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
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
Programs
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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]) -- 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);
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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
Comments