A295008 Numbers whose square has largest digit 8.

9, 22, 28, 29, 41, 59, 62, 72, 78, 90, 91, 92, 94, 104, 109, 122, 126, 128, 135, 151, 159, 168, 169, 178, 184, 191, 192, 195, 196, 202, 209, 220, 221, 232, 241, 242, 259, 261, 262, 268, 278, 279, 280, 284, 285, 289, 290, 291, 292, 294, 295, 296, 298, 322, 328, 329, 341, 344, 349

Offset: 1

Author

M. F. Hasler, Nov 12 2017

Keywords

• nonn
• base

Comments

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

Examples

			28 is in this sequence because 28^2 = 784 has 8 as largest digit.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A295018 (the corresponding squares), A277959 .. A277961 (same for digit 2 .. 4), A295005 .. A295009 (same for digit 5 .. 9).

Cf. A000290 (the squares).

Programs

Maple

select(t -> max(convert(t^2,base,10))=8, [$1..1000]); # Robert Israel, Nov 13 2017

Mathematica

Select[Range[400],Max[IntegerDigits[#^2]]==8&] (* Harvey P. Dale, Jun 02 2019 *)

PARI

select( is_A295008(n)=n&&vecmax(digits(n^2))==8 , [0..999]) \\ The "n&&" avoids an error message for n=0.

Python

def ok(n): return max(int(d) for d in str(n*n)) == 8
print(list(filter(ok, range(350)))) # Michael S. Branicky, Sep 22 2021

Formula

a(n) = sqrt(A295018(n)), where sqrt = A000196 or A000194 or A003059.