OFFSET
2,2
COMMENTS
Conjecture: a(n) > 0 for any n > 2.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 2..1000
EXAMPLE
The first terms, alongside an appropriate divisor d, in bases 10 and n, are:
n a(n) d n in base n d in base n
-- ---- ---- ----------- -----------
2 -1 N/A N/A N/A
3 64 32 2,1,0,1 1,0,1,2
4 36 18 2,1,0 1,0,2
5 16 8 3,1 1,3
6 700 350 3,1,2,4 1,3,4,2
7 36 12 5,1 1,5
8 42 21 5,2 2,5
9 64 16 7,1 1,7
10 3105 1035 3,1,0,5 1,0,3,5
11 45 15 4,1 1,4
12 594 198 4,1,6 1,4,6
13 105 21 8,1 1,8
14 130 65 9,4 4,9
15 168 56 11,3 3,11
16 945 315 3,11,1 1,3,11
PROG
(PARI) a(n) = {
if (n==2, return (-1));
for (k = 1, oo,
my (t = vecsort(digits(k, n)));
fordiv (k, d,
if (d < k && vecsort(digits(d, n))==t,
return (k); ); ); ); }
(Python)
from sympy import divisors
from sympy.ntheory import digits
from itertools import count
def a(n):
if n == 2:
return -1
for k in count(2*n):
divs, kdigs = divisors(k), sorted(digits(k, n)[1:])
for d in sorted(divs[:-1], reverse=True):
ddigs = sorted(digits(d, n)[1:])
if ddigs == kdigs:
return k
if len(ddigs) < len(kdigs):
break
print([a(n) for n in range(2, 52)]) # Michael S. Branicky, Apr 13 2025
CROSSREFS
KEYWORD
sign,base
AUTHOR
Rémy Sigrist, Apr 09 2025
STATUS
approved
