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A233322
Triangle read by rows: T(n,k) = number of palindromic partitions of n in which no part exceeds k, 1 <= k <= n.
2
1, 1, 2, 1, 1, 2, 1, 3, 3, 4, 1, 2, 3, 3, 4, 1, 4, 5, 6, 6, 7, 1, 2, 5, 5, 6, 6, 7, 1, 5, 7, 10, 10, 11, 11, 12, 1, 3, 7, 8, 10, 10, 11, 11, 12, 1, 6, 9, 14, 15, 17, 17, 18, 18, 19, 1, 3, 9, 11, 15, 15, 17, 17, 18, 18, 19, 1, 7, 12, 20, 22, 26, 26, 28, 28, 29, 29, 30
OFFSET
1,3
COMMENTS
See A025065 for a definition of palindromic partition.
LINKS
FORMULA
T(n,k) = Sum_{i=1..k} A233321(n,i).
T(n,k) = Sum_{i=0..(k+2*floor(n/2)-n)/2} A026820(floor(n/2)-i, k). - Andrew Howroyd, Oct 09 2017
EXAMPLE
Triangle begins:
1;
1, 2;
1, 1, 2;
1, 3, 3, 4;
1, 2, 3, 3, 4;
1, 4, 5, 6, 6, 7;
1, 2, 5, 5, 6, 6, 7;
1, 5, 7, 10, 10, 11, 11, 12;
1, 3, 7, 8, 10, 10, 11, 11, 12;
1, 6, 9, 14, 15, 17, 17, 18, 18, 19;
1, 3, 9, 11, 15, 15, 17, 17, 18, 18, 19;
1, 7, 12, 20, 22, 26, 26, 28, 28, 29, 29, 30;
...
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}]; a233322[n_] := {}; Do[Do[AppendTo[a233322[n], Sum[a233321[n][[j]], {j, k}]], {k, n}], {n, nmax}]; Table[a233322[n], {n, nmax}](* L. Edson Jeffery, Oct 09 2017 *)
PROG
(PARI) \\ here PartitionCount is A026820.
PartitionCount(n, maxpartsize)={my(t=0); forpart(p=n, t++, maxpartsize); t}
T(n, k)=sum(i=0, (k-n%2)\2, PartitionCount(n\2-i, k));
for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 09 2017
CROSSREFS
Cf. A025065, A026820; partial sums of row entries of A233321.
Cf. A233323, A233324 (palindromic compositions of n).
Sequence in context: A289495 A076302 A104524 * A233324 A268679 A128807
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