cp's OEIS Frontend

A353081 Numbers whose squares have the first two digits the same as the next two digits.

%I A353081 #15 Apr 24 2022 20:52:10
%S A353081 201,264,402,482,603,689,772,804,932,964,1005,1101,1146,1231,1557,
%T A353081 1798,1907,2010,2035,2084,2132,2202,2357,2582,2640,2659,2678,2734,
%U A353081 2878,3015,3114,3179,3334,3482,3624,3761,3893,4020,4021,4144,4264,4381,4495,4606,4714,4820,4924
%N A353081 Numbers whose squares have the first two digits the same as the next two digits.
%F A353081 201^2 = 40401 and 264^2 = 69696. Thus, both 201 and 264 are in this sequence.
%p A353081 q:= n-> (s-> is(s[1..2]=s[3..4]))(""||(n^2)):
%p A353081 select(q, [$32..10000])[];  # _Alois P. Heinz_, Apr 22 2022
%t A353081 Select[Range[32, 5000], Take[IntegerDigits[#^2], {1, 2}] ==  Take[IntegerDigits[#^2], {3, 4}] &]
%o A353081 (Python)
%o A353081 def ok(n): s = str(n**2); return len(s) > 3 and s[:2] == s[2:4]
%o A353081 print([k for k in range(5000) if ok(k)]) # _Michael S. Branicky_, Apr 22 2022
%o A353081 (PARI) do(n)=my(v=List()); for(a=1,9, for(b=0,9, my(N=10^(n-4), t=(1010*a+101*b)*N-1); for(k=sqrtint(t)+1,sqrtint(t+N), listput(v,k)))); Vec(v) \\ finds terms corresponding to n-digit squares; _Charles R Greathouse IV_, Apr 24 2022
%Y A353081 Cf. A123912, A353080.
%K A353081 nonn,base,easy
%O A353081 1,1
%A A353081 _Tanya Khovanova_, Apr 22 2022