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%I #14 Sep 23 2021 19:34:43
%S 5,15,35,39,45,50,55,65,71,105,112,115,145,150,155,185,188,205,211,
%T 229,235,335,350,365,368,388,389,390,450,461,485,495,500,501,502,505,
%U 550,579,585,595,635,650,652,665,671,710,711,715,718,729,735,745,1005,1015,1050
%N Numbers n such that the largest digit of n^2 is 5.
%H Michael S. Branicky, <a href="/A295005/b295005.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = sqrt(A295015(n)), where sqrt = A000196 or A000194 or A003059.
%e 39 is in this sequence because 39^2 = 1521 has 5 as largest digit.
%t Select[Sqrt[ #]&/@(FromDigits/@Select[Tuples[ Range[ 0,5],7],Max[#] == 5&]),IntegerQ] (* _Harvey P. Dale_, Sep 23 2021 *)
%o (PARI) select( is_A295005(n)=n&&vecmax(digits(n^2))==5 , [0..999]) \\ The "n&&" avoids an error message for n=0.
%o (Python)
%o def aupto(limit):
%o alst = []
%o for k in range(1, limit+1):
%o if max(str(k*k)) == "5": alst.append(k)
%o return alst
%o print(aupto(1050)) # _Michael S. Branicky_, May 15 2021
%Y Cf. A295015 (the corresponding squares), A277959 .. A277961 (same for digit 2 .. 4), A295006 .. A295009 (same for digit 6 .. 9).
%Y Cf. A000290 (the squares).
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Nov 12 2017
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