A295261 Partitions into parts with frequency less than or equal to their place in the list of summands.

1, 1, 1, 2, 2, 4, 4, 6, 8, 11, 12, 18, 22, 28, 34, 44, 54, 69, 82, 102, 125, 154, 185, 226, 271, 327, 393, 474, 562, 673, 797, 947, 1124, 1329, 1563, 1846, 2164, 2541, 2974, 3480, 4062, 4738, 5508, 6403, 7432, 8614, 9966, 11530, 13307, 15345, 17670, 20337

Offset: 0

Author

David S. Newman, Nov 18 2017

Keywords

• nonn

Comments

Let the summands of a partition be s(1) < s(2) < ... < s(k) and the frequency of s(i) be f(i). Then we count those partitions for which f(i) <= i.

Examples

			The partition 1+1 is not counted because its smallest part, 1, appears twice.
The partition 3+2+2+1 is counted because its smallest part, 1, appears once; its next smallest part, 2 appears twice (and 2 <= 2) and its third part, 3, appears 1 time (and 1 <= 3).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..2000

Crossrefs

Cf. A244395.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(n b(n, 1$2):
seq(a(n), n=0..60);  # Alois P. Heinz, Nov 18 2017

Mathematica

<< Combinatorica`;
nend = 30;
For[n = 1, n <= nend, n++, count[n] = 0;
  part = Partitions[n];
  For[i = 1, i <= Length[part], i++,
   t = Tally[part[[i]]];
   condition = True;
   For[j = 1, j <= Length[t], j++,
    If[t[[-j, 2]] > j, condition = False ]];
   If[condition, count[n]++]]];
Print[Table[count[i], {i, 1, nend}]]

Extensions

More terms from Alois P. Heinz, Nov 18 2017