +1)]: From Z: T(5, 0) = 2*5 - 10 + 10 - 5 + 1 = 6. From A: T(7, 3) = 35 + 35 = 70.
Boas-Buck column k=3 recurrence: T(7, 3) = (5/4)*(1 + 5 + 15 + 35) = 70. (End)
		

Crossrefs

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i) od;
  • Mathematica
    Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z] - Exp[z]), {z, k}] /. z -> 0, x], {k, 0, 11}]]
    CoefficientList[CoefficientList[Series[1/((1 - x)*(1 - x - x*y)), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
    
  • Sage
    Trow = lambda n: sum((x+1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else 0. (Partial sum of column k of A007318 (Pascal), or summation on the upper binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see A007318.) - Wolfdieter Lang, Aug 22 2012
E.g.f.: 1/x*((1 + x)*exp(t*(1 + x)) - exp(t)) = 1 + (2 + x)*t + (3 + 3*x + x^2)*t^2/2! + .... The infinitesimal generator for this triangle has the sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)*(1-x-x*y)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014: (Start)
[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as Bala's), which makes connections to the ubiquitous associated Laguerre polynomials of integer orders, for this case the Laguerre polynomials of order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n = exp[t*x/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3*t+t^2/2!)x^2 + (4+6*t+4*t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t) exp[(2+t)x] - 2 exp(2x)}/t.
(End)
From Tom Copeland, Apr 28 2014: (Start)
With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492. (End)
From Tom Copeland, Sep 17 2014: (Start)
A) U(x,s,t)= x^2/[(1-t*x)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 + 3s*t + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (t*x)^n L(n,-:xD:,-1)] U(x,s,0), where (:xD:)^k=x^k*(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1, related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) = (x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015
From Wolfdieter Lang, Nov 08 2018: (Start)
Recurrences from the A- and Z- sequences for the Riordan triangle (see the W. Lang link under A006232 with references), which are A(n) = A019590(n+1), [1, 1, repeat (0)] and Z(n) = (-1)^(n+1)*A054977(n), [2, repeat(-1, 1)]:
T(0, 0) = 1, T(n, k) = 0 for n < k, and T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and T(n, k) = T(n-1, k-1) + T(n-1, k), for n >= m >= 1.
Boas-Buck recurrence for columns (see the Aug 10 2017 remark in A036521 also for references):
T(n, k) = ((2 + k)/(n - k))*Sum_{j=k..n-1} T(j, k), for n >= 1, k = 0, 1, ..., n-1, and input T(n, n) = 1, for n >= 0, (the BB-sequences are alpha(n) = 2 and beta(n) = 1). (End)
T(n, k) = [x^k] Sum_{j=0..n} (x+1)^j. - Peter Luschny, Jul 09 2019

Extensions

Edited by Tom Copeland and N. J. A. Sloane, Dec 11 2007

A193842 Triangular array: the fission of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence ((x+2)^n: n >= 0). (Fission is defined at Comments.)

Original entry on oeis.org

1, 1, 4, 1, 7, 13, 1, 10, 34, 40, 1, 13, 64, 142, 121, 1, 16, 103, 334, 547, 364, 1, 19, 151, 643, 1549, 2005, 1093, 1, 22, 208, 1096, 3478, 6652, 7108, 3280, 1, 25, 274, 1720, 6766, 17086, 27064, 24604, 9841, 1, 28, 349, 2542, 11926, 37384, 78322, 105796
Offset: 0

