A385001 Irregular triangle read by rows: T(n,k) is the number of partitions of n with k designated summands, n >= 0, 0 <= k <= A003056(n).

1, 0, 1, 0, 3, 0, 4, 1, 0, 7, 3, 0, 6, 9, 0, 12, 15, 1, 0, 8, 30, 3, 0, 15, 45, 9, 0, 13, 67, 22, 0, 18, 99, 42, 1, 0, 12, 135, 81, 3, 0, 28, 175, 140, 9, 0, 14, 231, 231, 22, 0, 24, 306, 351, 51, 0, 24, 354, 551, 97, 1, 0, 31, 465, 783, 188, 3, 0, 18, 540, 1134, 330, 9

Offset: 0

Author

Omar E. Pol, Jul 17 2025

Keywords

• nonn
• look
• tabf

Comments

The divisor function sigma_1(n) = A000203(n) is also the number of partitions of n with only one designated summand, n >= 1.

When part i has multiplicity j > 0 exactly one part i is "designated".

The length of the row n is A002024(n+1) = 1 + A003056(n), hence the first positive element in column k is in the row A000217(k).

Alternating row sums give A329157.

Columns converge to A000716.

This triangle equals A060043 with reversed rows and an additional column 0.

Examples

			Triangle begins:
--------------------------------------------
   n\k:   0    1     2     3     4    5   6
--------------------------------------------
   0 |    1;
   1 |    0,   1;
   2 |    0,   3;
   3 |    0,   4,    1;
   4 |    0,   7,    3;
   5 |    0,   6,    9;
   6 |    0,  12,   15,    1;
   7 |    0,   8,   30,    3;
   8 |    0,  15,   45,    9;
   9 |    0,  13,   67,   22;
  10 |    0,  18,   99,   42,    1;
  11 |    0,  12,  135,   81,    3;
  12 |    0,  28,  175,  140,    9;
  13 |    0,  14,  231,  231,   22;
  14 |    0,  24,  306,  351,   51;
  15 |    0,  24,  354,  551,   97,   1;
  16 |    0,  31,  465,  783,  188,   3;
  17 |    0,  18,  540, 1134,  330,   9;
  18 |    0,  39,  681, 1546,  568,  22;
  19 |    0,  20,  765, 2142,  918,  51;
  20 |    0,  42,  945, 2835, 1452, 108;
  21 |    0,  32, 1040, 3758, 2233, 208,  1;
  ...
For n = 6 and k = 1 there are 12 partitions of 6 with only one designated summand as shown below:
   6'
   3'+ 3
   3 + 3'
   2'+ 2 + 2
   2 + 2'+ 2
   2 + 2 + 2'
   1'+ 1 + 1 + 1 + 1 + 1
   1 + 1'+ 1 + 1 + 1 + 1
   1 + 1 + 1'+ 1 + 1 + 1
   1 + 1 + 1 + 1'+ 1 + 1
   1 + 1 + 1 + 1 + 1'+ 1
   1 + 1 + 1 + 1 + 1 + 1'
So T(6,1) = 12, the same as A000203(6) = 12.
.
For n = 6 and k = 2 there are 15 partitions of 6 with two designated summands as shown below:
   5'+ 1'
   4'+ 2'
   4'+ 1'+ 1
   4'+ 1 + 1'
   3'+ 1'+ 1 + 1
   3'+ 1 + 1'+ 1
   3'+ 1 + 1 + 1'
   2'+ 2 + 1'+ 1
   2'+ 2 + 1 + 1'
   2 + 2'+ 1'+ 1
   2 + 2'+ 1 + 1'
   2'+ 1'+ 1 + 1 + 1
   2'+ 1 + 1'+ 1 + 1
   2'+ 1 + 1 + 1'+ 1
   2'+ 1 + 1 + 1 + 1'
So T(6,2) = 15, the same as A002127(6) = 15.
.
For n = 6 and k = 3 there is only one partition of 6 with three designated summands as shown below:
   3'+ 2'+ 1'
So T(6,3) = 1, the same as A002128(6) = 1.
There are 28 partitions of 6 with designated summands, so A077285(6) = 28.
.
		

Links

• Alois P. Heinz, Rows n = 0..1000, flattened

Crossrefs

Columns: A000007 (k=0), A000203 (k=1), A002127 (k=2), A002128 (k=3), A365664 (k=4), A365665 (k=5), A384926 (k=6).

Row sums give A077285.

Cf. A000217, A002024, A003056, A060043, A196020, A252117, A329157, A365434, A384999.

Cf. A000096, A000716, A116608, A293421, A384998, A384999.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1)+add(expand(b(n-i*j, i-1)*j*x), j=1..n/i)))
    end:
T:= n-> (p-> seq(coeff(p,x,i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..20);  # Alois P. Heinz, Jul 18 2025

Formula

From Alois P. Heinz, Jul 18 2025: (Start)

Sum_{k>=1} k * T(n,k) = A293421(n).

T(A000096(n),n) = A000716(n). (End)

G.f.: Product_{i>0} 1 + (y*x^i)/(1 - x^i)^2. - John Tyler Rascoe, Jul 23 2025

Conjecture: For fixed k >= 1, Sum_{j=1..n} T(j,k) ~ Pi^(2*k) * n^(2*k) / ((2*k)! * (2*k+1)!). - Vaclav Kotesovec, Aug 01 2025