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

1,1

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

Links

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

Formula

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

Example

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

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 *)

Prog

(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

Crossrefs

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

Cf. A000290 (the squares).

Sequence in context: A177458 A228009 A330477 * A154528 A130861 A049730

Adjacent sequences: A295005 A295006 A295007 * A295009 A295010 A295011

Keyword

nonn,base

Author

M. F. Hasler, Nov 12 2017

Status

approved