A382946 a(n) is the least positive integer k having a proper divisor d such that the base n expansions of k and d, without leading zeros, have, up to order, the same digits, or a(n) = -1 if no such k exists.
-1, 64, 36, 16, 700, 36, 42, 64, 3105, 45, 594, 105, 130, 168, 945, 120, 1666, 96, 266, 275, 2457, 231, 460, 351, 450, 273, 7938, 175, 7714, 280, 682, 1024, 308, 459, 7525, 741, 962, 665, 27300, 288, 17097, 560, 1290, 1265, 18540, 1035, 1974, 540, 952, 715
Offset: 2
Examples
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
Links
- Rémy Sigrist, Table of n, a(n) for n = 2..1000
Programs
-
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
Comments