A381266 a(n) = least positive integer m such that when m*(m+1) is written in base n, it contains every single digit exactly once, or 0 if no such number exists.

1, 0, 12, 34, 134, 0, 1477, 6891, 38627, 0, 891230, 4874690, 28507439, 0, 1078575795, 7002987575, 46916000817, 0, 2295911609450, 16720559375850, 124852897365573, 0, 7468470450367652, 59705969514613035, 487357094495846175, 0, 34452261762372201726, 297930994005481958694

Offset: 2

Author

Daniel Mondot and Ali Sada, Feb 18 2025

Keywords

• nonn
• base

Comments

It appears that for each base of the form 4k+3, no number can be found that satisfies the requirement.

From Chai Wah Wu, Mar 13 2025: (Start)

The above observation is true.

Theorem: if n==3 (mod 4), then a(n) = 0.

Proof: Since n^a == 1 (mod n-1), k == the digit sum of k in base n (mod n-1). Thus for a number k with every digit exactly once, k == n(n-1)/2 (mod n-1).

Suppose n==3 (mod 4), i.e. n=2q+1 for some odd q. Then n(n-1)/2 = 2q^2+q. Since n-1 = 2q, this means that n(n-1)/2 == q (mod n-1). As q is odd, m(m+1) is even and n-1 is even, this implies that m(m+1) <> q (mod n-1) and thus m(m+1) is not a number with every digit exactly once and the proof is complete.

Conjecture: a(n) = 0 if and only if n==3 (mod 4).

(End)

Examples

			1477 is 2705 in octal. 2705 * 2706 = 10247536 (base 8)
38627 * 38628 = 1492083756 (base 10)
see a381266.txt for more
		

Links

• Daniel Mondot, details for each term, in respective base.

Crossrefs

Cf. A381248.

Programs

Python

from itertools import count
from math import isqrt
from sympy.ntheory import digits
def A381266(n):
    k, l, d = (n*(n-1)>>1)%(n-1), n**n-(n**n-n)//(n-1)**2, tuple(range(n))
    clist = [i for i in range(n-1) if i*(i+1)%(n-1)==k]
    if len(clist) == 0:
        return 0
    s = (n**n-n)//(n-1)**2+n**(n-2)*(n-1)-1
    s = isqrt((s<<2)+1)-1>>1
    s += n-1-s%(n-1)
    if s%(n-1) <= max(clist):
        s -= n-1
    for a in count(s,n-1):
        if a*(a+1)>l:
            break
        for c in clist:
            m = a+c
            if m*(m+1)>l:
                break
            if tuple(sorted(digits(m*(m+1),n)[1:]))==d:
                return m
    return 0 # Chai Wah Wu, Mar 17 2025

Formula

a(n) = 0 if n == 3 (mod 4). - Chai Wah Wu, Mar 13 2025

Extensions

a(19)-a(29) from Chai Wah Wu, Mar 12 2025