A360822 Numbers whose squares have at most 2 digits less than 8.

1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 17, 22, 23, 27, 28, 29, 30, 31, 33, 43, 53, 63, 67, 77, 83, 91, 93, 94, 97, 99, 141, 167, 173, 197, 283, 293, 297, 298, 303, 313, 314, 316, 447, 583, 707, 767, 833, 836, 917, 943, 947, 1378, 2917, 2983, 3033, 5467, 9417, 9433, 29983, 31367, 94863

Offset: 1

Author

Dmitry Kamenetsky, Feb 22 2023

Keywords

• nonn
• base

Comments

From Michael S. Branicky, Feb 22 2023: (Start)

Conjecture: Sequence has 72 terms, with largest term 940206833.

No terms > 940206833 with less than 17 digits. (End)

Examples

			314641 is in the sequence, because 314641^2 = 98998958881 has only two digits that are less than 8.
		

Links

• Michael S. Branicky, Python program

Crossrefs

Cf. A360803.

Programs

Mathematica

Select[Range[10^5], Count[IntegerDigits[#^2], ?(#1 < 8 &)] < 3 &] (* _Amiram Eldar, Feb 22 2023 *)

PARI

isok(k) = #select(x->(x<8), digits(k^2)) <= 2; \\ Michel Marcus, Feb 22 2023

Python

def ok(n): return sum(1 for d in str(n**2) if d < "8") < 3
print([k for k in range(1, 10**5) if ok(k)]) # Michael S. Branicky, Feb 22 2023

Python

# see link for a faster version to find all terms

Python

from itertools import count, islice
def A360822_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:len(s:=str(n**2))<=s.count('8')+s.count('9')+2,count(max(startvalue,1)))
A360822_list = list(islice(A360822_gen(),62)) # Chai Wah Wu, Mar 11 2023