OFFSET

0,3

COMMENTS

Sum of terms in row n is 2^n (A000079). Sum of terms in column k is A001906(k+1) (the even-indexed Fibonacci numbers). Row n contains 1+floor(n/2) nonzero terms. Sum_{k=0..n} k*T(n,k) = (3n+1)*2^(n-2) = A066373(n+1) for n>=1.

Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1/2,-1/2,0,0,0,0,0, 0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 02 2008

From R. Bagula's comment in A053122 (cf. Damianou link), the columns of this array give the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014

Odd rows contain the Pascal triangle numbers A091042. See A034867 and A034839 for some relations to tan(x). - Tom Copeland, Oct 15 2014

LINKS

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

A. Collins et al., Binary words, n-color compositions and bisection of the Fibonacci numbers, Fib. Quarterly, 51 (2013), 130-136.

P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.

R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

FORMULA

T(n,k) = binomial(n+1,2k-n).

G.f.: 1/(1 - 2*t*z - t*(1-t)*z^2).

T(n,k) = A034867(n,n-k)

From Tom Copeland, Sep 30 2011: (Start)

With K(x,t) = 1/{d/dx{x/[t-1+1/(1-x)]}} = [t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, the g.f. of A119900 = K(x*t,t)-t+1.

From formulas in A134264: K(x,t)d/dx is a generator for A001263. A refinement of A119900 to partition polynomials is given by umbralizing

K(x,t) roughly as K(h.x,h_0) and precisely as in A134264 as

W(x)= 1/{d/dx[f(x)]}=1/{d/dx[x/h(x)]}. (End)

T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2). - Philippe Deléham, Oct 02 2011

From Tom Copeland, Dec 07 2015: (Start)

An alternate o.g.f. is (1/(x*t)) {-1 + 1 / [1 - (1/t)[x*t/(1-x*t)]^2]} = Sum_{n>0} x^(2(n-1)+1) t^(n-1) / (1-t*x)^(2n) = x + 2t x^2 + (t+3t^2) x^3 + ... .

The n-th diagonal has elements binomial(2n+1+k,k), starting with k=0 for the first non-vanishing element, with o.g.f. (1-x)^(-2(n+1)). The first few subdiagonals are shifted versions of A000292, A000389, and A000580. Cf. A049310.

See A034867 for the matrix representation for the infinitesimal generator K(x,t) d/dx for the Narayana polynomials. (End)

From Peter Bala, Aug 17 2016: (Start)

Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:

Sum_{k = 0..p} T(p,k)*S(2*k + 1,n) = (n*(n + 1)/2)^(p+1).

For example, for row 6 we find S(7,n) + 21*S(9,n) + 35*S(11,n) + 7*S(13,n) = (n*(n + 1)/2)^7.

There appears to be a similar result for the even power sums S(2*k,n) involving A207543. (End)

EXAMPLE

The binary word 1/0/01/01/1/1/01 has 7 strictly increasing runs.

T(5,3)=6 because we have 0/01/01, 01/0/01, 01/01/0, 01/1/01, 01/01/1 and 1/01/01 (the runs are separated by /).

Triangle starts:

1;

0,2;

0,1,3;

0,0,4,4;

0,0,1,10,5;

0,0,0,6,20,6;

MAPLE

T:=(n, k)->binomial(n+1, 2*k-n): for n from 0 to 12 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form

MATHEMATICA

Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 21 2016 *)

PROG

(PARI) for(n=0, 10, for(k=0, n, print1(binomial(n+1, 2*k-n), ", "))) \\ G. C. Greubel, Oct 22 2017

(Magma) /* triangle */ [[Binomial(n+1, 2*k-n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Oct 22 2017

CROSSREFS

Cf. A098158. - Philippe Deléham, Dec 02 2008

KEYWORD

nonn,tabl

AUTHOR

Emeric Deutsch, May 27 2006

EXTENSIONS

Keyword tabl added by Philippe Deléham, Jan 26 2010

STATUS

approved