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%I #26 Sep 13 2018 04:59:03
%S 1,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,2,1,1,2,1,1,0,3,0,1,1,3,2,1,0,4,0,
%T 1,1,1,4,3,1,1,0,5,0,3,1,1,5,4,3,1,1,0,6,0,6,0,1,1,6,5,6,3,1,0,7,0,10,
%U 0,1,1,1,7,6,10,6,1,1,0,8,0,15,0,4,1,1,8,7,15,10,4,1,1,0,9,0,21,0,10,0,1,1,9,8,21,15,10,4,1,0,10,0,28,0,20,0,1,1,1,10,9,28,21,20,10,1,1,0,11,0,36,0,35,0,5
%N Irregular triangle read by rows. For n >= 3 and 1 <= k <= floor(n/3), T(n,k) is the number of palindromic compositions of n into k parts of size at least 3.
%F T(n,k) = 0 if n is odd and k is even;
%F T(n,k) = binomial((n-1)/2-k,(k-1)/2) if n is odd and k is odd;
%F T(n,k) = binomial((n-2)/2-k,(k-1)/2) if n is even and k is odd;
%F T(n,k) = binomial((n-2)/2-k,(k-2)/2) if n is even and k is even.
%e For n=24 and k=3, T(24,3) = 8 = binomial((24-2)/2-3, (3-1)/2) = binomial(8,1).
%e The first entries of the irregular triangle formed by the values of T(n,k) are:
%e 1;
%e 1;
%e 1;
%e 1, 1;
%e 1, 0;
%e 1, 1;
%e 1, 0, 1;
%e 1, 1, 1;
%e 1, 0, 2;
%e 1, 1, 2, 1;
%e 1, 0, 3, 0;
%e 1, 1, 3, 2;
%e 1, 0, 4, 0, 1;
%e 1, 1, 4, 3, 1;
%e 1, 0, 5, 0, 3;
%e 1, 1, 5, 4, 3, 1;
%e 1, 0, 6, 0, 6, 0;
%e 1, 1, 6, 5, 6, 3;
%e 1, 0, 7, 0, 10, 0, 1;
%e 1, 1, 7, 6, 10, 6, 1;
%e 1, 0, 8, 0, 15, 0, 4;
%e 1, 1, 8, 7, 15, 10, 4, 1;
%e 1, 0, 9, 0, 21, 0, 10, 0;
%e 1, 1, 9, 8, 21, 15, 10, 4;
%e 1, 0, 10, 0, 28, 0, 20, 0, 1;
%e 1, 1, 10, 9, 28, 21, 20, 10, 1;
%e 1, 0, 11, 0, 36, 0, 35, 0, 5;
%t T[n_, k_] := If[Mod[n, 2] == 1 && Mod[k, 2] == 0, 0, Binomial[Quotient[n-1, 2] - k, Quotient[k-1, 2]]];
%t Table[T[n, k], {n, 3, 30}, {k, 1, Quotient[n, 3]}] // Flatten (* _Jean-François Alcover_, Sep 13 2018, from PARI *)
%o (PARI) T(n,k)=if(n%2==1&&k%2==0, 0, binomial((n-1)\2-k, (k-1)\2)); \\ _Andrew Howroyd_, Sep 07 2018
%Y Row sums of the triangle equal A226916(n+4).
%K nonn,tabf
%O 3,18
%A _Christian Barrientos_ and _Sarah Minion_, Jul 29 2018