A382050 a(n) = least positive integer m such that when m*(m+1) is written in base n, it is zeroless and contains every single nonzero digit exactly once, or 0 if no such number exists.

0, 0, 5, 0, 79, 0, 650, 2716, 17846, 0, 277166, 1472993, 8233003, 0, 286485314, 1797613432, 11675780880, 0, 538954048563, 3821844010905, 27824692448867, 0, 1587841473665581, 12417635018180828, 99246128296767625, 0, 6742930364132819544, 57228575814672196977, 494789896551823383745, 0, 38997607084561562847324

Offset: 2

Author

Chai Wah Wu, Mar 13 2025

Keywords

• nonn
• base

Comments

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 zeroless number k with every nonzero 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 zeroless number with every nonzero digit exactly once.

Examples

			a(9) = 2716. 2716*2717 = 7379372 which is 14786532 in base 9.
		

Crossrefs

Cf. A381266.

Programs

Python

from itertools import count
from math import isqrt
from sympy.ntheory import digits
def A382050(n):
    k, l, d = (n*(n-1)>>1)%(n-1), n**(n-1)-(n**(n-1)-1)//(n-1)**2, tuple(range(1,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**2 + n - 1)//((n - 1)**2)
    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).

Conjecture: a(n) > 0 if n > 5 and n <> 3 (mod 4).