A219181 Number of partitions of n into the maximal possible number of distinct prime parts or 0 if there are no such partitions.

1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 1, 3, 1, 3, 2, 4, 2, 4, 2, 5, 2, 1, 4, 1, 4, 1, 4, 1, 6, 2, 6, 1, 6, 2, 8, 4, 10, 2, 1, 5, 1, 6, 1, 5, 2, 6, 2, 10, 1, 9, 1, 11, 4, 15, 3, 14, 3, 1, 6, 1, 6, 1, 5, 1, 10, 1, 11

Offset: 0

Author

Alois P. Heinz, Nov 13 2012

Keywords

• nonn
• look

Comments

a(n) is the last element of row n of triangle A219180 or 0 if the row is empty. a(n) = 0 iff n in {1,4,6}.

Examples

			a(18) = 2 because there are 2 partitions of 18 into 3 distinct prime parts ([2,3,13], [2,5,11]) but no partitions of 18 into more than 3 distinct prime parts.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..3500

Programs

Maple

with(numtheory):
b:= proc(n, i) option remember;
      `if`(n=0, [1], `if`(i<1, [], zip((x, y)->x+y, b(n, i-1),
       [0, `if`(ithprime(i)>n, [], b(n-ithprime(i), i-1))[]], 0)))
    end:
a:= proc(n) local l; l:=b(n,pi(n));
       while nops(l)>0 and l[-1]=0 do
          l:= subsop(-1=NULL, l)
       od;
       `if`(nops(l)=0, 0, l[-1])
    end:
seq(a(n), n=0..100);

Mathematica

zip = With[{m = Max[Length[#1], Length[#2]]}, PadRight[#1, m] + PadRight[#2, m]]&; b[n_, i_] := b[n, i] = If[n == 0, {1}, If[i<1, {}, zip[b[n, i-1], Join[{0}, If[Prime[i]>n, {}, b[n-Prime[i], i-1]]]]]]; a[n_] := (l = b[n, PrimePi[n]]; While[Length[l]>0 && l[[-1]] == 0, l = ReplacePart[l, -1 -> Nothing]]; If[Length[l] == 0, 0, l[[-1]]]);  Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 12 2017, translated from Maple *)

Formula

a(n) = A219180(n,A024936(n)) if A024936(n) >= 0, a(n) = 0 else.