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%I #30 Oct 10 2017 12:08:34
%S 1,1,1,1,0,1,1,2,0,1,1,1,1,0,1,1,3,1,1,0,1,1,1,3,0,1,0,1,1,4,2,3,0,1,
%T 0,1,1,2,4,1,2,0,1,0,1,1,5,3,5,1,2,0,1,0,1,1,2,6,2,4,0,2,0,1,0,1,1,6,
%U 5,8,2,4,0,2,0,1,0,1,1,3,8,3,7,1,3,0,2,0,1,0,1,1,7,7,11,4,7,1,3,0,2,0,1,0,1
%N Triangle read by rows: T(n,k) = number of palindromic partitions of n in which the largest part is equal to k, 1 <= k <= n.
%C A partition of n is said to be "palindromic" if its parts can be arranged to form a palindrome in at least one way (cf. A025065).
%H Andrew Howroyd, <a href="/A233321/b233321.txt">Table of n, a(n) for n = 1..1275</a>
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 0, 1;
%e 1, 2, 0, 1;
%e 1, 1, 1, 0, 1;
%e 1, 3, 1, 1, 0, 1;
%e 1, 1, 3, 0, 1, 0, 1;
%e 1, 4, 2, 3, 0, 1, 0, 1;
%e 1, 2, 4, 1, 2, 0, 1, 0, 1;
%e 1, 5, 3, 5, 1, 2, 0, 1, 0, 1;
%e 1, 2, 6, 2, 4, 0, 2, 0, 1, 0, 1;
%e ...
%t (* run this first: *)
%t Needs["Combinatorica`"];
%t (* run the following in a different cell: *)
%t a233321[n_] := {}; Do[Do[a = Partitions[n]; count = 0; Do[If[Max[a[[j]]] == k, x = Permutations[a[[j]]]; Do[If[x[[m]] == Reverse[x[[m]]], count++; Break[]], {m, Length[x]}]], {j, Length[a]}]; AppendTo[a233321[n], count], {k, n}], {n, nmax}]; Table[a233321[n], {n, nmax}](* _L. Edson Jeffery_, Oct 09 2017 *)
%o (PARI) \\ here V(n,k) is A233322.
%o PartitionCount(n,maxpartsize)={my(t=0); forpart(p=n, t++, maxpartsize); t}
%o V(n,k)=sum(i=0, (k-n%2)\2, PartitionCount(n\2-i, k));
%o T(n,k)=V(n,k)-V(n,k-1);
%o for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ _Andrew Howroyd_, Oct 09 2017
%Y Cf. A025065 (row sums), A233322.
%Y Cf. A233323-A233324 (palindromic compositions of n).
%K nonn,tabl
%O 1,8
%A _L. Edson Jeffery_, Dec 10 2013
%E Corrected row 7 as communicated by _Andrew Howroyd_. - _L. Edson Jeffery_, Oct 09 2017
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