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
Examples
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; ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275
Programs
-
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 *) -
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
Extensions
Corrected row 7 as communicated by Andrew Howroyd. - L. Edson Jeffery, Oct 09 2017
Comments