2,3

It appears that for each base of the form 4k+3, no number can be found that satisfies the requirement.

From Chai Wah Wu, Mar 13 2025: (Start)

The above observation is true.

Theorem: if n==3 (mod 4), then a(n) = 0.

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 exactly once, k == 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 exactly once and the proof is complete.

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

(End)

Daniel Mondot, details for each term, in respective base.

a(n) = 0 if n == 3 (mod 4). - Chai Wah Wu, Mar 13 2025

1477 is 2705 in octal. 2705 * 2706 = 10247536 (base 8)

38627 * 38628 = 1492083756 (base 10)

see a381266.txt for more

(Python)

from itertools import count

from math import isqrt

from sympy.ntheory import digits

def A381266(n):

k, l, d = (n*(n-1)>>1)%(n-1), n**n-(n**n-n)//(n-1)**2, tuple(range(n))

clist = [i for i in range(n-1) if i*(i+1)%(n-1)==k]

if len(clist) == 0:

return 0

s = (n**n-n)//(n-1)**2+n**(n-2)*(n-1)-1

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

Cf. A381248.

Sequence in context: A199114 A009760 A078194 * A034510 A083101 A133294

Adjacent sequences: A381263 A381264 A381265 * A381267 A381268 A381269

nonn,base

Daniel Mondot and Ali Sada, Feb 18 2025

a(19)-a(29) from Chai Wah Wu, Mar 12 2025

approved