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A119900
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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.
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15
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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T(n,k) = binomial(n+1,2k-n).
G.f.: 1/(1 - 2*t*z - t*(1-t)*z^2).
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
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)
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)
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EXAMPLE
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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;
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MAPLE
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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
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Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 21 2016 *)
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PROG
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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
(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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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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