A384881 Triangle read by rows where T(n,k) is the number of integer partitions of n with k maximal runs of consecutive parts decreasing by 1.

1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 1, 3, 0, 1, 0, 2, 2, 2, 0, 1, 0, 2, 3, 3, 2, 0, 1, 0, 2, 5, 3, 2, 2, 0, 1, 0, 1, 8, 4, 4, 2, 2, 0, 1, 0, 3, 5, 10, 4, 3, 2, 2, 0, 1, 0, 2, 9, 9, 9, 5, 3, 2, 2, 0, 1, 0, 2, 11, 13, 9, 9, 4, 3, 2, 2, 0, 1

Offset: 0

Author

Gus Wiseman, Jun 25 2025

Keywords

• nonn
• easy
• tabl

Examples

			The partition (5,4,2,1,1) has maximal runs ((5,4),(2,1),(1)) so is counted under T(13,3) = 23.
Row n = 9 counts the following partitions:
  9    63    333    6111    33111   411111   3111111   111111111
  54   72    441    22221   51111   2211111  21111111
  432  81    522    42111   222111
       621   531    321111
       3321  711
             3222
             4221
             4311
             5211
             32211
Triangle begins:
  1
  0  1
  0  1  1
  0  2  0  1
  0  1  3  0  1
  0  2  2  2  0  1
  0  2  3  3  2  0  1
  0  2  5  3  2  2  0  1
  0  1  8  4  4  2  2  0  1
  0  3  5 10  4  3  2  2  0  1
  0  2  9  9  9  5  3  2  2  0  1
  0  2 11 13  9  9  4  3  2  2  0  1
  0  2 13 15 17  8 10  4  3  2  2  0  1
  0  2 14 23 16 17  8  9  4  3  2  2  0  1
  0  2 16 26 26 19 16  9  9  4  3  2  2  0  1
  0  4 13 37 32 26 19 16  8  9  4  3  2  2  0  1
		

Links

• John Tyler Rascoe, Rows n = 0..100, flattened

Crossrefs

Row sums are A000041.

Column k = 1 is A001227.

For distinct parts instead of maximal runs we have A116608.

The strict case appears to be A116674.

For anti-runs instead of runs we have A268193.

Partitions with distinct runs of this type are counted by A384882, gapless A384884.

For prime indices see A385213, A287170, A066205, A356229.

A007690 counts partitions with no singletons, complement A183558.

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

Cf. A000009, A000217, A008284, A047966, A047993, A098859, A325325, A375136, A384887.

Programs

Mathematica

Table[Length[Select[IntegerPartitions[n],Length[Split[#,#1==#2+1&]]==k&]],{n,0,10},{k,0,n}]

PARI

tri(n) = {(n*(n+1)/2)}
B_list(N) = {my(v = vector(N, i, 0)); v[1] = q*t; for(m=2,N, v[m] = t * (q^tri(m) + sum(i=1,m-1, q^tri(i) * v[m-i] * (q^((m-i)*(i-1))/(1 - q^(m-i)) - q^((m-i)*i) + O('q^(N-tri(i)+1)))))); v}
A_qt(max_row) = {my(N = max_row+1, B = B_list(N), g = 1 + sum(m=1,N, B[m]/(1 - q^m)) + O('q^(N+1))); vector(N, n, Vecrev(polcoeff(g, n-1)))} \\ John Tyler Rascoe, Aug 18 2025

Formula

G.f.: 1 + Sum_{m>0} B(m,q,t)/(1 - q^m) where B(m,q,t) = t * (q^tri(m) + Sum_{i=1..m-1} q^tri(i) * B(m-i,q,t) * ((q^((m-i)*(i-1))/(1 - q^(m-i))) - q^((m-i)*i))) and tri(n) = A000217(n). - John Tyler Rascoe, Aug 18 2025