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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.

%I A382054 #19 Mar 17 2025 22:15:38
%S A382054 0,0,14,54,0,616,2251,12069,0,251085,1348305,7619403,0,269717049,
%T A382054 1698727527,11061795398,0,513383208454,3648738866370,26618719297968,0,
%U A382054 1524495582671125,11941193897016731,95578593301936475,0,6510865478836888683,55324396705324796861,478855818873249715068,0,37817609915967014967822
%N 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.
%C A382054 Theorem:
%C A382054 a(n) = 0 if n == 3 (mod 4).
%C A382054 Proof:
%C A382054 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).
%C A382054 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.
%C A382054 Conjecture: a(n) > 0 if n > 4 and n <> 3 (mod 4).
%F A382054 a(n) = 0 if n == 3 (mod 4).
%e A382054 a(9) = 2251 since 2251*2252 = 5069252 which is 10475632 in base 9.
%o A382054 (Python)
%o A382054 from itertools import count
%o A382054 from math import isqrt
%o A382054 from sympy.ntheory import digits
%o A382054 def A382054(n):
%o A382054     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))
%o A382054     clist = [i for i in range(n-1) if i*(i+1)%(n-1)==k]
%o A382054     if len(clist) == 0:
%o A382054         return 0
%o A382054     s = (n**(n - 1) + (n - 1)**2*(n**(n - 3)*(n - 1) - 1) - 1)//(n - 1)**2
%o A382054     s = isqrt((s<<2)+1)-1>>1
%o A382054     s += n-1-s%(n-1)
%o A382054     if s%(n-1) <= max(clist):
%o A382054         s -= n-1
%o A382054     for a in count(s,n-1):
%o A382054         if a*(a+1)>l:
%o A382054             break
%o A382054         for c in clist:
%o A382054             m = a+c
%o A382054             if m*(m+1)>l:
%o A382054                 break
%o A382054             if tuple(sorted(digits(m*(m+1),n)[1:]))==d:
%o A382054                 return m
%o A382054     return 0 # _Chai Wah Wu_, Mar 17 2025
%Y A382054 Cf. A381266, A382050, A382054.
%K A382054 nonn,base
%O A382054 3,3
%A A382054 _Chai Wah Wu_, Mar 13 2025