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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.

%I A244395 #34 Jun 06 2021 09:00:39
%S A244395 1,1,1,2,3,4,5,8,11,15,20,26,34,46,59,78,101,129,163,209,261,329,412,
%T A244395 517,641,798,986,1216,1493,1829,2229,2721,3303,4000,4841,5841,7034,
%U A244395 8458,10144,12137,14512,17306,20596,24483,29045,34391,40680,48032,56627,66666
%N 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.
%H A244395 Alois P. Heinz, <a href="/A244395/b244395.txt">Table of n, a(n) for n = 0..2000</a>
%e A244395 For n=6 the partitions counted are: 6, 51, 42, 411, 321.
%p A244395 b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1, 0,
%p A244395       b(n, i-1, t) +add(b(n-i*j, i-1, t+1), j=1..min(t, n/i))))
%p A244395     end:
%p A244395 a:= n-> b(n$2, 1):
%p A244395 seq(a(n), n=0..60);  # _Alois P. Heinz_, Jul 03 2014
%t A244395 nend = 20;
%t A244395 For[n = 1, n <= nend, n++,
%t A244395 count[n] = 0;
%t A244395 Ip = IntegerPartitions[n];
%t A244395 For[i = 1, i <= Length[Ip], i++,
%t A244395 m = Max[Ip[[i]]];
%t A244395 condition = True;
%t A244395 Tip = Tally[Ip[[i]]];
%t A244395 For[j = 1, j <= Length[Tip], j++,
%t A244395 condition = condition && (Tip[[j]][[2]] <= j)];
%t A244395 If[condition, count[n]++ (* ; Print[Ip[[i]]] *)]];
%t A244395 ]
%t A244395 Table[count[i], {i, 1, nend}]
%t A244395 (* Second program: *)
%t A244395 b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i < 1, 0,
%t A244395     b[n, i-1, t] + Sum[b[n-i*j, i-1, t+1], {j, 1, Min[t, n/i]}]]];
%t A244395 a[n_] := b[n, n, 1];
%t A244395 a /@ Range[0, 60] (* _Jean-François Alcover_, Jun 06 2021, after _Alois P. Heinz_ *)
%Y A244395 Cf. A100471, A100883, A244393, A295261.
%K A244395 nonn
%O A244395 0,4
%A A244395 _David S. Newman_, Jul 03 2014
%E A244395 More terms from _Alois P. Heinz_, Jul 03 2014