A382054 a(n) = least positive integer m such that when m*(m+1) is written in base n, it does not contain the digit n-1 and contains every single digit from 0 to n-2 exactly once, or 0 if no such number exists.

0, 0, 14, 54, 0, 616, 2251, 12069, 0, 251085, 1348305, 7619403, 0, 269717049, 1698727527, 11061795398, 0, 513383208454, 3648738866370, 26618719297968, 0, 1524495582671125, 11941193897016731, 95578593301936475, 0, 6510865478836888683, 55324396705324796861, 478855818873249715068, 0, 37817609915967014967822

Offset: 3

Author

Chai Wah Wu, Mar 13 2025

Keywords

• nonn
• base

Comments

Theorem:

a(n) = 0 if n == 3 (mod 4).

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 from 0 to n-2 exactly once, k == (n-2)(n-1)/2 == 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 from 0 to n-2 exactly once and the proof is complete.

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

Examples

			a(9) = 2251 since 2251*2252 = 5069252 which is 10475632 in base 9.
		

Crossrefs

Cf. A381266, A382050, A382054.

Programs

Python

from itertools import count
from math import isqrt
from sympy.ntheory import digits
def A382054(n):
    k, l, d= (n*(n-1)>>1)%(n-1), (n - n**n + (n - 1)**2*(n**n + n**(n - 1)*(1 - n)))//(n - 1)**2, tuple(range(n-1))
    clist = [i for i in range(n-1) if i*(i+1)%(n-1)==k]
    if len(clist) == 0:
        return 0
    s = (n**(n - 1) + (n - 1)**2*(n**(n - 3)*(n - 1) - 1) - 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).