A105260 - OEIS

%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