OFFSET
1,1
COMMENTS
7 = (1^2 + 2^2 + 4^2)/3
10 = (1^2 + 2^2 + 5^2)/3
14 = (1^2 + 4^2 + 5^2)/3
First case with 2 ways: 23 = (1^2 + 2^2 + 8^2)/3 = (2^2 + 4^2 + 7^2)/3
42 has 3 sets of triples {a,b,c} such that 42= (a^2 + b^2 + c^2)/3: {1,2,11}, {1,5,10}, {3,6,9}
63 has 4 sets of triples {a,b,c}: {2,4,13}, {2,8,11}, {3,6,12}, {5,8,10}, etc.
LINKS
Zak Seidov, Table of n, a(n) for n = 1..1000
MATHEMATICA
Take[Select[Mean/@Subsets[Range[20]^2, {3}], IntegerQ]//Union, 70] (* Harvey P. Dale, Aug 16 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Aug 16 2012
STATUS
approved