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A194005
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Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.
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9
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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, 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, 56, 35, 35, 15, 6, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
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OFFSET
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0,5
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COMMENTS
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This triangle is a companion to Parks' triangle A103631.
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.
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.
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
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LINKS
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FORMULA
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a(n,k) = binomial(floor((2*n+1-k)/2), n-k).
a(n,k) = sum(A103631(n1,k), n1=k..n), 0<=k<=n and n>=0.
a(n,k) = sum(binomial(floor((2*n1-k-1)/2), n1-k), n1=k..n).
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
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EXAMPLE
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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.
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)]]).
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)).
These a(k) values lead to d(k) = 1 and subsequently to b(k) = 1 and this confirms our initial assumption, see the comments.
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MAPLE
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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);
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);
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MATHEMATICA
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Flatten[Table[Binomial[Floor[(2n+1-k)/2], n-k], {n, 0, 20}, {k, 0, n}]] (* Harvey P. Dale, Apr 15 2012 *)
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PROG
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(Haskell)
a194005 n k = a194005_tabl !! n !! k
a194005_row n = a194005_tabl !! n
a194005_tabl = [1] : [1, 1] : f [1] [1, 1] where
f row' row = rs : f row rs where
rs = zipWith (+) ([0, 1] ++ row') (row ++ [0])
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CROSSREFS
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Interesting diagonals: T(n,n-4) = A189976(n+5) and T(n,n-5) = A189980(n+6)
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KEYWORD
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AUTHOR
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STATUS
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approved
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