A382055 a(n) = least positive integer m such that when m*(m+1) is written in base n, it does not contain the digits 0 or n-1 and contains every single digit from 1 to n-2 exactly once, or 0 if no such number exists.

0, 2, 6, 19, 0, 420, 924, 3672, 0, 78880, 431493, 2173950, 0, 71583429, 436726936, 2750336517, 0, 120521201887, 833996387274, 5932255141224, 0, 324116744376715, 2483526997445916, 19463766853506024, 0, 1274294107710603710, 10627079743009611713, 90335862784009245081, 0

Offset: 3

Author

Chai Wah Wu, Mar 13 2025

Keywords

• nonn
• base

Examples

			a(9) = 924. 924*925 = 854700 which is 1542376 in base 9.
		

Crossrefs

Cf. A381266, A382050, A382054.

Programs

Python

from itertools import count
from math import isqrt
from sympy.ntheory import digits
def A382055(n):
    k, l, d= (n*(n-1)>>1)%(n-1), (-n**(n - 1) + (n - 1)**2*(n**(n - 2)*(1 - n) + n**(n - 1)) + 1)//(n - 1)**2, tuple(range(1,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**2 + n + n**n - (n - 1)**3 - 1)//(n*(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). For a proof of this see A382054 for the proof of a similar result which also holds in this case.

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