A383412 Lexicographically earliest sequence of integers >= 2 such that whenever a(k_1) = ... = a(k_m) with k_1 < ... < k_m, the sum k_1 + ... + k_m can be computed without carries in base a(k_1).

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

Offset: 0

Author

Rémy Sigrist, Apr 26 2025

Keywords

• nonn
• base

Comments

This sequence is a variant of A279125 and A336206 exploiting several bases.

This sequence is unbounded:

- by contradiction, suppose that a(n) <= B for some B >= 2,

- let U_B = {1 + B!*k, k >= 0},

- the base b expansion of any term of U_B ends with a digit 1 in any base b in the interval 2..B,

- by the pigeonhole principle, for some b in the interval 2..B, we have a(u) = b for infinitely many terms of U_B,

- however we can at most add b-1 such terms in base b, a contradiction.

Examples

			The first terms, in decimal and in base a(n), alongside the corresponding sums of indices k <= n such that a(k) = a(n) in base a(n), are:
  n   a(n)  n in base a(n)  Sums in base a(n)
  --  ----  --------------  -----------------
   0     2  0               0
   1     2  1               1
   2     2  1,0             1,1
   3     3  1,0             1,0
   4     2  1,0,0           1,1,1
   5     3  1,2             2,2
   6     4  1,2             1,2
   7     5  1,2             1,2
   8     2  1,0,0,0         1,1,1,1
   9     3  1,0,0           1,2,2
  10     5  2,0             3,2
  11     6  1,5             1,5
  12     6  2,0             3,5
  13     7  1,6             1,6
  14     7  2,0             3,6
  15     8  1,7             1,7
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A131577, A279125, A336206, A375776.

Programs

PARI

{   t = [0, 0];
    for (n = 0, 73,
        for (b = 2, oo,
            if (#t < b,
                t = concat(t, vector(#t)););
            if (sumdigits(t[b]+n, b) == sumdigits(t[b], b) + sumdigits(n, b),
                print1 (b", "); t[b] += n; break;););); }

Formula

a(n) = 2 iff n belongs to A131577.