A046854 Triangle read by rows: T(n, k) = binomial(floor((n+k)/2), k), n >= k >= 0.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 3, 4, 1, 1, 1, 3, 6, 4, 5, 1, 1, 1, 4, 6, 10, 5, 6, 1, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 1, 6, 21, 35, 70, 56, 84, 36, 45, 10, 11, 1, 1

Offset: 0

Author

Wouter Meeussen

Keywords

• nonn
• tabl
• easy

Comments

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004

Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005

As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008

From Emeric Deutsch, Jun 18 2010: (Start)

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.

T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)

From L. Edson Jeffery, Mar 01 2011: (Start)

This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table

1

1

1 1

1 1

1 2 1

1 2 1

1 3 3 1

1 3 3 1

...,

then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with

1;

1, -1;

1, -1, -1;

1, -2, -1, 1;

1, -2, -3, 1, 1;

...

(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.

For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)

Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011

Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012

Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014

From Jeremy Dover, Jun 07 2016: (Start)

T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.

T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Examples

			Triangle begins:
  1;
  1 1;
  1 1 1;
  1 2 1 1;
  1 2 3 1 1;
  1 3 3 4 1 1;
  ...
		

References

• J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1978. [Emeric Deutsch, Jun 18 2010]

Links

• Nathaniel Johnston, Rows 0..100, flattened
• Robert G. Donnelly, Molly W. Dunkum, and Rachel McCoy, Olry Terquem's forgotten problem, arXiv:2303.05949 [math.HO], 2023.
• Jeremy M. Dover, Some Notes on Pairs in Binary Strings, arXiv:1609.00980 [math.CO], 2016.
• Dominique Foata and Guo-Niu Han, Multivariable Tangent and Secant q-derivative Polynomials, arXiv:1304.2486 [math.CO], 2013; see Fig. 10.1.
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Reflected version of A065941, which is considered the main entry. A deficient version is in A030111.

Cf. A066170, A097806, A134513, A168561, A187660.

Cf. A055244. - Gary W. Adamson, Dec 23 2008

Programs

GAP

Flat(List([0..16], n-> List([0..n], k-> Binomial(Int((n+k)/2), k) ))); # G. C. Greubel, Jul 13 2019

Haskell

a046854 n k = a046854_tabl !! n !! k
a046854_row n = a046854_tabl !! n
a046854_tabl = [1] : f [1] [1,1] where
   f us vs = vs : f vs  (zipWith (+) (us ++ [0,0]) ([0] ++ vs))
-- Reinhard Zumkeller, Apr 24 2013

Magma

[Binomial(Floor((n+k)/2), k): k in [0..n], n in [0..16]]; // G. C. Greubel, Jul 13 2019

Maple

A046854:= proc(n,k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n,k),k=0..n),n=0..16); # Nathaniel Johnston, Jun 30 2011

Mathematica

Table[Binomial[Floor[(n+k)/2], k], {n,0,16}, {k,0,n}]//Flatten
(* next program: Polynomials *)
z = 12; f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1;
t = Table[Factor[f[x, n]], {n, 1, z}]
u = Flatten[CoefficientList[Numerator[t], x]] (* this sequence *)
v = Flatten[CoefficientList[Denominator[t], x]]
(* Clark Kimberling, Oct 13 2014 *)

PARI

T(n,k) = binomial((n+k)\2, k); \\ G. C. Greubel, Jul 13 2019

Sage

[[binomial(floor((n+k)/2), k) for k in (0..n)] for n in (0..16)] # G. C. Greubel, Jul 13 2019

Formula

T(n,k) = binomial(floor((n+k)/2), k).

G.f.: (1+x)/(1-x*y-x^2). - Ralf Stephan, Feb 13 2005

Triangle = A097806 * A168561, as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007

T(n,k) = A065941(n,n-k) = abs(A130777(n,k)) = abs(A066170(n,k)) = abs(A187660(n,k)). - Johannes W. Meijer, Aug 08 2011

For n > 1: T(n, k) = T(n-1, k-1) + T(n-2, k), 0 < k < n-1. - Reinhard Zumkeller, Apr 24 2013

T(n,k) = A168561(n,k) + A168561(n-1,k). - R. J. Mathar, Feb 10 2024