A371511 a(n) is the smallest prime such that its representation in base n contains each of the digits 0,1,...,n-2 at least once and does not contain the digit n-1.

3, 73, 683, 8521, 123323, 2140069, 43720693, 1012356487, 26411157737, 749149003087, 23459877380431, 798411310382011, 29471615863458281, 1158045600182881261, 48851274656431280857, 2193475267557861578041, 104737172422274885174411, 5257403213296398892278377

Offset: 3

Author

Chai Wah Wu, Apr 10 2024

Keywords

• nonn
• base

Comments

Conjecture: for n>3, a(n) has digit sum 2+(n-2)(n-1)/2 if n is of the form 4k+3 and has digit sum 1+(n-2)(n-1)/2 otherwise.

Examples

			The corresponding base-n representations are:
n   a(n) in base n
------------------------
3   10
4   1021
5   10213
6   103241
7   1022354
8   10123645
9   101236457
10  1012356487
11  10223456798
12  10123459a867
13  1012345678a9b
14  1012345678c9ab
15  1022345678a9cdb
16  10123456789acbed
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 3..387
• Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024.

Crossrefs

Cf. A185122, A371194.

Programs

Python

from math import gcd
from sympy import nextprime
from sympy.ntheory import digits
def A371511(n):
    m, j = n, 0
    if n > 3:
        for j in range(1,n-1):
            if gcd((n*(n-1)>>1)+j,n-1) == 1:
                 break
    if j == 0:
        for i in range(2,n-1):
            m = n*m+i
    elif j == 1:
        for i in range(1,n-1):
            m = n*m+i
    else:
        for i in range(2,1+j):
            m = n*m+i
        for i in range(j,n-1):
            m = n*m+i
    m -= 1
    while True:
        s = digits(m:=nextprime(m), n)[1:]
        if n-1 not in s and len(set(s))==n-1:
            return m

Formula

For n>3, a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-1). If n = 4k+3 for k>0, then a(n) >= (n^(n-1)-n)/(n-1)^2 + n^(n-1) + n^(n-3) .