OFFSET
1,5
COMMENTS
For n>3, odd n have (many) more partitions than even n.
LINKS
Zak Seidov, Table of n, a(n) for n = 1..200
EXAMPLE
a(4)=1 because 16=2+3+11,
a(6)=3 because 36=2+3+31=2+5+29=2+11+13.
MATHEMATICA
Table[Count[Union/@IntegerPartitions[n^2, {3}], _?(Length[#]==3&&AllTrue[ #, PrimeQ]&)], {n, 60}] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, May 02 2018 *)
PROG
(PARI) a(n)=my(s); n*=n; forprime(p=n\3, n-4, forprime(q=(n-p)\2+1, min(n-p, p-1), if(isprime(n-p-q), s++))); s \\ Charles R Greathouse IV, Aug 27 2012
(Haskell)
a183168 n = z (drop (fromInteger (mod n 2)) a000040_list) (n ^ 2) 3 where
z _ m 1 = if m <= 0 then 0 else a010051 m
z (p:ps) m c = if m <= 2*p then 0 else z ps (m - p) (c - 1) + z ps m c
-- Reinhard Zumkeller, Aug 28 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Aug 27 2012
STATUS
approved