OFFSET
1,1
COMMENTS
We can easily prove that, except for the first term, all terms are of the form 100*m^2 + 100*m + 25 where mod(m, 10) is one of the numbers 1, 3, 6 or 8. Also we can show that all numbers of the form ((5 * 10^n - 5)/3)^2 where n is a natural number, are in the sequence. - Farideh Firoozbakht, Dec 09 2008
LINKS
Zhao Hui Du, Table of n, a(n) for n = 1..45
FORMULA
a(n) = A030487(n)^2. - M. F. Hasler, Dec 23 2012
MATHEMATICA
Flatten[Table[Select[FromDigits/@Tuples[{2, 5, 7}, n], IntegerQ[Sqrt[#]] &], {n, 17}]] (* The program takes a long time to run *) (* Harvey P. Dale, Jan 18 2015 *)
Select[(5Range[1, 9999, 2])^2, Complement[IntegerDigits[#], {2, 5, 7}] == {} &] (* Alonso del Arte, Feb 19 2020 *)
PROG
(PARI) fromTernary(n, d)=sum(i=0, d-1, [2, 5, 7][(n\3^i)%3+1]*10^i)
v=List([25]); for(d=0, 16, for(n=0, 3^d-1, if(issquare(t=225+1000*fromTernary(n, d)), listput(v, t); print1(t", ")))); Vec(v) \\ Charles R Greathouse IV, Dec 22 2012
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Dec 11 1999
EXTENSIONS
Extended and corrected by author, May 08 2000
a(17)-a(19) from Farideh Firoozbakht, Dec 09 2008
STATUS
approved