A194005 Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.

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

Offset: 0

Author

Johannes W. Meijer and A. Hirschberg (a.hirschberg(AT)tue.nl), Aug 11 2011

Keywords

• nonn
• easy
• tabl

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

Examples

			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;
		

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, pp. 257-271, with corrections.
• 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.
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Cf. A065941 and A103631.

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)

Interesting diagonals: T(n,n-4) = A189976(n+5) and T(n,n-5) = A189980(n+6)

Cf. A052509.

Programs

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

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
A194005(n,k):= add(A103631(n1,k), n1=k..n) od: od:
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 *)

Formula

T(n,k) = binomial(floor((2*n+1-k)/2), n-k).

T(n,k) = sum(A103631(n1,k), n1=k..n), 0<=k<=n and n>=0.

T(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