Views

Author

Clark Kimberling, Aug 07 2011

Keywords

Comments

Suppose that p = p(n)*x^n + p(n-1)*x^(n-1) + ... + p(1)*x + p(0) is a polynomial and that Q is a sequence of polynomials:
...
q(k,x) = t(k,0)*x^k + t(k,1)*x^(k-1) + ... + t(k,k-1)*x + t(k,k),
...
for k = 0, 1, 2, ... The Q-downstep of p is the polynomial given by
...
D(p) = p(n)*q(n-1,x) + p(n-1)*q(n-2,x) + ... + p(1)*q(0,x). (Note that p(0) does not appear. "Q-downstep" as just defined differs slightly from "Q-downstep" as defined for a different purpose at A193649.)
...
Now suppose that P = (p(n,x): n >= 0) and Q = (q(n,x): n >= 0) are sequences of polynomials, where n indicates degree. The fission of P by Q, denoted by P^^Q, is introduced here as the sequence W = (w(n,x): n >= 0) of polynomials defined by w(0,x) = 1 and w(n,x) = D(p(n+1,x)).
...
Strictly speaking, ^^ is an operation on sequences of polynomials. However, if P and Q are regarded as numerical triangles (of coefficients of polynomials), then ^^ can be regarded as an operation on numerical triangles. In this case, row n of P^^Q, for n > 0, is given by the matrix product P(n+1)*QQ(n), where P(n+1) =(p(n+1,n+1), p(n+1,n), ..., p(n+1,2), p(n+1,1)) and QQ(n) is the (n+1)-by-(n+1) matrix given by
...
q(n,0) .. q(n,1)............. q(n,n-1) .... q(n,n)
0 ....... q(n-1,0)........... q(n-1,n-2)... q(n-1,n-1)
0 ....... 0.................. q(n-2,n-3) .. q(n-2,n-2)
...
0 ....... 0.................. q(1,0) ...... q(1,1)
0 ....... 0 ................. 0 ........... q(0,0).
Here, the polynomial q(k,x) is taken to be
q(k,0)*x^k + q(k,1)x^(k-1) + ... + q(k,k)*x + q(k,k);
i.e., "q" is used instead of "t".
...
Example: Let p(n,x) = (x+1)^n and q(n,x) = (x+2)^n. Then
...
w(0,x) = 1 by the definition of W,
w(1,x) = D(p(2,x)) = 1*(x+2) + 2*1 = x + 4,
w(2,x) = D(p(3,x)) = 1*(x^2+4*x+4) + 3*(x+2) + 3*1 = x^2 + 7*x + 13,
w(3,x) = D(p(4,x)) = 1*(x^3+6*x^2+12*x+8) + 4*(x^2+4x+4) + 6*(x+2) + 4*1 = x^3 + 10*x^2 + 34*x + 40.
...
From these first 4 polynomials in the sequence P^^Q, we can write the first 4 rows of P^^Q when P, Q, and P^^Q are regarded as triangles:
1
1...4
1...7....13
1...10...34...40
...
In the following examples, r(P^^Q) is the mirror of P^^Q, obtained by reversing the rows of P^^Q. Let u denote the polynomial x^n + x^(n-1) + ... + x + 1.
...
..P........Q...........P^^Q........r(P^^Q)
(x+1)^n....(x+2)^n.....A193842.....A193843
(x+1)^n....(x+1)^n.....A193844.....A193845
(x+2)^n....(x+1)^n.....A193846.....A193847
(2x+1)^n...(x+1)^n.....A193856.....A193857
(x+1)^n....(2x+1)^n....A193858.....A193859
(x+1)^n.......u........A054143.....A104709
..u........(x+1)^n.....A074909.....A074909
..u...........u........A002260.....A004736
(x+2)^n.......u........A193850.....A193851
..u.........(x+2)^n....A193844.....A193845
(2x+1)^n......u........A193860.....A193861
..u.........(2x+1)^n...A115068.....A193862
...
Regarding A193842,
col 1 ...... A000012
col 2 ...... A016777
col 3 ...... A081271
w(n,n) ..... A003462
w(n,n-1) ... A014915

Examples

			First six rows, for 0 <= k <= n and 0 <= n <= 5:
  1
  1...4
  1...7....13
  1...10...34....40
  1...13...64....142...121
  1...16...103...334...547...364
		

Crossrefs

Cf. A193722 (fusion of P by Q), A193649 (Q-residue), A193843 (mirror of A193842).

