|
%I #6 Aug 21 2016 15:51:01
%S 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,
%T 70,6,1,16,91,220,210,56,1,1,18,120,364,495,252,28,1,20,153,560,1001,
%U 792,210,8,1,22,190,816,1820,2002,924,120,1,1,24,231,1140,3060,4368,3003
%N Triangle read by rows: T(n,k)=C(2n-2k,k), n>=0, 0<=k<=floor(2n/3).
%D E. Deutsch, Math. Magazine, vol. 75, No. 3, 2002, p. 228, problem 1623.
%F T(n, k)=C(2n-2k, k), n>=0, 0<=k<=floor(2n/3). G.f.=1/[1-z(1+tz)^2].
%F T(n,k) = A102547(2*n,k). - _R. J. Mathar_, Aug 21 2016
%e Triangle begins:
%e 1;
%e 1;
%e 1,2;
%e 1,4,1;
%e 1,6,6;
%e 1,8,15,4;
%e Row n contains 1+floor(2n/3) terms.
%p 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
%Y Row sums yield A002478.
%K nonn,tabf
%O 0,4
%A _Emeric Deutsch_, Apr 14 2005
|
