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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.

%I A119900 #73 Jan 05 2025 19:51:38
%S A119900 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,
%T A119900 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,
%U A119900 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
%N 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.
%C A119900 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.
%C A119900 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
%C A119900 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
%C A119900 Odd rows contain the Pascal triangle numbers A091042. See A034867 and A034839 for some relations to tan(x). - _Tom Copeland_, Oct 15 2014
%H A119900 G. C. Greubel, <a href="/A119900/b119900.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H A119900 A. Collins et al., <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/51-2/CollinsDedricksonWang.pdf">Binary words, n-color compositions and bisection of the Fibonacci numbers</a>, Fib. Quarterly, 51 (2013), 130-136.
%H A119900 P. Damianou, <a href="http://arxiv.org/abs/1110.6620">On the characteristic polynomials of Cartan matrices and Chebyshev polynomials</a>, arXiv:1110.6620 [math.RT], 2014.
%H A119900 R. Zielinski, <a href="https://arxiv.org/abs/1608.04006">Induction and Analogy in a Problem of Finite Sums</a> arXiv:1608.04006 [math.GM], 2016.
%F A119900 T(n,k) = binomial(n+1,2k-n).
%F A119900 G.f.: 1/(1 - 2*t*z - t*(1-t)*z^2).
%F A119900 T(n,k) = A034867(n,n-k)
%F A119900 From _Tom Copeland_, Sep 30 2011: (Start)
%F A119900 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.
%F A119900 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
%F A119900   K(x,t) roughly as K(h.x,h_0) and precisely as in A134264 as
%F A119900   W(x)= 1/{d/dx[f(x)]}=1/{d/dx[x/h(x)]}. (End)
%F A119900 T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2). - _Philippe Deléham_, Oct 02 2011
%F A119900 From _Tom Copeland_, Dec 07 2015: (Start)
%F A119900 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 + ... .
%F A119900 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.
%F A119900 See A034867 for the matrix representation for the infinitesimal generator K(x,t) d/dx for the Narayana polynomials. (End)
%F A119900 From _Peter Bala_, Aug 17 2016: (Start)
%F A119900 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:
%F A119900 Sum_{k = 0..p} T(p,k)*S(2*k + 1,n) = (n*(n + 1)/2)^(p+1).
%F A119900 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.
%F A119900 There appears to be a similar result for the even power sums S(2*k,n) involving A207543. (End)
%e A119900 The binary word 1/0/01/01/1/1/01 has 7 strictly increasing runs.
%e A119900 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 /).
%e A119900 Triangle starts:
%e A119900   1;
%e A119900   0,2;
%e A119900   0,1,3;
%e A119900   0,0,4,4;
%e A119900   0,0,1,10,5;
%e A119900   0,0,0,6,20,6;
%p A119900 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
%t A119900 Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Aug 21 2016 *)
%o A119900 (PARI) for(n=0,10, for(k=0,n, print1(binomial(n+1, 2*k-n), ", "))) \\ _G. C. Greubel_, Oct 22 2017
%o A119900 (Magma) /* triangle */ [[Binomial(n+1, 2*k-n): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, Oct 22 2017
%Y A119900 Cf. A000079, A001906, A066373, A034867.
%Y A119900 Cf. A098158. - _Philippe Deléham_, Dec 02 2008
%Y A119900 Cf. A000292, A000389, A000580, A001263, A049310, A053122, A134264, A207543.
%K A119900 nonn,tabl
%O A119900 0,3
%A A119900 _Emeric Deutsch_, May 27 2006
%E A119900 Keyword tabl added by _Philippe Deléham_, Jan 26 2010