A360803 Numbers whose squares have a digit average of 8 or more.

3, 313, 94863, 298327, 987917, 3162083, 9893887, 29983327, 99477133, 99483667, 197483417, 282753937, 314623583, 315432874, 706399164, 773303937, 894303633, 947047833, 948675387, 989938887, 994927133, 994987437, 998398167, 2428989417, 2754991833, 2983284917, 2999833327

Offset: 1

Author

Dmitry Kamenetsky, Feb 21 2023

Keywords

• nonn
• base

Comments

This sequence is infinite. For example, numbers floor(30*100^k - (5/3)*10^k) beginning with 2 followed by k 9s, followed by 8 and k 3s, have a square whose digit average converges to (but never equals) 8.25. [Corrected and formula added by M. F. Hasler, Apr 11 2023]

Only a few examples are known whose square has a digit average of 8.25 and above: 3^2 = 9, 707106074079263583^2 = 499998999999788997978888999589997889 (digit average 8.25), 94180040294109027313^2 = 8869879989799999999898984986998979999969 (digit average 8.275).

This is the union of A164772 (digit average = 8) and A164841 (digit average > 8). - M. F. Hasler, Apr 11 2023

Examples

			94863 is in the sequence, because 94863^2 = 8998988769, which has a digit average of 8.1 >= 8.
		

Links

• Michael S. Branicky, Python program for OEIS A360822 and searching for squares with < given number of digits < 8
• Matthieu Dufour and Silvia Heubach, Squares with Large Digit Average, 2020.
• Dmitry Kamenetsky, The first 68 terms and their digit average

Crossrefs

Cf. A004159, A185679.

Cf. A164772 (digit average = 8), A164841 (digit average > 8).

Programs

PARI

isok(k) = my(d=digits(k^2)); vecsum(d)/#d >= 8; \\ Michel Marcus, Feb 22 2023

Python

def ok(n): d = list(map(int, str(n**2))); return sum(d) >= 8*len(d)
print([k for k in range(10**6) if ok(k)]) # Michael S. Branicky, Feb 22 2023