Programs

  • Magma
    [ (&+[3^(k-j)*Binomial(n-j,k-j): j in [0..k]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020
  • Maple
    fission := proc(p, q, n) local d, k;
    p(n+1,0)*q(n,x)+add(coeff(p(n+1,x),x^k)*q(n-k,x), k=1..n);
    seq(coeff(%,x,n-k), k=0..n) end:
    A193842_row := n -> fission((n,x) -> (x+1)^n, (n,x) -> (x+2)^n, n);
    for n from 0 to 5 do A193842_row(n) od; # Peter Luschny, Jul 23 2014
    # Alternatively:
    p := (n,x) -> add(x^k*(1+3*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
  • Mathematica
    (* First program *)
    z = 10;
    p[n_, x_] := (x + 1)^n;
    q[n_, x_] := (x + 2)^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}]]  (* A193842 *)
    TableForm[Table[h[n], {n, 0, z}]]  (* A193843 *)
    Flatten[Table[h[n], {n, -1, z}]]
    (* Second program *)
    Table[SeriesCoefficient[((x+3)^(n+1) -1)/(x+2), {x,0,n-k}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)
  • PARI
    T(n,k) = sum(j=0,k, 3^(k-j)*binomial(n-j,k-j)); \\ G. C. Greubel, Feb 18 2020
    
  • Sage
    from mpmath import mp, hyp2f1
    mp.dps = 100; mp.pretty = True
    def T(n,k):
        return 3^k*binomial(n,k)*hyp2f1(1,-k,-n,1/3)-0^(n-k)//2
    for n in range(7):
        print([int(T(n,k)) for k in (0..n)]) # Peter Luschny, Jul 23 2014
    
  • Sage
    # Second program using the 'fission' operation.
    def fission(p, q, n):
        F = p(n+1,0)*q(n,x)+add(expand(p(n+1,x)).coefficient(x,k)*q(n-k,x) for k in (1..n))
        return [expand(F).coefficient(x,n-k) for k in (0..n)]
    A193842_row = lambda k: fission(lambda n,x: (x+1)^n, lambda n,x: (x+2)^n, k)
    for n in range(7): A193842_row(n) # Peter Luschny, Jul 23 2014
    

Formula

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = Sum_{i = 0..k} 3^(k-i)*binomial(n-i,k-i).
O.g.f.: 1/((1 - x*t)*(1 - (1 + 3*x)*t)) = 1 + (1 + 4*x)*t + (1 + 7*x + 13*x^2)*t^2 + ....
The n-th row polynomial is R(n,x) = (1/(2*x + 1))*((3*x + 1)^(n+1) - x^(n+1)). (End)
T(n,k) = T(n-1,k) + 4*T(n-1,k-1) - T(n-2,k-1) - 3*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 4, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 17 2014
T(n,k) = 3^k * C(n,k) * hyp2F1(1, -k, -n, 1/3) with or without the additional term -0^(n-k)/2 depending on the exact definition of the hypergeometric function used. Compare formulas 15.2.5 and 15.2.6 in the DLMF reference. - Peter Luschny, Jul 23 2014

Extensions

Name and Comments edited by Petros Hadjicostas, Jun 05 2020

A115068 Triangle read by rows: T(n,k) = number of elements in the Coxeter group D_n with descent set contained in {s_k}, for 0<=k<=n-1.

Original entry on oeis.org

1, 2, 2, 4, 6, 3, 8, 16, 12, 4, 16, 40, 40, 20, 5, 32, 96, 120, 80, 30, 6, 64, 224, 336, 280, 140, 42, 7, 128, 512, 896, 896, 560, 224, 56, 8, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1024, 5632, 14080, 21120
Offset: 1

Views

Author

Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006

Keywords

Comments

A115068 is the fission of the polynomial sequence (p(x,n)) by the polynomial sequence ((2x+1)^n), where p(n,x)=x^n+x^(n-1)+...+x+1, n>=0. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011

Examples

			First six rows:
1
2...2
4...6....3
8...16...12...4
16..40...40...20...5
32..96...120..80...30...6
		

References

  • A. Bjorner and F. Brenti, Combinatorics of Coxeter Groups, Springer, New York, 2005.
  • J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge University Press, Cambridge, 1990.

