A105260 Triangle read by rows: T(n,k)=C(2n-2k,k), n>=0, 0<=k<=floor(2n/3).
1, 1, 1, 2, 1, 4, 1, 1, 6, 6, 1, 8, 15, 4, 1, 10, 28, 20, 1, 1, 12, 45, 56, 15, 1, 14, 66, 120, 70, 6, 1, 16, 91, 220, 210, 56, 1, 1, 18, 120, 364, 495, 252, 28, 1, 20, 153, 560, 1001, 792, 210, 8, 1, 22, 190, 816, 1820, 2002, 924, 120, 1, 1, 24, 231, 1140, 3060, 4368, 3003
Offset: 0
Examples
Triangle begins: 1; 1; 1,2; 1,4,1; 1,6,6; 1,8,15,4; Row n contains 1+floor(2n/3) terms.
References
- E. Deutsch, Math. Magazine, vol. 75, No. 3, 2002, p. 228, problem 1623.
Crossrefs
Row sums yield A002478.
Programs
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Maple
T:=(n,k)->binomial(2*n-2*k,k): for n from 0 to 14 do seq(T(n,k),k=0..floor(2*n/3)) od;# yields sequence in triangular form
Formula
T(n, k)=C(2n-2k, k), n>=0, 0<=k<=floor(2n/3). G.f.=1/[1-z(1+tz)^2].
T(n,k) = A102547(2*n,k). - R. J. Mathar, Aug 21 2016