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%I #29 Sep 22 2021 07:45:42
%S 9,22,28,29,41,59,62,72,78,90,91,92,94,104,109,122,126,128,135,151,
%T 159,168,169,178,184,191,192,195,196,202,209,220,221,232,241,242,259,
%U 261,262,268,278,279,280,284,285,289,290,291,292,294,295,296,298,322,328,329,341,344,349
%N Numbers whose square has largest digit 8.
%C Includes a*10^n+b for n >= 2 and [a,b] in {[4, 1], [9, 1], [2, 2], [9, 2], [1, 4], [6, 4], [9, 4], [8, 5], [4, 6], [9, 6], [5, 8], [8, 8], [9, 8], [1, 9], [2, 9], [4, 9], [6, 9], [8, 9], [9, 9]}. - _Robert Israel_, Nov 13 2017
%H Robert Israel, <a href="/A295008/b295008.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = sqrt(A295018(n)), where sqrt = A000196 or A000194 or A003059.
%e 28 is in this sequence because 28^2 = 784 has 8 as largest digit.
%p select(t -> max(convert(t^2,base,10))=8, [$1..1000]); # _Robert Israel_, Nov 13 2017
%t Select[Range[400],Max[IntegerDigits[#^2]]==8&] (* _Harvey P. Dale_, Jun 02 2019 *)
%o (PARI) select( is_A295008(n)=n&&vecmax(digits(n^2))==8 , [0..999]) \\ The "n&&" avoids an error message for n=0.
%o (Python)
%o def ok(n): return max(int(d) for d in str(n*n)) == 8
%o print(list(filter(ok, range(350)))) # _Michael S. Branicky_, Sep 22 2021
%Y Cf. A295018 (the corresponding squares), A277959 .. A277961 (same for digit 2 .. 4), A295005 .. A295009 (same for digit 5 .. 9).
%Y Cf. A000290 (the squares).
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Nov 12 2017
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