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Number of partitions of n such that (number of distinct parts) = number of 2's.
2

%I #14 Jun 05 2021 06:17:06

%S 1,0,1,0,0,1,1,2,2,2,4,4,6,9,10,12,19,21,24,36,44,49,66,81,100,123,

%T 144,180,229,265,317,391,473,566,675,798,968,1154,1354,1621,1926,2241,

%U 2675,3170,3691,4345,5113,5956,7002,8182,9503,11095,12919,14976,17446

%N Number of partitions of n such that (number of distinct parts) = number of 2's.

%H Alois P. Heinz, <a href="/A239961/b239961.txt">Table of n, a(n) for n = 0..1000</a>

%e a(10) counts these 4 partitions : 622, 3322, 32221, 22111111.

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

%p b(n, i-1)+`if`(i=2, 0, add(b(n-2-i*j, i-1), j=1..(n-2)/i))))

%p end:

%p a:= n-> `if`(n=0, 1, b(n-2$2)):

%p seq(a(n), n=0..70); # _Alois P. Heinz_, Apr 03 2014

%t z = 54; d[p_] := d[p] = Length[DeleteDuplicates[p]]; Table[Count[IntegerPartitions[n], p_ /; d[p] == Count[p, 2]], {n, 0, z}]

%t (* Second program: *)

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

%t If[i == 2, 0, Sum[b[n-2-i*j, i-1], {j, 1, (n-2)/i}]]]];

%t a[n_] := If[n == 0, 1, b[n-2, n-2]];

%t a /@ Range[0, 70] (* _Jean-François Alcover_, Jun 05 2021, after _Alois P. Heinz_ *)

%Y Cf. A239960.

%K nonn,easy

%O 0,8

%A _Clark Kimberling_, Mar 30 2014