%I #20 Aug 05 2024 14:22:22
%S 1,4,81,121,144,441,484,841,1444,8281,11881,14884,28224,48841,114244,
%T 128881,142884,221841,228484,848241,1121481,1281424,1418481,2184484,
%U 2214144,8282884,11142244,11282881,18241441,18818244,18844281,21242881,21818241,28281124,82428241,121242121,121484484,124121881
%N Squares whose digits are powers of 2.
%C Intersection of A000290 and A028846.
%C Note that in contrast to this sequence, which contains 102 terms up to 10^12, the analogous sequence of cubes (A272826) may contain only 3 in total.
%C Moreover, the similar sequences for the fourth and fifth perfect powers seem to contain only two terms (1, 81) in the case of the former and only one term (1) in the case of the latter. Higher powers also appear to produce sequences with one (mostly) or two terms only.
%C Unlike the analogous sequence for cubes, this sequence is heuristically infinite. - _Charles R Greathouse IV_, May 08 2016
%C This sequence is infinite because it contains the squares of the numbers of the forms 10*(10^k-1)/3+8 and 100*(10^k-1)/3+59. - _Giovanni Resta_, May 09 2016
%C Additionally, this sequence contains the squares of the numbers of the form 1000*(10^k-1)/3 + 809 for k > 2. For k > 2, numbers of the form (1000*(10^k-1)/3 + 809)^2 contains all digits that are powers of 2. - _Altug Alkan_, May 14 2016
%H Giovanni Resta, <a href="/A272884/b272884.txt">Table of n, a(n) for n = 1..10000</a>
%e 144 is a term as its digits are only powers of 2 and it is a square, 144 = 12^2.
%t Select[Range[12000]^2, SubsetQ[{1, 2, 4, 8}, IntegerDigits@#] &]
%t Select[Flatten[Table[FromDigits/@Tuples[{1,2,4,8},n],{n,9}]],IntegerQ[Sqrt[#]]&] (* _Harvey P. Dale_, Aug 05 2024 *)
%o (PARI) is(n)=issquare(n) && #setintersect(Set(digits(n)), [0,3,5,6,7,9])==0 \\ _Charles R Greathouse IV_, May 08 2016
%Y Cf. A000290 (squares), A028846 (numbers whose digits are powers of 2), A272826 (similar sequence for cubes).
%K nonn,base
%O 1,2
%A _Waldemar Puszkarz_, May 08 2016