A233321 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.

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, 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, 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

Offset

1,8

Comments

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).

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Example

Triangle begins:
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, 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;
...

Mathematica

(* run this first: *)
Needs["Combinatorica`"];
(* run the following in a different cell: *)
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 *)

Prog

(PARI) \\ here V(n, k) is A233322.
PartitionCount(n, maxpartsize)={my(t=0); forpart(p=n, t++, maxpartsize); t}
V(n, k)=sum(i=0, (k-n%2)\2, PartitionCount(n\2-i, k));
T(n, k)=V(n, k)-V(n, k-1);
for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 09 2017

Crossrefs

Cf. A025065 (row sums), A233322.

Cf. A233323-A233324 (palindromic compositions of n).

Sequence in context: A373335 A178798 A318277 * A233323 A115381 A115382

Adjacent sequences: A233318 A233319 A233320 * A233322 A233323 A233324

Keyword

nonn,tabl

Author

L. Edson Jeffery, Dec 10 2013

Extensions

Corrected row 7 as communicated by Andrew Howroyd. - L. Edson Jeffery, Oct 09 2017

Status

approved