OFFSET
2,1
COMMENTS
Number of digits in n^k is equal to floor(1 + k*log_10(n)).
LINKS
V. Raman, Table of n, a(n) for n = 2..10000
MATHEMATICA
Table[k = 1; While[d = IntegerDigits[n^k]; ! MemberQ[Partition[Differences[d], 2, 1], {0, 0}], k++]; Length[d], {n, 2, 100}] (* T. D. Noe, Oct 03 2012 *)
spn3[n_]:=Module[{p=1}, While[SequenceCount[IntegerDigits[n^p], {x_, x_, x_}]<1, p++]; IntegerLength[n^p]]; Array[spn3, 80, 2] (* Harvey P. Dale, Nov 10 2022 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
V. Raman, Sep 27 2012
STATUS
approved