|
|
A226171
|
|
Smallest base in which n is not Niven (or zero if n is Niven in every base).
|
|
2
|
|
|
0, 0, 2, 0, 2, 0, 2, 6, 2, 4, 2, 8, 2, 2, 2, 6, 2, 8, 2, 7, 5, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 6, 2, 3, 2, 8, 2, 2, 2, 12, 2, 3, 2, 2, 2, 2, 2, 14, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 8, 2, 2, 2, 6, 2, 3, 2, 3, 3, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 8, 5, 2, 2, 5, 2, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
Niven numbers (in base b) are divisible by the sum of their digits (in base b).
Questions: are 1, 2, 4 and 6 the only zeros in this sequence? Where are the records or high water marks?
1,2,4,6 are the only numbers that are Niven in every base.
Proof: Suppose n is Niven in every base, then consider the base-b representations of n for (n/2) < b <= n. These are all 2-digit numbers with 1 as 1st digit and (n-b) as last digit. Then 1+n-b is a divisor of n for all b, meaning that all numbers between 1 up to n/2 are divisors of n. Clearly there are no such numbers larger than 6.
a(n) < 60 for n < 10^13.
(End)
|
|
LINKS
|
|
|
EXAMPLE
|
The sum of digits of 24 in bases 1 through 14 are: 24, 2, 4, 3, 8, 4, 6, 3, 8, 6, 4, 2, 12, 11. 24 is divisible by all these numbers except the last one; therefore a(24) = 14.
|
|
MATHEMATICA
|
Table[b = 2; While[s = Total[IntegerDigits[n, b]]; s < n && Mod[n, s] == 0, b++]; If[s == n, b = 0]; b, {n, 100}] (* T. D. Noe, May 30 2013 *)
|
|
PROG
|
(PARI) a(n) = {for (b=2, n-1, if (frac(n/sumdigits(n, b)), return(b)); ); 0; } \\ Michel Marcus, Oct 23 2018
|
|
CROSSREFS
|
Cf. A225427 (least Niven number for all bases from 1 to n).
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|