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

0,3

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.

Links

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

Example

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

Prog

(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

Crossrefs

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

Sequence in context: A101933 A117127 A278148 * A033763 A033803 A035531

Adjacent sequences: A136621 A136622 A136623 * A136625 A136626 A136627

Keyword

nonn,tabf

Author

Alford Arnold, Jan 17 2008

Extensions

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

Status

approved