|
Search: id:A158911
|
|
|
| A158911 |
|
numbers n such that 10^n divided thru the number of digits of 10^n is an integer. |
|
+0
1
|
|
| 0, 1, 3, 4, 7, 9, 15, 19, 24, 31, 39, 49, 63, 79, 99, 127, 159, 199, 255, 319, 399, 511, 639, 799, 1023, 1279, 1599, 2047, 2559, 3199, 4095, 5119, 6399, 8191, 10239, 12799, 16383, 20479
(list; graph; listen)
|
|
|
OFFSET
|
1,3
|
|
|
COMMENT
|
n = (2^a)*(5^b)-1 , integers a,b>=0. The generalized question : for which n is K^n (written in base 10) divided thru the number of digits of K^n an integer ? So we look after n for which (K^n)/(floor(n*log(K)/log(10))+1) should be an integer. This is true only if (log(K)/log(10) is a rational number, thus the sequence which gives the number of digits of K^n (written in base 10) is periodic. For this sequence we have an easy solution for n and this because log(10)/log(10) is a rational number and the sequence which gives the number of digits of 10^n (written in base 10) is periodic. For the sequence A158520 there is not a solution as log(2)/log(10) is an irrational number and the sequence which gives the number of digits of 2^n written in base 10 is not periodic.
|
|
CROSSREFS
|
Cf. A158520, A034887, A129344,
Sequence in context: A163468 A069183 A119907 this_sequence A086772 A086336 A108796
Adjacent sequences: A158908 A158909 A158910 this_sequence A158912 A158913 A158914
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Ctibor O. Zizka (c.zizka(AT)email.cz), Mar 30 2009
|
|
|
Search completed in 0.002 seconds
|
