A116595 Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts that appear exactly once (n>=0, k>=0).

1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 2, 4, 4, 2, 1, 2, 8, 4, 1, 6, 8, 6, 2, 5, 12, 10, 3, 9, 16, 12, 4, 1, 7, 23, 19, 6, 1, 16, 24, 25, 10, 2, 11, 40, 33, 14, 3, 22, 45, 41, 22, 5, 20, 59, 63, 27, 6, 1, 33, 72, 73, 42, 10, 1, 28, 99, 101, 53, 14, 2, 51, 108, 127, 75, 21, 3, 42, 153, 167

Offset: 0

Author

Emeric Deutsch, Feb 18 2006

Keywords

• nonn
• tabf

Comments

Row n has 1+floor([sqrt(1+8n)-1]/2) terms. Row sums yield the partition numbers (A000041). T(n,0)=A007690(n). Sum(k*T(n,k),k>=0)=A024786(n+1).

Examples

			T(7,2) = 4 because we have [6,1], [5,2], [4,3], [3,2,1,1].
Triangle starts:
1;
0, 1;
1, 1;
1, 1,  1;
2, 2,  1;
1, 4,  2;
4, 4,  2,  1;
2, 8,  4,  1;
6, 8,  6,  2;
5, 12, 10, 3;
9, 16, 12, 4, 1;
		

Links

• Alois P. Heinz, Rows n = 0..500, flattened

Crossrefs

Cf. A000041, A007690, A024786.

Programs

Maple

g:=product(1+t*x^j+x^(2*j)/(1-x^j),j=1..40): gser:=simplify(series(g,x=0,23)): P[0]:=1: for n from 1 to 21 do P[n]:=sort(coeff(gser,x^n)) od: for n from 0 to 21 do seq(coeff(P[n],t,j),j=0..floor((sqrt(1+8*n)-1)/2)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; local j; if n=0 then 1
      elif i<1 then 0 else []; for j from 0 to n/i do zip((x, y)
      ->x+y, %, [`if`(j=1, 0, [][]), b(n-i*j, i-1)], 0) od; %[] fi
    end:
T:= n-> b(n, n):
seq(T(n), n=0..30);  # Alois P. Heinz, Nov 07 2012

Mathematica

b[n_, i_] := b[n, i] = Module[{j, pc}, If[n == 0, pc = {1}, If[i<1, pc = {0}, pc = {}; For[j = 0, j <= n/i, j++, pc = Plus @@ PadRight[{pc, If[j == 1, {0}, {}] ~Join~ b[n-i*j, i-1]}]]; pc]]]; T[n_] := b[n, n]; Table[T[n], {n, 0, 30}] // Flatten (* Jean-François Alcover, Jan 31 2014, after Alois P. Heinz *)

Formula

G.f.: product(1+tx^j+x^(2j)/(1-x^j), j=1..infinity).

More generally, g.f. for the number of partitions of n having exactly k parts that appear exactly m times is product((t-1)*x^(m*j)+1/(1-x^j), j=1..infinity). - Vladeta Jovovic, Feb 21 2006