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

3,2

Links

Table of n, a(n) for n=3..31.

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).

Example

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

Prog

(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

Crossrefs

Cf. A381266, A382050, A382054.

Sequence in context: A262971 A253380 A128123 * A227884 A186769 A213400

Adjacent sequences: A382052 A382053 A382054 * A382056 A382057 A382058

Keyword

nonn,base

Author

Chai Wah Wu, Mar 13 2025

Status

approved