A103631 Triangle read by rows: T(n,k) = abs(qStirling2(n,k,q)) for q = -1, with 0 <= k <= n.
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
Offset: 0
Examples
From _Paul Barry_, 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)
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) pp. 694-702.
Crossrefs
Cf. A103633 (signed version).
Programs
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Haskell
a103631 n k = a103631_tabl !! n !! k a103631_row n = a103631_tabl !! n a103631_tabl = [1] : [0,1] : f [1] [0,1] where f xs ys = zs : f ys zs where zs = zipWith (+) ([0,0] ++ xs) (ys ++ [0]) -- Reinhard Zumkeller, May 07 2012 -
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
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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
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 *) Table[Binomial[Floor[(2*n - k - 1)/2], n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 27 2016 *) 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]; qStirling2[n_, 0, _] := KroneckerDelta[n, 0]; qStirling2[0, k_, _] := KroneckerDelta[0, k]; qStirling2[, , _] = 0; Table[Abs[qStirling2[n, k, -1]], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *) -
Sage
from sage.combinat.q_analogues import q_stirling_number2 for n in (0..9): print([abs(q_stirling_number2(n,k).substitute(q=-1)) for k in [0..n]]) # Peter Luschny, Mar 09 2020
Formula
T(n,k) = binomial(floor((2*n-k-1)/2), n-k).
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
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
G.f.: (1+(y-1)*x)/(1-x-y^2*x^2). - Philippe Deléham, Mar 09 2012
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
Extensions
New name from Peter Luschny, Mar 09 2020
Comments