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%I #61 Jan 30 2022 17:34:18
%S 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,
%T 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,
%U 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
%N Triangle read by rows: T(n,k) = abs(qStirling2(n,k,q)) for q = -1, with 0 <= k <= n.
%C Previous name: An invertible triangle whose row sums are F(n+1).
%C Triangle inverse has general term (-1)^(n-k)*binomial(floor(n/2),n-k). Diagonal sums are A103632.
%C 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 Deléham_, Oct 08 2005
%C Row sums are Fibonacci numbers (A000045).
%C Another version of triangle in A065941. - _Philippe Deléham_, Jan 01 2009
%C From _Johannes W. Meijer_, Aug 11 2011: (Start)
%C The T(n,k) coefficients appear in appendix 2 of Parks'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.
%C The T(n,k) triangle is related to a linear (n+1)-th order differential equation with coefficients a(n,k), see triangle A194005.
%C Parks's triangle appears to be an appropriate name for the triangle given above. (End)
%H Reinhard Zumkeller, <a href="/A103631/b103631.txt">Rows n = 0..150 of triangle, flattened</a>
%H Henry W. Gould, <a href="http://www.fq.math.ca/Scanned/3-4/gould.pdf">A Variant of Pascal's Triangle</a>, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271.
%H P. C. Parks, <a href="http://dx.doi.org/10.1017/S030500410004072X"> A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov </a>, Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) pp. 694-702.
%F T(n,k) = binomial(floor((2*n-k-1)/2), n-k).
%F A polynomial recursion which produces this triangle: p(x, n) = p(x, n - 1) + x^2*p(x, n - 2). - _Roger L. Bagula_, Apr 27 2008
%F Sum_{k=0..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. - _Philippe Deléham_, Jun 12 2009
%F G.f.: (1+(y-1)*x)/(1-x-y^2*x^2). - _Philippe Deléham_, Mar 09 2012
%F T(n,k) = T(n-1,k) + T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - _Philippe Deléham_, Mar 09 2012
%e From _Paul Barry_, Oct 02 2009: (Start)
%e Triangle begins:
%e 1,
%e 0, 1,
%e 0, 1, 1,
%e 0, 1, 1, 1,
%e 0, 1, 1, 2, 1,
%e 0, 1, 1, 3, 2, 1,
%e 0, 1, 1, 4, 3, 3, 1,
%e 0, 1, 1, 5, 4, 6, 3, 1,
%e 0, 1, 1, 6, 5, 10, 6, 4, 1,
%e 0, 1, 1, 7, 6, 15, 10, 10, 4, 1
%e Production matrix is:
%e 0, 1,
%e 0, 1, 1,
%e 0, 0, 0, 1,
%e 0, 0, 0, 1, 1,
%e 0, 0, 0, 0, 0, 1,
%e 0, 0, 0, 0, 0, 1, 1,
%e 0, 0, 0, 0, 0, 0, 0, 1,
%e 0, 0, 0, 0, 0, 0, 0, 1, 1,
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 (End)
%p From _Johannes W. Meijer_, Aug 11 2011: (Start)
%p 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);
%p 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)
%t 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]; (* with *) Table[ExpandAll[p[x, n]], {n, 0, 10}]; (* or *) a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] (* _Roger L. Bagula_, Apr 27 2008 *)
%t Table[Binomial[Floor[(2*n - k - 1)/2], n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Aug 27 2016 *)
%t qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k - 1) qStirling2[n - 1, k - 1, q] + Sum[q^j, {j, 0, k - 1}] qStirling2[n - 1, k, q];
%t qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
%t qStirling2[0, k_, _] := KroneckerDelta[0, k];
%t qStirling2[_, _, _] = 0;
%t Table[Abs[qStirling2[n, k, -1]], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 10 2020 *)
%o (Haskell)
%o a103631 n k = a103631_tabl !! n !! k
%o a103631_row n = a103631_tabl !! n
%o a103631_tabl = [1] : [0,1] : f [1] [0,1] where
%o f xs ys = zs : f ys zs where
%o zs = zipWith (+) ([0,0] ++ xs) (ys ++ [0])
%o -- _Reinhard Zumkeller_, May 07 2012
%o (Magma) /* As triangle: */ [[Binomial(Floor((2*n-k-1)/2), n-k): k in [0..n]]: n in [0..15]]; // _Vincenzo Librandi_, Aug 28 2016
%o (Sage)
%o from sage.combinat.q_analogues import q_stirling_number2
%o for n in (0..9):
%o print([abs(q_stirling_number2(n,k).substitute(q=-1)) for k in [0..n]])
%o # _Peter Luschny_, Mar 09 2020
%Y Cf. A103633 (signed version).
%K nonn,tabl,easy
%O 0,14
%A _Paul Barry_, Feb 11 2005
%E New name from _Peter Luschny_, Mar 09 2020
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