%I #38 Nov 19 2022 00:53:49
%S 0,1,10,11,100,101,110,1000,1001,1010,1100,10000,10001,10010,10100,
%T 11000,100000,100001,100010,100100,101000,110000,1000000,1000001,
%U 1000010,1000100,1001000,1010000,1100000,10000000,10000001,10000010,10000100,10001000,10010000
%N Numbers that when raised to the fourth power and written backwards give squares.
%C It seems that the numbers contain only the digits 0 and 1, and that the reversed fourth power and the square root of the reversed fourth power are both palindromes.
%C If the above comment is correct, and also if (as it appears) no more than two ones are among the digits of any term, this Mathematica program quickly generates the terms of the sequence: Flatten[Table[Select[ FromDigits/@Permutations[PadRight[PadRight[{},k,1],8,0]],IntegerQ[ Sqrt[ IntegerReverse[#^4]]]&],{k,0,2}]]//Sort - _Harvey P. Dale_, May 05 2020
%e 101 is in the sequence because 101^4 = 104060401 and 104060401 = 10201^2.
%e 110 is in the sequence because 110^4 = 146410000 and 14641 = 121^2.
%t Select[Range[0,10^7],IntegerQ[Sqrt[IntegerReverse[#^4]]]&] (* _Harvey P. Dale_, May 05 2020 *)
%o (PARI) revint(n) = m=n%10; n\=10; while(n>0, m=m*10+n%10; n\=10); m
%o s=[]; for(i=0, 1000000, if(issquare(revint(i^4)), s=concat(s, i))); s
%o (Magma) [n: n in [0..10^7] | IsSquare(Seqint(Reverse(Intseq(n^4))))]; // _Bruno Berselli_, Dec 27 2013
%o (Python)
%o from itertools import count, islice
%o from sympy import integer_nthroot
%o def A234472_gen(startvalue=0): # generator of terms >= startvalue
%o return filter(lambda n:integer_nthroot(int(str(n**4)[::-1]),2)[1], count(max(startvalue,0)))
%o A234472_list = list(islice(A234472_gen(),10)) # _Chai Wah Wu_, Nov 18 2022
%Y Cf. A102859, A043681.
%K nonn,base,nice
%O 1,3
%A _Colin Barker_, Dec 26 2013