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A317489 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.

%I A317489 #26 Sep 13 2018 04:59:03
%S A317489 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 A317489 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 A317489 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 A317489 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 A317489 T(n,k) = 0 if n is odd and k is even;
%F A317489 T(n,k) = binomial((n-1)/2-k,(k-1)/2) if n is odd and k is odd;
%F A317489 T(n,k) = binomial((n-2)/2-k,(k-1)/2) if n is even and k is odd;
%F A317489 T(n,k) = binomial((n-2)/2-k,(k-2)/2) if n is even and k is even.
%e A317489 For n=24 and k=3, T(24,3) = 8 = binomial((24-2)/2-3, (3-1)/2) = binomial(8,1).
%e A317489 The first entries of the irregular triangle formed by the values of T(n,k) are:
%e A317489   1;
%e A317489   1;
%e A317489   1;
%e A317489   1,  1;
%e A317489   1,  0;
%e A317489   1,  1;
%e A317489   1,  0,  1;
%e A317489   1,  1,  1;
%e A317489   1,  0,  2;
%e A317489   1,  1,  2,  1;
%e A317489   1,  0,  3,  0;
%e A317489   1,  1,  3,  2;
%e A317489   1,  0,  4,  0,  1;
%e A317489   1,  1,  4,  3,  1;
%e A317489   1,  0,  5,  0,  3;
%e A317489   1,  1,  5,  4,  3,  1;
%e A317489   1,  0,  6,  0,  6,  0;
%e A317489   1,  1,  6,  5,  6,  3;
%e A317489   1,  0,  7,  0, 10,  0,  1;
%e A317489   1,  1,  7,  6, 10,  6,  1;
%e A317489   1,  0,  8,  0, 15,  0,  4;
%e A317489   1,  1,  8,  7, 15, 10,  4,  1;
%e A317489   1,  0,  9,  0, 21,  0, 10,  0;
%e A317489   1,  1,  9,  8, 21, 15, 10,  4;
%e A317489   1,  0, 10,  0, 28,  0, 20,  0,  1;
%e A317489   1,  1, 10,  9, 28, 21, 20, 10,  1;
%e A317489   1,  0, 11,  0, 36,  0, 35,  0,  5;
%t A317489 T[n_, k_] := If[Mod[n, 2] == 1 && Mod[k, 2] == 0, 0, Binomial[Quotient[n-1, 2] - k, Quotient[k-1, 2]]];
%t A317489 Table[T[n, k], {n, 3, 30}, {k, 1, Quotient[n, 3]}] // Flatten (* _Jean-François Alcover_, Sep 13 2018, from PARI *)
%o A317489 (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 A317489 Row sums of the triangle equal A226916(n+4).
%K A317489 nonn,tabf
%O A317489 3,18
%A A317489 _Christian Barrientos_ and _Sarah Minion_, Jul 29 2018