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%I #48 Feb 07 2019 19:31:52
%S 1,1,1,1,2,1,1,3,2,1,1,4,3,3,1,1,5,4,6,3,1,1,6,5,10,6,4,1,1,7,6,15,10,
%T 10,4,1,1,8,7,21,15,20,10,5,1,1,9,8,28,21,35,20,15,5,1,1,10,9,36,28,
%U 56,35,35,15,6,1,1,11,10,45,36,84,56,70,35,21,6,1
%N Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.
%C This triangle is a companion to Parks' triangle A103631.
%C The coefficients of triangle A103631(n,k) appear in appendix 2 of Park’s remarkable article “A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov” if we assume that the b(n) coefficients are all equal to 1, see the second Maple program.
%C The a(n,k) coefficients of the triangle given above are related to the coefficients of a linear (n+1)-th order differential equation for the case b(n)=1, see the examples.
%C a(n,k) is also the number of symmetric binary strings of odd length n with Hamming weight k>0 and no consecutive 1's. - _Christian Barrientos_ and _Sarah Minion_, Feb 27 2018
%H Reinhard Zumkeller, <a href="/A194005/b194005.txt">Rows n = 0..150 of triangle, flattened</a>
%H Henry W. Gould, <a href="http://www.fq.math.ca/3-4.html"> A Variant of Pascal's Triangle </a>, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
%H P.C. Parks, <a href="https://doi.org/10.1017/S030500410004072X"> A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov </a>, Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694-702.
%H Chris Zheng, Jeffrey Zheng, <a href="https://doi.org/10.1007/978-981-13-2282-2_4">Triangular Numbers and Their Inherent Properties</a>, Variant Construction from Theoretical Foundation to Applications, Springer, Singapore, 51-65.
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F a(n,k) = binomial(floor((2*n+1-k)/2), n-k).
%F a(n,k) = sum(A103631(n1,k), n1=k..n), 0<=k<=n and n>=0.
%F a(n,k) = sum(binomial(floor((2*n1-k-1)/2), n1-k), n1=k..n).
%F T(n,0) = T(n,n) = 1, T(n,k) = T(n-2,k-2) + T(n-1,k), 0 < k < n. - _Reinhard Zumkeller_, Nov 23 2012
%e For the 5th-order linear differential equation the coefficients a(k) are: a(0) = 1, a(1) = a(4,0) = 1, a(2) = a(4,1) = 4, a(3) = a(4,2) = 3, a(4) = a(4,3) = 3 and a(5) = a(4,4) = 1.
%e The corresponding Hurwitz matrices A(k) are, see Parks: A(5) = Matrix([[a(1),a(0),0,0,0], [a(3),a(2),a(1),a(0),0], [a(5),a(4),a(3),a(2),a(1)], [0,0,a(5),a(4),a(3)], [0,0,0,0,a(5)]]), A(4) = Matrix([[a(1),a(0),0,0], [a(3),a(2),a(1),a(0)], [a(5),a(4),a(3),a(2)], [0,0,a(5),a(4)]]), A(3) = Matrix([[a(1),a(0),0], [a(3),a(2),a(1)], [a(5),a(4),a(3)]]), A(2) = Matrix([[a(1),a(0)], [a(3),a(2)]]) and A(1) = Matrix([[a(1)]]).
%e The values of b(k) are, see Parks: b(1) = d(1), b(2) = d(2)/d(1), b(3) = d(3)/(d(1)*d(2)), b(4) = d(1)*d(4)/(d(2)*d(3)) and b(5) = d(2)*d(5)/(d(3)*d(4)).
%e These a(k) values lead to d(k) = 1 and subsequently to b(k) = 1 and this confirms our initial assumption, see the comments.
%p A194005 := proc(n,k): binomial(floor((2*n+1-k)/2),n-k) end: for n from 0 to 11 do seq(A194005(n,k), k=0..n) od; seq(seq(A194005(n,k), k=0..n), n=0..11);
%p nmax:=11: for n from 0 to nmax+1 do b(n):=1 od: A103631 := proc(n,k) option remember: local j: if k=0 and n=0 then b(1) elif k=0 and n>=1 then 0 elif k=1 then b(n+1) elif k=2 then b(1)*b(n+1) elif k>=3 then expand(b(n+1)*add(procname(j,k-2), j=k-2..n-2)) fi: end: for n from 0 to nmax do for k from 0 to n do A194005(n,k):= add(A103631(n1,k), n1=k..n) od: od: seq(seq(A194005(n,k),k=0..n), n=0..nmax);
%t Flatten[Table[Binomial[Floor[(2n+1-k)/2],n-k],{n,0,20},{k,0,n}]] (* _Harvey P. Dale_, Apr 15 2012 *)
%o (Haskell)
%o a194005 n k = a194005_tabl !! n !! k
%o a194005_row n = a194005_tabl !! n
%o a194005_tabl = [1] : [1,1] : f [1] [1,1] where
%o f row' row = rs : f row rs where
%o rs = zipWith (+) ([0,1] ++ row') (row ++ [0])
%o -- _Reinhard Zumkeller_, Nov 22 2012
%Y Cf. A065941 and A103631.
%Y Triangle sums (see A180662): A000071 (row sums; alt row sums), A075427 (Kn22), A000079 (Kn3), A109222(n+1)-1 (Kn4), A000045 (Fi1), A034943 (Ca3), A001519 (Gi3), A000930 (Ze3)
%Y Interesting diagonals: T(n,n-4) = A189976(n+5) and T(n,n-5) = A189980(n+6)
%Y Cf. A052509.
%K nonn,easy,tabl
%O 0,5
%A _Johannes W. Meijer_ & A. Hirschberg (a.hirschberg(AT)tue.nl), Aug 11 2011
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