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