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%I #14 Apr 25 2019 20:37:14
%S 0,0,0,0,0,0,1,1,2,3,3,4,5,6,6,8,7,9,9,10,10,12,11,14,13,14,13,16,13,
%T 18,15,18,16,20,18,23,20,25,23,26,22,28,23,30,23,30,23,32,26,32,27,34,
%U 28,37,28,36,29,40,31,43,28,42,32,44,32,46,32,46,35,46,35,50,34,51,37,53,36,59,36,57,41
%N Number of partitions of n into 3 prime powers (not including 1).
%H Robert Israel, <a href="/A307727/b307727.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F a(n) = [x^n y^3] Product_{k>=1} 1/(1 - y*x^A246655(k)).
%F a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} [omega(i) * omega(j) * omega(n-i-j) == 1], where omega(n) is the number of distinct prime factors of n and [==] is the Iverson bracket. - _Wesley Ivan Hurt_, Apr 25 2019
%e a(11) = 4 because we have [7, 2, 2], [5, 4, 2], [5, 3, 3] and [4, 4, 3].
%p f:= proc(n,k,pmax) option remember;
%p local t,p,j;
%p if n = 0 then return `if`(k=0, 1, 0) fi;
%p if k = 0 then return 0 fi;
%p if n > k*pmax then return 0 fi;
%p t:= 0:
%p for p in A246655 do
%p if p > pmax then return t fi;
%p t:= t + add(procname(n-j*p, k-j, min(p-1,n-j*p)),j=1..min(k,floor(n/p)))
%p od;
%p t
%p end proc:
%p seq(f(n,3,n),n=0..80) # _Robert Israel_, Apr 25 2019
%t Array[Count[IntegerPartitions[#, {3}], _?(AllTrue[#, PrimePowerQ] &)] &, 81, 0]
%Y Cf. A000961, A023894, A068307, A246655, A280243, A307726.
%K nonn,look
%O 0,9
%A _Ilya Gutkovskiy_, Apr 24 2019
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