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

Author

David S. Newman, Jul 03 2014

Keywords

• nonn

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

Crossrefs

Cf. A100471, A100883, A244393, A295261.

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