A382050 -id:A382050 - cp's OEIS frontend

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.

0, 0, 14, 54, 0, 616, 2251, 12069, 0, 251085, 1348305, 7619403, 0, 269717049, 1698727527, 11061795398, 0, 513383208454, 3648738866370, 26618719297968, 0, 1524495582671125, 11941193897016731, 95578593301936475, 0, 6510865478836888683, 55324396705324796861, 478855818873249715068, 0, 37817609915967014967822

Offset: 3

Chai Wah Wu, Mar 13 2025

Theorem:
a(n) = 0 if n == 3 (mod 4).
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 from 0 to n-2 exactly once, k == (n-2)(n-1)/2 == 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 from 0 to n-2 exactly once and the proof is complete.
Conjecture: a(n) > 0 if n > 4 and n <> 3 (mod 4).

			a(9) = 2251 since 2251*2252 = 5069252 which is 10475632 in base 9.

Cf. A381266, A382050, A382054.

from itertools import count
from math import isqrt
from sympy.ntheory import digits
def A382054(n):
    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))
    clist = [i for i in range(n-1) if i*(i+1)%(n-1)==k]
    if len(clist) == 0:
        return 0
    s = (n**(n - 1) + (n - 1)**2*(n**(n - 3)*(n - 1) - 1) - 1)//(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

a(n) = 0 if n == 3 (mod 4).

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.

Original entry on oeis.org

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

Chai Wah Wu, Mar 13 2025

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

Cf. A381266, A382050, A382054.

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

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).

cp's OEIS Frontend

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.

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