A094305 Triangle read by rows: T(n,k) = ((n+1)(n+2)/2) * binomial(n,k) (0 <= k <= n).
1, 3, 3, 6, 12, 6, 10, 30, 30, 10, 15, 60, 90, 60, 15, 21, 105, 210, 210, 105, 21, 28, 168, 420, 560, 420, 168, 28, 36, 252, 756, 1260, 1260, 756, 252, 36, 45, 360, 1260, 2520, 3150, 2520, 1260, 360, 45, 55, 495, 1980, 4620, 6930, 6930, 4620, 1980, 495, 55, 66
Offset: 0
Examples
Triangle begins: 1 3 3 6 12 6 10 30 30 10 15 60 90 60 15 21 105 210 210 105 21 ... The n-th row is the product of the n-th triangular number and the n-th row of Pascal's triangle. The fifth row is (15,60,90,60,15) or 15*{1,4,6,4,1}.
References
- A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 152.
Links
- Reinhard Zumkeller, Rows n = 0..100 of table, flattened
- Mircea Merca, A Special Case of the Generalized Girard-Waring Formula, J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
Crossrefs
Programs
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Haskell
a094305 n k = a094305_tabl !! n !! k a094305_row n = a094305_tabl !! n a094305_tabl = zipWith (map . (*)) (tail a000217_list) a007318_tabl -- Reinhard Zumkeller, Jul 30 2013
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Maple
A094305:= proc(n,k) (n+1)*(n+2)/2 * binomial(n,k); end;
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Mathematica
nn=10; f[list_]:=Select[list,#>0&];a=1/(1-x-y x); Map[f,CoefficientList[Series[a^3,{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Apr 06 2013 *) Flatten[Table[((n+1)(n+2))/2 Binomial[n,k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 31 2014 *)
Formula
T(n,k) = Sum_{i=1..k+1} (-1)^(i+1)*i^2*binomial(n+2,k+i+1)*binomial(n+2,k-i+1). - Mircea Merca, Apr 05 2012
O.g.f.: 1/(1 - x - y*x)^3. - Geoffrey Critzer, Apr 06 2013
Extensions
Edited by Ralf Stephan, Feb 04 2005
Further comments from David Wasserman, Oct 04 2007
Further editing by N. J. A. Sloane, Oct 07 2007
Comments