Crossrefs

Programs

  • Haskell
    a115068 n k = a115068_tabl !! (n-1) !! (k-1)
    a115068_row n = a115068_tabl !! (n-1)
    a115068_tabl = iterate (\row -> zipWith (+) (row ++ [1]) $
                                    zipWith (+) (row ++ [0]) ([0] ++ row)) [1]
    -- Reinhard Zumkeller, Jul 22 2013
  • Mathematica
    z = 11;
    p[0, x_] := 1; p[n_, x_] := x*p[n - 1, x] + 1;
    q[n_, x_] := (2 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}]]  (* A115068 *)
    TableForm[Table[h[n], {n, 0, z}]]
    Flatten[Table[h[n], {n, -1, z}]]   (* A193862 *)

Formula

T(n,k)=binomial(n,k)*2^(n-k-1).
T(n,1) = 2^(n-1), T(n,n) = n, for n > 1: T(n,k) = T(n-1,k-1) + 2*T(n-1,k), 1 < k < n. - Reinhard Zumkeller, Jul 22 2013

A185342 Triangle of successive recurrences in columns of A117317(n).

Original entry on oeis.org

2, 4, -4, 6, -12, 8, 8, -24, 32, -16, 10, -40, 80, -80, 32, 12, -60, 160, -240, 192, -64, 14, -84, 280, -560, 672, -448, 128, 16, -112, 448, -1120, 1792, -1792, 1024, -256, 18, -144, 672, -2016, 4032, -5376, 4608, -2304, 512, 20, -180, 960, -3360, 8064
Offset: 0

Views

Author

Paul Curtz, Jan 26 2012

Keywords

Comments

A117317 (A):
1
2 1
4 5 1
8 16 9 1
16 44 41 14 1
32 112 146 85 20 1
64 272 456 377 155 27 1
have for their columns successive signatures
(2) (4,-4) (6,-12,8) (8,-24, 32, -16) (10,-40,80,-80,32) i.e. a(n).
Take based on abs(A133156) (B):
1
2 0
4 1 0
8 4 0 0
16 12 1 0 0
32 32 6 0 0 0
64 80 24 1 0 0 0.
The recurrences of successive columns are also a(n). a(n) columns: A005843(n+1), A046092(n+1), A130809, A130810, A130811, A130812, A130813.

Examples

			Triangle T(n,k),for 1<=k<=n, begins :
2                                         (1)
4    -4                                   (2)
6   -12   8                               (3)
8   -24  32   -16                         (4)
10  -40  80   -80   32                    (5)
12  -60 160  -240  192   -64              (6)
14  -84 280  -560  672  -448  128         (7)
16 -112 448 -1120 1792 -1792 1024 -256    (8)
Successive rows can be divided by A171977.
		

Crossrefs

Cf. For (A): A053220, A056243. For (B): A000079, A001787, A001788, A001789. For A193862: A115068 (a Coxeter group). For (2): A014480 (also (3),(4),(5),..); also A053220 and A001787.
Cf. A007318.

Programs

  • Mathematica
    Table[(-1)*Binomial[n, k]*(-2)^k, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Jun 27 2017 *)
  • PARI
    for(n=1,20, for(k=1,n, print1((-2)^(k+1)*binomial(n,k)/2, ", "))) \\ G. C. Greubel, Jun 27 2017

Formula

Take A133156(n) without 1's or -1's double triangle (C)=
2
4
8 -4
16 -12
32 -32 6
64 -80 24
128 -192 80 -8
256 -448 240 -40
512 -1024 672 -160 10;
a(n) is increasing odd diagonals and increasing (sign changed) even diagonals. Rows sum of (C) = A201629 (?) Another link between Chebyshev polynomials and cos( ).
Absolute values: A013609(n) without 1's. Also 2*A193862 = (2*A002260)*A135278.
T(n,k) = T(n-1,k) - 2*T(n-1,k-1) for k>1, T(n,1) = 2*n = 2*T(n-1,1) - T(n-2,1). - Philippe Deléham, Feb 11 2012
T(n,k) = (-1)* Binomial(n,k)*(-2)^k, 1<=k<=n. - Philippe Deléham, Feb 11 2012

A334009 Triangle read by rows: T(n, k) = binomial(n + k - 1, 2*k - 1) * 4^(k - 1) * n/k, 1 <= k <= n.

