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%I #18 Nov 19 2017 08:02:11
%S 1,1,1,2,2,4,4,6,8,11,12,18,22,28,34,44,54,69,82,102,125,154,185,226,
%T 271,327,393,474,562,673,797,947,1124,1329,1563,1846,2164,2541,2974,
%U 3480,4062,4738,5508,6403,7432,8614,9966,11530,13307,15345,17670,20337
%N Partitions into parts with frequency less than or equal to their place in the list of summands.
%C 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.
%H Alois P. Heinz, <a href="/A295261/b295261.txt">Table of n, a(n) for n = 0..2000</a>
%e The partition 1+1 is not counted because its smallest part, 1, appears twice.
%e 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).
%p b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(n<i, 0,
%p add(b(n-i*j, i+1, t+`if`(j=0, 0, 1)), j=0..min(t, n/i))))
%p end:
%p a:= n-> b(n, 1$2):
%p seq(a(n), n=0..60); # _Alois P. Heinz_, Nov 18 2017
%t << Combinatorica`;
%t nend = 30;
%t For[n = 1, n <= nend, n++, count[n] = 0;
%t part = Partitions[n];
%t For[i = 1, i <= Length[part], i++,
%t t = Tally[part[[i]]];
%t condition = True;
%t For[j = 1, j <= Length[t], j++,
%t If[t[[-j, 2]] > j, condition = False ]];
%t If[condition, count[n]++]]];
%t Print[Table[count[i], {i, 1, nend}]]
%Y Cf. A244395.
%K nonn
%O 0,4
%A _David S. Newman_, Nov 18 2017
%E More terms from _Alois P. Heinz_, Nov 18 2017
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