A119900 Triangle read by rows: T(n,k) is the number of binary words of length n with k strictly increasing runs, for 0<=k<=n.
1, 0, 2, 0, 1, 3, 0, 0, 4, 4, 0, 0, 1, 10, 5, 0, 0, 0, 6, 20, 6, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 8, 56, 56, 8, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 10, 120, 252, 120, 10, 0, 0, 0, 0, 0, 1, 55, 330, 462, 165, 11, 0, 0, 0, 0, 0, 0, 12, 220, 792, 792, 220, 12, 0, 0, 0, 0, 0, 0, 1, 78
Offset: 0
Examples
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;
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.
Crossrefs
Programs
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Magma
/* triangle */ [[Binomial(n+1, 2*k-n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Oct 22 2017
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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
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Mathematica
Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 21 2016 *)
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PARI
for(n=0,10, for(k=0,n, print1(binomial(n+1, 2*k-n), ", "))) \\ G. C. Greubel, Oct 22 2017
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)
Extensions
Keyword tabl added by Philippe Deléham, Jan 26 2010
Comments