OFFSET
1,1
COMMENTS
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
LINKS
Michel Marcus, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = sqrt(A191486(n)).
EXAMPLE
72335^2 = 5232352225 = A191486(23).
MATHEMATICA
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 *)
PROG
(PARI) is(n)=#setintersect(Set(digits(n^2)), [0, 1, 4, 6, 8, 9])==0 \\ Charles R Greathouse IV, Sep 16 2016
(Python)
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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Zak Seidov, Aug 15 2016
EXTENSIONS
More terms from Michel Marcus, Aug 17 2016
STATUS
approved