

A046854


Triangle in which kth entry of row n is binomial(floor(n/2+k/2),k), n>=0, n >= k >= 0.


50



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
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OFFSET

0,8


COMMENTS

Row sums are fibonacci(n+2). Diagonal sums are A016116.  Paul Barry, Jul 07 2004
Riordan array (1/(1x), x/(1x^2)). Matrix inverse is A106180.  Paul Barry, Apr 24 2005
As an infinite lower triangular matrix * [1,2,3,...] = A055244: (1, 1, 3, 6, 12, 23,...). [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 rth 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 kth 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^(Nk)]=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_(r1)(t)Q_(r2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N1)(phi_j), where phi_j=2*(1)^(j1)*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^32*x^2x+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)^(j1)*cos(j*Pi/7)]^21, j=1,2,3. (End)
Given the signed polynomials (+++,...) of the triangle, the largest root of the nth 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 nth 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,n1). 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 0s and no pairs of consecutive 1s.
T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0s and no pairs of consecutive 1s. (End)


REFERENCES

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


LINKS

Nathaniel Johnston, Rows 0..100, flattened
Jeremy M. Dover, Some Notes on Pairs in Binary Strings, arXiv:1609.00980 [math.CO], 2016.
Index entries for triangles and arrays related to Pascal's triangle


FORMULA

T(n,k) = binomial(floor(n/2+k/2),k)
G.f.: (1+x)/(1xyx^2).  Ralf Stephan, Feb 13 2005
Triangle = A097806 * A049310, as infinite lower triangular matrices.  Gary W. Adamson, Oct 28 2007
T(n,k) = A065941(n,nk) = 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(n1, k1) + T(n2, k), 0 < k < n1.  Reinhard Zumkeller, Apr 24 2013


EXAMPLE

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


MAPLE

A046854:= proc(n, k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n, k), k=0..n), n=0..11); # Nathaniel Johnston, Jun 30 2011


MATHEMATICA

Table[ Binomial[ Floor[ n/2 +k/2 ], k ], {n, 0, 16}, {k, 0, n} ]
(* 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]] (* A046854 *)
v = Flatten[CoefficientList[Denominator[t], x]]
(* Clark Kimberling, Oct 13 2014 *)


PROG

(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


CROSSREFS

Reflected version of A065941, which is considered the main entry. A deficient version is in A030111.
Cf. A066170, A097806, A049310, A187660.
Cf. A055244 [From Gary W. Adamson, Dec 23 2008]
Sequence in context: A225631 A267482 A130777 * A187660 A066170 A184957
Adjacent sequences: A046851 A046852 A046853 * A046855 A046856 A046857


KEYWORD

nonn,tabl,easy


AUTHOR

Wouter Meeussen


STATUS

approved



