0,1

a1! + a2! + a3! = z^2.

These appear to be the only solutions. 8 and 27 appear to be the only cubes

that are the sum of 3 factorials. Again, it appears that 2 and 3 are the only

powers of n satisfying a1!+a2!+a3! = z^n. The complete list of solutions is

a1 a2 a3 z^2

0 0 2 4

0 1 2 4

0 2 3 9

0 4 4 49

0 5 6 841

1 1 2 4

1 2 3 9

1 4 4 49

1 5 6 841

3 3 4 36

4 5 7 5184

d = 50; a = Union[ Flatten[ Table[a! + b! + c!, {a, 1, d}, {b, a, d}, {c, b, d}]]]; l = Length[a]; Do[ If[ IntegerQ[ Sqrt[ a[[i]]]], Print[ a[[i]]]], {i, 1, l}]

(PARI) sum3factsq(n) = { for(a1=1, n, for(a2=a1, n, for(a3=a2, n, z = a1!+a2!+a3!; if(issquare(z), print1(z" ")) ) ) ) }

Sequence in context: A081069 A053967 A028945 this_sequence A086541 A053965 A058444

Adjacent sequences: A082872 A082873 A082874 this_sequence A082876 A082877 A082878

easy,nonn

Cino Hilliard (hillcino368(AT)gmail.com), May 25 2003