A362264 Numbers > 9 with increasingly large digit average of their square, in base 10.

10, 11, 12, 13, 17, 63, 83, 313, 94863, 3162083, 994927133

Offset: 0

Author

M. F. Hasler, Apr 13 2023

Keywords

• nonn
• base
• more
• hard

Comments

The single-digit number 3, whose square is 9, has the highest possible digit average, therefore this "trivial solution" is excluded. However, the sequence could be defined as "numbers > 3 ..." in which case it would start 4, 6, 7, 63, ... see examples.

It is conjectured but not known that there are only finitely many numbers whose square has a digit average above 8.3.

Can it be proved or disproved that all terms > 17 end in a digit 3?

Next terms might be 707106074079263583 (da = 8.25) and 94180040294109027313 (da = 8.275), but there might be other terms in between.

Examples

			The respective digit averages are:
   n  |    a(n)   |       a(n)^2     | #digits | sum(digits) | digit average
  ----+-----------+------------------+---------+-------------+------------------
   -  |      4    |          16      |    2    |       7     |    7/2 = 3.5
   -  |      6    |          36      |    2    |       9     |    9/2 = 4.5
   -  |      7    |          49      |    2    |      13     |   13/2 = 6.5
   0  |     10    |         100      |    3    |       1     |    1/3 = 0.333...
   1  |     11    |         121      |    3    |       4     |    4/3 = 1.333...
   2  |     12    |         144      |    3    |       9     |     3  = 3.0
   3  |     13    |         169      |    3    |      16     |   16/3 = 3.333...
   4  |     17    |         289      |    3    |      19     |   19/3 = 6.333...
   5  |     63    |        3969      |    4    |      27     |   27/4 = 6.75
   6  |     83    |        6889      |    4    |      31     |   31/4 = 7.75
   7  |    313    |       97969      |    5    |      40     |     8  = 8.0
   8  |   94863   |     8998988769   |   10    |      81     |  81/10 = 8.1
   9  |  3162083  |   9998768898889  |   13    |     106     | 106/13 = 8.15...
  10  | 994927133 |989879999979599689|   18    |     148     |   74/9 = 8.222...
		

Crossrefs

Cf. A164841, A068947, A068809.

Programs

PARI

m=0; for(k=10,oo, vecsum(d=digits(k^2))>m*#d && !print1(k", ") && m=vecsum(d)/#d)