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%I #17 Apr 24 2023 09:40:34
%S 10,11,12,13,17,63,83,313,94863,3162083,994927133
%N Numbers > 9 with increasingly large digit average of their square, in base 10.
%C 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.
%C It is conjectured but not known that there are only finitely many numbers whose square has a digit average above 8.3.
%C Can it be proved or disproved that all terms > 17 end in a digit 3?
%C Next terms might be 707106074079263583 (da = 8.25) and 94180040294109027313 (da = 8.275), but there might be other terms in between.
%e The respective digit averages are:
%e n | a(n) | a(n)^2 | #digits | sum(digits) | digit average
%e ----+-----------+------------------+---------+-------------+------------------
%e - | 4 | 16 | 2 | 7 | 7/2 = 3.5
%e - | 6 | 36 | 2 | 9 | 9/2 = 4.5
%e - | 7 | 49 | 2 | 13 | 13/2 = 6.5
%e 0 | 10 | 100 | 3 | 1 | 1/3 = 0.333...
%e 1 | 11 | 121 | 3 | 4 | 4/3 = 1.333...
%e 2 | 12 | 144 | 3 | 9 | 3 = 3.0
%e 3 | 13 | 169 | 3 | 16 | 16/3 = 3.333...
%e 4 | 17 | 289 | 3 | 19 | 19/3 = 6.333...
%e 5 | 63 | 3969 | 4 | 27 | 27/4 = 6.75
%e 6 | 83 | 6889 | 4 | 31 | 31/4 = 7.75
%e 7 | 313 | 97969 | 5 | 40 | 8 = 8.0
%e 8 | 94863 | 8998988769 | 10 | 81 | 81/10 = 8.1
%e 9 | 3162083 | 9998768898889 | 13 | 106 | 106/13 = 8.15...
%e 10 | 994927133 |989879999979599689| 18 | 148 | 74/9 = 8.222...
%o (PARI) m=0; for(k=10,oo, vecsum(d=digits(k^2))>m*#d && !print1(k", ") && m=vecsum(d)/#d)
%Y Cf. A164841, A068947, A068809.
%K nonn,base,more,hard
%O 0,1
%A _M. F. Hasler_, Apr 13 2023