OFFSET
1,2
COMMENTS
Let # denote "number of" and c(p) = conjugate of partitionp. Then
A240057(n) = # p such that min(p) not = max(c(p));
A039899(n) = # p such that min(p) < max(c(p));
A039900(n) = # p such that min(p) <= max(c(p));
A006141(n) = # p such that min(p) = max(c(p));
A003114(n) = # p such that min(p) > max(c(p));
A003016(n) = # p such that min(p) >= max(c(p));
A064173(n) = # p such that max(p) < max(c(p));
A064174(n) = # p such that max(p) <= max(c(p));
A047993(n) = # p such that max(p) = max(c(p)).
See A240178 for related sequences. - Clark Kimberling, Apr 11 2014
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..1000
EXAMPLE
a(9) = 28 counts all the 30 partitions of 9 except 333 and 2211111.
MAPLE
b:= proc(n, i) option remember; `if`(n<0, 0, `if`(n=0, 1,
`if`(i<1, 0, b(n, i-1)+`if`(i>n, 0, b(n-i, i)))))
end:
a:= n->combinat[numbpart](n)-add(b(n-j^2, j-1), j=0..isqrt(n)):
seq(a(n), n=1..50); # Alois P. Heinz, Apr 03 2014
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n];
t1 = Table[Count[f[n], p_ /; Max[p] < Count[p, Max[p]]], {n, 0, z}] (* A003106 *)
t2 = Table[Count[f[n], p_ /; Max[p] <= Count[p, Max[p]]], {n, 0, z}] (* A003114 *)
t3 = Table[Count[f[n], p_ /; Max[p] == Count[p, Max[p]]], {n, 0, z}] (* A006141 *)
tt = Table[Count[f[n], p_ /; Max[p] != Count[p, Max[p]]], {n, 0, z}] (* A240057 *)
t4 = Table[Count[f[n], p_ /; Max[p] > Count[p, Max[p]]], {n, 0, z}] (* A039899 *)
t5 = Table[Count[f[n], p_ /; Max[p] >= Count[p, Max[p]]], {n, 0, z}] (* A039900 *)
(* second program: *)
b[n_, i_] := b[n, i] = If[n < 0, 0, If[n == 0, 1, If[i < 1, 0, b[n, i - 1] + If[i > n, 0, b[n - i, i]]]]];
a[n_] := PartitionsP[n] - Sum[b[n - j^2, j - 1], {j, 0, Sqrt[n]}];
Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Aug 30 2016, after Alois P. Heinz *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 02 2014
STATUS
approved