OFFSET
0,5
COMMENTS
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
LINKS
Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle , The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
P.C. Parks, A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694-702.
Chris Zheng, Jeffrey Zheng, Triangular Numbers and Their Inherent Properties, Variant Construction from Theoretical Foundation to Applications, Springer, Singapore, 51-65.
FORMULA
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
EXAMPLE
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.
'
Triangle starts:
[0] 1;
[1] 1, 1;
[2] 1, 2, 1;
[3] 1, 3, 2, 1;
[4] 1, 4, 3, 3, 1;
[5] 1, 5, 4, 6, 3, 1;
[6] 1, 6, 5, 10, 6, 4, 1;
[7] 1, 7, 6, 15, 10, 10, 4, 1;
[8] 1, 8, 7, 21, 15, 20, 10, 5, 1;
[9] 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
MAPLE
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
seq(seq(A194005(n, k), k=0..n), n=0..nmax);
MATHEMATICA
Flatten[Table[Binomial[Floor[(2n+1-k)/2], n-k], {n, 0, 20}, {k, 0, n}]] (* Harvey P. Dale, Apr 15 2012 *)
PROG
(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])
-- Reinhard Zumkeller, Nov 22 2012
CROSSREFS
KEYWORD
AUTHOR
Johannes W. Meijer & A. Hirschberg (a.hirschberg(AT)tue.nl), Aug 11 2011
STATUS
approved