Original entry on oeis.org

1, 4, 4, 9, 24, 16, 16, 80, 128, 64, 25, 200, 560, 640, 256, 36, 420, 1792, 3456, 3072, 1024, 49, 784, 4704, 13440, 19712, 14336, 4096, 64, 1344, 10752, 42240, 90112, 106496, 65536, 16384, 81, 2160, 22176, 114048, 329472, 559104, 552960, 294912, 65536, 100
Offset: 1

Views

Author

Michael Somos, Apr 12 2020

Keywords

Comments

Let P(n, x) := Sum_{k=1..n} T(n, k)*x^k. Then P(n, P(m, x)) = P(n*m, x) for all n and m in Z.
The r=4 case of the Logistic Map is 4*x*(1 - x) = -P(1, -x). The r=2 case leads to A193862.

Examples

			First four rows:
.1
.4...4
.9..24..16
16..80.128..64
		

Crossrefs

Programs

  • Mathematica
    T[ n_, k_] := If[k == 0, 0, Binomial[n + k - 1, 2 k - 1] 4^(k - 1) n / k];
  • PARI
    {T(n, k) = if(k == 0, 0, binomial(n + k - 1, 2*k - 1) * 4^(k - 1) * n/k)};

Formula

P(n, x) = sinh(n * arcsinh(sqrt(x)))^2 = (hypergeom([-n, n], [1/2], -x) - 1)/2 are the row polynomials.
G.f.: Sum_{n, m} T(n, k) * x^k * y^n = x * y * (1 + y) / ((1 - y) * (1 - (2 + 4*x)*y + y^2)).
Row sums are A001108.
T(n, k) = (-1)^n * (-4)^(k-1) * A039598(-k - 1, n - 1) for all n in Z if k<0.
T(n, k) = -(-1)^(n+k) * A123588(n,k) if 1 <= k <= n.

A276985 Triangle read by rows: T(n,k) = number of k-dimensional elements in an n-dimensional cross-polytope, n>=1, 0<=k

Original entry on oeis.org

2, 4, 4, 6, 12, 8, 8, 24, 32, 16, 10, 40, 80, 80, 32, 12, 60, 160, 240, 192, 64, 14, 84, 280, 560, 672, 448, 128, 16, 112, 448, 1120, 1792, 1792, 1024, 256, 18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512, 20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120
Offset: 1

Views

Author

Felix Fröhlich, Sep 24 2016

Keywords

Comments

It appears that this is 2*A193862 (but with a different offset) and that the sum of terms of the n-th row is A024023(n) = 3^n - 1. - Michel Marcus, Sep 29 2016

Examples

			T(4, 1..4) = 8, 24, 32, 16, because the 16-cell has 8 0-faces (vertices), 24 1-faces (edges), 32 2-faces (faces) and 16 3-faces (cells).
Triangle starts
2
4, 4
6, 12, 8
8, 24, 32, 16
10, 40, 80, 80, 32
12, 60, 160, 240, 192, 64
14, 84, 280, 560, 672, 448, 128
16, 112, 448, 1120, 1792, 1792, 1024, 256
18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512
20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120, 1024
		

References

  • H. S. M. Coxeter, Regular Polytopes, Third Edition, Dover Publications, 1973, ISBN 9780486141589.

Crossrefs

Cf. A038207 (hypercube), A135278 (simplex).
Rows: A005843(n), A046092(n), A130809(n+2), A130810(n+3).
Columns: A000079(n), A001787(n), A001788(n), A001789(n+3).

Programs

  • Mathematica
    Table[2^(k + 1) Binomial[n, k + 1], {n, 10}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Sep 25 2016 *)
  • PARI
    T(n, k) = 2^(k+1)*binomial(n, k+1)
    trianglerows(n) = for(x=1, n, for(y=0, x-1, print1(T(x, y), ", ")); print(""))
    trianglerows(10) \\ print initial 10 rows of triangle

Formula

T(n,k) = 2^(k+1) * binomial(n, k+1) (cf. Coxeter, 1973, formula 7.22).
T(n,k) = A182059(n,k) = A013609(n,k) . - R. J. Mathar, May 03 2017
G.f.: 2*x/((1 - x)*(1 - x - 2*x*y)). - Stefano Spezia, Jul 17 2025
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