A338295 -id:A338295 - cp's OEIS frontend

A333970 Irregular triangle read by rows where the n-th row lists the bases 2<=b<=n+1 where n in base b contains the digit b-1.

2, 2, 3, 2, 4, 2, 5, 2, 3, 6, 2, 3, 7, 2, 3, 4, 8, 2, 3, 9, 2, 5, 10, 2, 11, 2, 3, 4, 6, 12, 2, 4, 13, 2, 4, 7, 14, 2, 3, 4, 5, 15, 2, 3, 4, 8, 16, 2, 3, 17, 2, 3, 6, 9, 18, 2, 3, 19, 2, 3, 4, 5, 10, 20, 2, 3, 5, 7, 21, 2, 3, 5, 11, 22, 2, 3, 5, 23, 2, 3, 4, 5, 6, 8, 12, 24

Offset: 1

Devansh Singh, Sep 03 2020

If a number n has base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)) contains digit b-1, where b = q*(k+1)/k, k>=1 , and Sum_{i>=0} ((A(i)(mod b-q))*((b-q)^i)) > 0 then there exists  n' < n such that that n' in base b-q = b' contains digit b'-1 at the same place as n in base b and 0 <= (A(i)-A'(i))/b' <= (k+1)-((A'(i)+1)/b') (A'(i) is digit of n' in base b')for all i>=0.*
This condition is necessary and sufficient.
Proof that Condition is Necessary:
Since b-1 = b-q+q-1 and b' = q/k (as b = q*(k+1)/k). Therefore (b-1) (mod b') = (b'+q-1) (mod b') = (q-1) (mod b') = b'-1 :-(1).
n in base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)).Then n = Sum_{i>=0} (A(i)*(b^i)) = Sum_{i>=0} (A(i)*((b-q+q)^i)).
n = Sum_{i>=0} (A(i)*(b'^i)) +
    Sum_{i>=1} (A(i)*(b^i - b'^i))
   = Sum_{i>=0} (A'(i)*(b'^i)) + Sum_{i>=0} ((A(i)-A'(i))* (b'^i)) + Sum_{i>=1} (A(i)*(b^i - b'^i)),
         where A'(i) = A(i) (mod b').
Now n-Sum_{i>=0} ((A(i)-A'(i))*(b'^i))
       - Sum_{i>=1} (A(i)*(b^i - b'^i))
       = Sum_{i>=0} (A'(i)*(b'^i)).
Since A'(j) = A(j) (mod b') =  (b-1) (mod b') = b'-1(due to equation (1) above and A(j) = b-1.
Hence there exists n'  = Sum_{i>=0} (A'(i)*(b'^i)) > 0 containing digit b'-1 in base b'.
Table of n/b with cell containing T(n, b) = (n', b') for q = b/2. n' = Sum_{i>=0} (A'(i)*(b'^i))
n/b|  4  |  6  |  8  |  10 | 12
3  |(1,2)|     |     |     |
4  |     |     |     |     |
5  |     |(2,3)|     |     |
6  |     |     |     |     |
7  |(3,2)|     |(3,4)|     |
8  |     |     |     |     |
9  |     |     |     |(4,5)|
10 |     |     |     |     |
11 |(1,2)|(5,3)|     |     |(5,6)
Example: For table n/b in comments containing (n',b') in its cells.
For n = 7:
In base b = 4, n = 13 :- q = b' = 4/2 = 2, and n' = (3 mod (2))*(2)^0 + (1 mod(2))*(2)^1 = 1+2 = 3.
In base b = 8, n = 7 :- q = b' = 8/2 = 4, and n' = (7 mod (4))*(4)^0 = 3.
There are no other bases b >= 4 except 4, 8 for n = 7.
(n, b) maps to (0, 1) if b is prime. Following this and comment in A337536 we can say that all of the terms of A337536 will map to (0, 1) only, except A337536(2).
For above (n, b) -> (n', b') one possible (n, b) pair for (n', b') is { Sum_{i>=0} ((A'(i)+b') *((2*b')^i)), 2*b'}.

			Triangle begins
  Row    Bases
  n=1:   2
  n=2:   2  3
  n=3:   2  4
  n=4:   2  5
  n=5:   2  3  6
  n=6:   2  3  7
  n=7:   2  3  4  8
  n=8:   2  3  9
  n=9:   2  5  10
  n=10:  2  11

Rémy Sigrist, Table of n, a(n) for n = 1..10027 (rows for n = 1..1014, flattened)

Cf. A337535 (second column), A338295 (penultimate column), A337496 (row widths), A337536 (width 2), A337143 (width 3).

Rows containing bases 3..11 respectively: A074940, A337250, A337572, A333656, A337141, A337239, A338090, A011539, A095778.

row(n) = {my(list = List()); for (b=2, n+1, if (vecmax(digits(n, b)) == b-1, listput(list, b));); Vec(list);} \\ Michel Marcus, Sep 11 2020

More terms from Michel Marcus, Sep 11 2020

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.

Original entry on oeis.org

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

Devansh Singh, Apr 18 2021

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

Cf. A000040, A338295.

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

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

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

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)

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

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A333970 Irregular triangle read by rows where the n-th row lists the bases 2<=b<=n+1 where n in base b contains the digit b-1.

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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.

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