A295005 - OEIS

%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