A096144 Triangle T(n,k) = number of partitions of n in which the least part occurs exactly k times, k=1..n.

1, 1, 1, 2, 0, 1, 2, 2, 0, 1, 4, 1, 1, 0, 1, 4, 3, 2, 1, 0, 1, 7, 3, 2, 1, 1, 0, 1, 8, 6, 2, 3, 1, 1, 0, 1, 12, 5, 6, 2, 2, 1, 1, 0, 1, 14, 11, 5, 4, 3, 2, 1, 1, 0, 1, 21, 11, 8, 5, 4, 2, 2, 1, 1, 0, 1, 24, 17, 11, 9, 4, 5, 2, 2, 1, 1, 0, 1, 34, 20, 15, 9, 8, 4, 4, 2, 2, 1, 1, 0, 1, 41, 30, 18, 14, 9, 7, 5, 4, 2, 2, 1, 1, 0, 1

Offset: 1

Author

Vladeta Jovovic, Jul 24 2004

Keywords

• easy
• nonn
• tabl

Comments

Reversed rows converge to A002865. - Joerg Arndt, Jul 07 2014

T(n,k) is the number of partitions of n for which the difference between the two largest distinct parts is k (in partitions having only 1 part, we assume that 0 is also a part). This follows easily from the definition by taking the conjugate partitions. Example: T(7,2) = 3 because we have [3,1,1,1,1], [3,3,1], and [4,2,1]. - Emeric Deutsch, Dec 05 2015.

Examples

			Triangle starts:
01:  1
02:  1 1
03:  2 0 1
04:  2 2 0 1
05:  4 1 1 0 1
06:  4 3 2 1 0 1
07:  7 3 2 1 1 0 1
08:  8 6 2 3 1 1 0 1
09: 12 5 6 2 2 1 1 0 1
10: 14 11 5 4 3 2 1 1 0 1
11: 21 11 8 5 4 2 2 1 1 0 1
12: 24 17 11 9 4 5 2 2 1 1 0 1
13: 34 20 15 9 8 4 4 2 2 1 1 0 1
14: 41 30 18 14 9 7 5 ...
T(7,2)=3 because we have: 5+1+1, 3+2+2, 3+2+1+1. - _Geoffrey Critzer_, Jun 20 2014
		

Links

• Alois P. Heinz, Rows n = 1..141, flattened

Crossrefs

Cf. A002865 (first column), A096373 (second column), A000041 (row sums).

T(2n,n) gives A232697(n). - Alois P. Heinz, Jun 20 2014

Programs

Maple

b:= proc(n, i) option remember; `if`(i=1, x^n,
      `if`(irem(n, i, 'k')=0, x^k, 0)+
       add(b(n-i*j, i-1), j=0..(n-1)/i))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)):
seq(T(n), n=1..20);  # Alois P. Heinz, Jun 20 2014

Mathematica

nn=20;Table[Take[Map[Drop[#,1]&,Drop[CoefficientList[Series[Sum[y x^k/(1-y x^k) Product[1/(1- x^j),{j,k+1,nn}],{k,1,nn}],{x,0,nn}],{x,y}],1]][[i]],i],{i,1,nn}]//Grid (* Geoffrey Critzer, Jun 20 2014 *)

Formula

G.f. for k-th column: sum(m>=1, x^(k*m)/prod(i>=m+1, 1-x^i ) ).