A244395 Number of partitions of n in which the largest summand has frequency 1, the next largest summand has frequency at most 2, the third largest has frequency at most 3, etc.
1, 1, 1, 2, 3, 4, 5, 8, 11, 15, 20, 26, 34, 46, 59, 78, 101, 129, 163, 209, 261, 329, 412, 517, 641, 798, 986, 1216, 1493, 1829, 2229, 2721, 3303, 4000, 4841, 5841, 7034, 8458, 10144, 12137, 14512, 17306, 20596, 24483, 29045, 34391, 40680, 48032, 56627, 66666
Offset: 0
Keywords
Examples
For n=6 the partitions counted are: 6, 51, 42, 411, 321.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..2000
Programs
-
Maple
b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1, 0, b(n, i-1, t) +add(b(n-i*j, i-1, t+1), j=1..min(t, n/i)))) end: a:= n-> b(n$2, 1): seq(a(n), n=0..60); # Alois P. Heinz, Jul 03 2014 -
Mathematica
nend = 20; For[n = 1, n <= nend, n++, count[n] = 0; Ip = IntegerPartitions[n]; For[i = 1, i <= Length[Ip], i++, m = Max[Ip[[i]]]; condition = True; Tip = Tally[Ip[[i]]]; For[j = 1, j <= Length[Tip], j++, condition = condition && (Tip[[j]][[2]] <= j)]; If[condition, count[n]++ (* ; Print[Ip[[i]]] *)]]; ] Table[count[i], {i, 1, nend}] (* Second program: *) b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i < 1, 0, b[n, i-1, t] + Sum[b[n-i*j, i-1, t+1], {j, 1, Min[t, n/i]}]]]; a[n_] := b[n, n, 1]; a /@ Range[0, 60] (* Jean-François Alcover, Jun 06 2021, after Alois P. Heinz *)
Extensions
More terms from Alois P. Heinz, Jul 03 2014