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

0,4

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.

Links

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

Example

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).

Maple

b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(n<i, 0,
      add(b(n-i*j, i+1, t+`if`(j=0, 0, 1)), j=0..min(t, n/i))))
    end:
a:= 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}]]

Crossrefs

Cf. A244395.

Sequence in context: A227135 A162417 A240012 * A293627 A264393 A362307

Adjacent sequences: A295258 A295259 A295260 * A295262 A295263 A295264

Keyword

nonn

Author

David S. Newman, Nov 18 2017

Extensions

More terms from Alois P. Heinz, Nov 18 2017

Status

approved