A030487 -id:A030487 - cp's OEIS frontend

A030485 Squares composed of digits {2, 5, 7}.

25, 225, 7225, 27225, 55225, 2772225, 227557225, 277722225, 27777222225, 72272257225, 2777772222225, 25772527522225, 277777722222225, 2775552752755225, 27522257555772225, 27777777222222225, 77525222275255225, 257727727257277225, 722555225555275225, 2275752775775227225

Offset: 1

Patrick De Geest, Dec 11 1999

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

Zhao Hui Du, Table of n, a(n) for n = 1..45
Patrick De Geest, Squares containing at most three distinct digits, Index entries for related sequences

Subsequence of A191486. Also subsequence of A017330. Cf. A030487.

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 *)

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

a(n) = A030487(n)^2. - M. F. Hasler, Dec 23 2012

Extended and corrected by author, May 08 2000

a(17)-a(19) from Farideh Firoozbakht, Dec 09 2008

A275971 Numbers n such that the decimal digits of n^2 are all prime.

Original entry on oeis.org

5, 15, 85, 165, 235, 485, 1665, 1885, 4835, 5765, 7585, 15085, 15885, 16665, 18365, 18915, 22885, 27115, 27885, 50235, 57665, 58115, 72335, 85635, 87885, 150915, 166665, 182415, 194235, 194365, 229635, 240365, 268835, 503515, 507665, 524915, 568835, 570415, 577515, 581165

Offset: 1

Zak Seidov, Aug 15 2016

Apparently 5, 235 and 72335 are the only terms using digits {2,3,5,7}.
a(n)/5 = {1, 3, 17, 33, 47, 97, 333, 377, 967, 1153, 1517, 3017, 3177, 3333, ...}; terms b(n) that have n 3's must be in the sequence since (5 b(n))^2 yields the decimal number 2 followed by (n-1) 7's then n 2's, and ending in 5 (i.e., 225, 27225, 2772225). Thus 5 b(n) = {15, 165, 1665, 16665, etc.} appears in this sequence. - Michael De Vlieger, Aug 15 2016
All terms are odd multiples of 5 (A017329), i.e., must end in 5, which is the only digit whose square ends in a prime digit. The sequence contains A030487 as an infinite proper subsequence which in turn contains all numbers of the form (5*10^n-5)/3 (these are the above 5 b(n)) as a proper subsequence. - M. F. Hasler, Sep 16 2016

			72335^2 = 5232352225 = A191486(23).

Michel Marcus, Table of n, a(n) for n = 1..1000

Cf. A191486, A017329, A030487.

w = Boole@! PrimeQ@ # & /@ RotateLeft@ Range[0, 9]; Sqrt@ Select[Range[10^6]^2, Total@ Pick[DigitCount@ #, w, 1] == 0 &] (* Michael De Vlieger, Aug 15 2016 *)

is(n)=#setintersect(Set(digits(n^2)), [0, 1, 4, 6, 8, 9])==0 \\ Charles R Greathouse IV, Sep 16 2016

def aupto(limit):
  alst = []
  for k in range(1, limit+1):
    if set(str(k*k)) <= set("2357"): alst.append(k)
  return alst
print(aupto(10**6)) # Michael S. Branicky, May 15 2021

a(n) = sqrt(A191486(n)).

More terms from Michel Marcus, Aug 17 2016

cp's OEIS Frontend

A030485 Squares composed of digits {2, 5, 7}.

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