A343541 For n > 1, a(n) is the largest base b <= prime(n)-1 such that the digits of prime(n)-1 in base b contain the digit b-1.

2, 2, 3, 2, 4, 3, 3, 5, 4, 6, 2, 2, 7, 7, 4, 8, 8, 6, 6, 9, 9, 2, 3, 10, 5, 6, 6, 5, 11, 8, 3, 12, 12, 5, 3, 13, 13, 13, 5, 6, 6, 14, 14, 10, 10, 15, 15, 5, 5, 11, 11, 16, 16, 2, 3, 4, 5, 17, 17, 17, 10, 18, 18, 18, 18, 13, 13, 19, 19, 19

Offset: 2

Author

Devansh Singh, Apr 18 2021

Keywords

• nonn
• base
• look

Links

• Robert Israel, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A000040, A338295.

Programs

Maple

f:= proc(n) local p,b,L;
    p:= ithprime(n);
    for b from floor((1 + sqrt(4*p - 3))/2) by -1 do
      L:= convert(p-1,base,b);
      if member(b-1,L) then return b fi
    od;
end proc:
map(f, [$2 .. 100]); # Robert Israel, Dec 10 2024

Mathematica

Table[Max@Select[Range[2,Prime@n-1],MemberQ[IntegerDigits[Prime@n-1,#],#-1]&],{n,2,71}] (* Giorgos Kalogeropoulos, Nov 22 2021 *)

PARI

a(n) = my(q=prime(n)-1); forstep(b=q, 2, -1, if (vecmax(digits(q, b)) == b-1, return (b))); \\ Michel Marcus, Apr 19 2021

Python

import sympy
def a_n(N):
    a_n=[2]
    for i in sympy.primerange(5, N+1):
        a_n.append(A338295(i-1))
    print(a_n)
def A338295(n):
    checker=0
    for b in range(n//2, 1,-1):
        checker=main_base_check(n, b)
        if checker!=0:
            break
    return checker
def main_base_check(m, b):
    while m!=0:
        if m%b == b-1:
            return b
        m = m//b
    return 0
a_n(500)

Formula

a(n) <= (1 + sqrt(4*prime(n) - 3))/2 for all n. Prime(n), which is 111 in some base Q, has a(n) = Q+1. Example: 31 = 6*5 + 1 and it is 111 in base 5. - Devansh Singh, Nov 22 2021