A268193 Triangle read by rows: T(n,k) (n>=1, k>=0) is the number of partitions of n which have k distinct parts i such that i+1 is also a part.

1, 2, 2, 1, 4, 1, 4, 3, 8, 2, 1, 8, 6, 1, 13, 7, 2, 15, 11, 4, 22, 15, 4, 1, 24, 24, 7, 1, 37, 26, 12, 2, 40, 42, 16, 3, 57, 50, 22, 6, 64, 72, 33, 6, 1, 89, 84, 46, 11, 1, 98, 122, 60, 15, 2, 135, 141, 82, 24, 3, 149, 198, 106, 32, 5, 199, 231, 144, 45, 8, 224, 309, 187, 61, 10, 1

Offset: 1

Author

Emeric Deutsch, Feb 13 2016

Keywords

• nonn
• look
• tabf

Comments

T(n,k) = number of partitions of n having k singleton parts other than the largest part. Example: T(5,1) = 3 because we have [4,1'], [3,2'], [2,2,1'] (the counted singletons are marked). These partitions are connected by conjugation to those in the definition.

From Gus Wiseman, Jul 10 2025: (Start)

Also the number of integer partitions of n with k maximal subsequences of consecutive parts not decreasing by 1 (anti-runs). For example, row n = 8 counts partitions with the following anti-runs:

((8)) ((3,3),(2)) ((3),(2,2),(1))

((4,4)) ((4),(3,1)) ((3),(2),(1,1,1))

((5,3)) ((5,2),(1))

((6,2)) ((4,2),(1,1))

((7,1)) ((2,2,2),(1,1))

((4,2,2)) ((2,2),(1,1,1,1))

((6,1,1)) ((2),(1,1,1,1,1,1))

((2,2,2,2))

((3,3,1,1))

((5,1,1,1))

((4,1,1,1,1))

((3,1,1,1,1,1))

((1,1,1,1,1,1,1,1))

(End)

Examples

			T(5,1) = 3 because we have [3,2], [2,2,1], and [2,1,1,1].
T(9,2) = 4 because we have [3,2',1,1,1,1'], [3,2,2',1,1'], [3,3,2',1'], and [4,3',2'] (the i's are marked).
Triangle starts:
  1;
  2;
  2,1;
  4,1;
  4,3;
  8,2,1;
  8,6,1;
From _Gus Wiseman_, Jul 11 2025: (Start)
Row n = 8 counts the following partitions by number of singleton parts other than the largest part:
  (8)                (5,3)        (4,3,1)
  (4,4)              (6,2)        (5,2,1)
  (4,2,2)            (7,1)
  (6,1,1)            (3,3,2)
  (2,2,2,2)          (3,2,2,1)
  (3,3,1,1)          (4,2,1,1)
  (5,1,1,1)          (3,2,1,1,1)
  (2,2,2,1,1)
  (4,1,1,1,1)
  (2,2,1,1,1,1)
  (3,1,1,1,1,1)
  (2,1,1,1,1,1,1)
  (1,1,1,1,1,1,1,1)
(End)
		

Links

• Alois P. Heinz, Rows n = 1..800, flattened

Crossrefs

Row sums are A000041.

Row lengths are A003056.

For distinct parts instead of anti-runs we have A116608.

Column k = 1 is A116931.

For runs instead of anti-runs we have A384881.

The strict case is A384905.

The corresponding rank statistic is A356228, non-strict version A384906.

The proper case is A385814, runs A385815.

A007690 counts partitions with no singletons, complement A183558.

A034296 counts flat or gapless partitions, ranks A066311 or A073491.

Cf. A000009, A008284, A024786, A047993, A098859, A116674, A287170, A325325, A384882, A385213.

Programs

Maple

g := add(x^j*mul(1+t*x^i+x^(2*i)/(1-x^i), i = 1 .. j-1)/(1-x^j), j = 1 .. 80): gser := simplify(series(g, x = 0, 27)): for n from 0 to 25 do P[n] := sort(coeff(gser, x, n)) end do: for n to 25 do seq(coeff(P[n], t, k), k = 0 .. degree(P[n])) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i, t) option remember; expand(`if`(n=0, 1,
      `if`(i<1, 0, add(b(n-i*j, i-1, t or j>0)*
      `if`(t and j=1, x, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, false)):
seq(T(n), n=1..20);  # Alois P. Heinz, Feb 13 2016

Mathematica

b[n_, i_, t_] := b[n, i, t] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1, t || j > 0]*If[t && j == 1, x, 1], {j, 0, n/i}]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, False]]; Table[T[n], {n, 1, 20}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)
Table[Length[Select[IntegerPartitions[n],Length[Split[#,#1!=#2+1&]]==k&]],{n,0,10},{k,0,n}] (* Delete zeros for A268193. Gus Wiseman, Jul 10 2025 *)

Formula

T(n,0) = A116931(n).

Sum_{k>=1} T(n, k) = A000041(n) (the partition numbers).

Sum_{k>=1} k*T(n,k) = A024786(n-1).

G.f.: G(t,x) = Sum_{j>=1} ((x^j/(1-x^j))*Product_{i=1..j-1} (1 + tx^i + x^{2i}/(1-x^i))).