OFFSET
1,1
COMMENTS
If a term has a zero in it, its digit product is 0. Thus it is trivial to include cubes with one or more zeros.
Is this sequence finite?
Replacing "squares" with "cubes", this sequence would only consist of {4} for n < 10^8. 4 is believed to be the only number to satisfy this property with cubes.
If it exists, a(20) > 10^10.
a(80) > 10^27. - Hiroaki Yamanouchi, Mar 16 2015
LINKS
Hiroaki Yamanouchi, Table of n, a(n) for n = 1..79
EXAMPLE
46692 + 4*6*6*9*2 = 49284 = 222^2 and 46692 - 4*6*6*9*2 = 210^2. So 46692 is a member of this sequence.
PROG
(PARI) for(n=0, 10^7, d=digits(n); p=prod(i=1, #d, d[i]); if(p&&issquare(n-p)&&issquare(n+p), print1(n, ", ")))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Derek Orr, Mar 13 2015
EXTENSIONS
a(12)-a(19) from Michel Marcus, Mar 14 2015
a(20)-a(25) from Hiroaki Yamanouchi, Mar 16 2015
STATUS
approved