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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.
3

%I #34 Jun 06 2021 09:00:39

%S 1,1,1,2,3,4,5,8,11,15,20,26,34,46,59,78,101,129,163,209,261,329,412,

%T 517,641,798,986,1216,1493,1829,2229,2721,3303,4000,4841,5841,7034,

%U 8458,10144,12137,14512,17306,20596,24483,29045,34391,40680,48032,56627,66666

%N 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 Alois P. Heinz, <a href="/A244395/b244395.txt">Table of n, a(n) for n = 0..2000</a>

%e For n=6 the partitions counted are: 6, 51, 42, 411, 321.

%p b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1, 0,

%p b(n, i-1, t) +add(b(n-i*j, i-1, t+1), j=1..min(t, n/i))))

%p end:

%p a:= n-> b(n$2, 1):

%p seq(a(n), n=0..60); # _Alois P. Heinz_, Jul 03 2014

%t nend = 20;

%t For[n = 1, n <= nend, n++,

%t count[n] = 0;

%t Ip = IntegerPartitions[n];

%t For[i = 1, i <= Length[Ip], i++,

%t m = Max[Ip[[i]]];

%t condition = True;

%t Tip = Tally[Ip[[i]]];

%t For[j = 1, j <= Length[Tip], j++,

%t condition = condition && (Tip[[j]][[2]] <= j)];

%t If[condition, count[n]++ (* ; Print[Ip[[i]]] *)]];

%t ]

%t Table[count[i], {i, 1, nend}]

%t (* Second program: *)

%t b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i < 1, 0,

%t b[n, i-1, t] + Sum[b[n-i*j, i-1, t+1], {j, 1, Min[t, n/i]}]]];

%t a[n_] := b[n, n, 1];

%t a /@ Range[0, 60] (* _Jean-François Alcover_, Jun 06 2021, after _Alois P. Heinz_ *)

%Y Cf. A100471, A100883, A244393, A295261.

%K nonn

%O 0,4

%A _David S. Newman_, Jul 03 2014

%E More terms from _Alois P. Heinz_, Jul 03 2014