A239312 Number of condensed integer partitions of n.
1, 1, 1, 2, 3, 3, 5, 6, 9, 10, 14, 16, 23, 27, 33, 41, 51, 62, 75, 93, 111, 134, 159, 189, 226, 271, 317, 376, 445, 520, 609, 714, 832, 972, 1129, 1304, 1520, 1753, 2023, 2326, 2692, 3077, 3540, 4050, 4642, 5298, 6054, 6887, 7854, 8926, 10133, 11501, 13044
Offset: 0
Keywords
Examples
a(5) = 3 gives the number of partitions of 5 that result from condensations as shown here: 5 -> 5, 41 -> 41, 32 -> 32, 311 -> 32, 221 -> 41, 2111 -> 32, 11111 -> 5. From _Gus Wiseman_, Mar 12 2024: (Start) The a(1) = 1 through a(9) = 10 condensed partitions: (1) (2) (3) (4) (5) (6) (7) (8) (9) (2,1) (2,2) (3,2) (3,3) (4,3) (4,4) (5,4) (3,1) (4,1) (4,2) (5,2) (5,3) (6,3) (5,1) (6,1) (6,2) (7,2) (3,2,1) (3,2,2) (7,1) (8,1) (4,2,1) (3,3,2) (4,3,2) (4,2,2) (4,4,1) (4,3,1) (5,2,2) (5,2,1) (5,3,1) (6,2,1) (End)
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..100 (first 84 terms from Manfred Scheucher)
- Manfred Scheucher, Python Script
Crossrefs
Programs
-
Maple
b:= proc(n,i) option remember; `if`(n=0, {[]}, `if`(i=1, {[n]}, {seq(map(x-> `if`(j=0, x, sort([x[], i*j])), b(n-i*j, i-1))[], j=0..n/i)})) end: a:= n-> nops(b(n$2)): seq(a(n), n=0..50); # Alois P. Heinz, Jul 01 2019
-
Mathematica
u[n_, k_] := u[n, k] = Map[Total, Split[IntegerPartitions[n][[k]]]]; t[n_] := t[n] = DeleteDuplicates[Table[Sort[u[n, k]], {k, 1, PartitionsP[n]}]]; Table[Length[t[n]], {n, 0, 30}] Table[Length[Select[IntegerPartitions[n], Length[Select[Tuples[Divisors/@#],UnsameQ@@#&]]>0&]], {n,0,30}] (* Gus Wiseman, Mar 12 2024 *)
Extensions
Typo in definition corrected by Manfred Scheucher, May 29 2015
Name edited by Gus Wiseman, Mar 13 2024
Comments