A136624 Irregular triangle read by rows: classify each numeric partition by sum of its parts and by the size of the staircase Ferrers board required to contain it. The triangle gives the number of partitions in each class, cf. A136102 and A136103.

1, 1, 2, 1, 2, 3, 3, 1, 2, 2, 6, 7, 6, 4, 1, 2, 2, 4, 8, 12, 15, 17, 14, 10, 5, 1, 2, 2, 4, 6, 12, 15, 23, 30, 39, 42, 40, 35, 25, 15, 6, 1, 2, 2, 4, 6, 10, 16, 23, 29, 42, 56, 71, 88, 103, 112, 114, 102, 86, 65, 41, 21, 7, 1, 2, 2, 4, 6, 10, 14, 24, 31, 43

Offset: 0

Author

Alford Arnold, Jan 17 2008

Keywords

• nonn
• tabf

Comments

Sequences A136102 and A136103 encode the numeric partitions by least prime signature and the Ferrers boards by 1 2 12 360 75600 174636000 ... A006939.

Examples

			Starting a new row each time we are required to use a larger Ferrer board the triangle begins:
  1
  ..1
  .....2...1
  .........2...3...3...1
  .............2...2...6...7...6...4...1
  .................2...2...4...8..12..15..17..14..10...5...1
  .....................2...2...4...6..12..15..23..30..39..42..40..35..25..15..6..1
		

Links

• John Tyler Rascoe, Rows n = 0..16, flattened

Crossrefs

Cf. A000041 (column sums), A000108, A006939, A025487, A071724 (row sums), A136102, A136103, A136625.

Programs

PARI

d(s,n) = {my(v = setminus([1..n],s), r=[], c=1); for(i=2, #v, if(v[i]==v[i-1]+1, c++ , r=concat(r, c); c=1)) ; return(concat(r, c))}
tri(n) = {n*(n+1)/2}
S(n) = {my(R = x^tri(n)); if(n<1, return(1), for(i=1,n-1, forsubset([n,i], s, my(u=d(vecextract([1..n],s),n)); R+=(x^(tri(n)-sum(j=1,#u, tri(u[j]))))*prod(j=1,#u, sum(z=0,u[j]-1, S(z))))); return(R))}
A136624(row_n) = {Vecrev(S(row_n)/x^(row_n))} \\ John Tyler Rascoe, Feb 25 2025

Extensions

a(26) onwards from John Tyler Rascoe, Feb 25 2025