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A103631
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Triangle read by rows: an invertible triangle whose row sums are F(n+1).
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9
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1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 2, 1, 0, 1, 1, 4, 3, 3, 1, 0, 1, 1, 5, 4, 6, 3, 1, 0, 1, 1, 6, 5, 10, 6, 4, 1, 0, 1, 1, 7, 6, 15, 10, 10, 4, 1, 0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 0, 1, 1, 11
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,14
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COMMENTS
| Triangle inverse has general term (-1)^(n-k)*binomial(floor(n/2),n-k)}. Diagonal sums are A103632.
Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 08 2005
Row sums are Fibonacci numbers (A000045).
Another version of triangle in A065941. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jan 01 2009]
From Johannes W. Meijer, Aug 11 2011: (Start)
The T(n,k) coefficients 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(r) coefficients are all equal to 1; see the second Maple program.
The T(n,k) triangle is related to a linear (n+1)-th order differential equation with coefficients a(n,k), see triangle A194005.
Park’s triangle appears to be an appropriate name for the triangle given above. (End)
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LINKS
| 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.
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FORMULA
| T(n,k) = binomial(floor((2*n-k-1)/2), n-k)
Sum_{k, 0<=k<=n}T(n,k)*x^k = A152163(n), A000007(n), A000045(n+1), A026597(n), A122994(n+1), A158608(n), A122995(n+1), A158797(n), A122996(n+1), A158798(n), A158609(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jun 12 2009]
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EXAMPLE
| Rows begin {1}, {0,1}, {0,1,1}, {0,1,1,1}, {0,1,1,2,1},..
Contribution from Paul Barry (pbarry(AT)wit.ie), Oct 02 2009: (Start)
Triangle begins
1,
0, 1,
0, 1, 1,
0, 1, 1, 1,
0, 1, 1, 2, 1,
0, 1, 1, 3, 2, 1,
0, 1, 1, 4, 3, 3, 1,
0, 1, 1, 5, 4, 6, 3, 1,
0, 1, 1, 6, 5, 10, 6, 4, 1,
0, 1, 1, 7, 6, 15, 10, 10, 4, 1
Production matrix is
0, 1,
0, 1, 1,
0, 0, 0, 1,
0, 0, 0, 1, 1,
0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1 (End)
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MAPLE
| From Johannes W. Meijer, Aug 11 2011: (Start)
A103631 := proc(n, k): binomial(floor((2*n-k-1)/2), n-k) end: seq(seq(A103631(n, k), k=0..n), n=0..12);
nmax:=12: 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 seq(A103631(n, k), k=0..n) od: seq(seq(A103631(n, k), k=0..n), n=0..nmax); # (End)
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MATHEMATICA
| A polynomial recursion which produces this triangle: p(x, n) = p(x, n - 1) + x^2*p(x, n - 2). - Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Apr 27 2008
p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = x; p[x, 2] = x + x^2; p[x_, n_] := p[x, n] = p[x, n - 1] + x^2*p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] - Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Apr 27 2008
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CROSSREFS
| Cf. A103633.
Sequence in context: A131255 A198295 A133607 * A192517 A083856 A081718
Adjacent sequences: A103628 A103629 A103630 * A103632 A103633 A103634
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Feb 11 2005
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