A344888 a(n) is the least base k >= 2 where n is an undulating number (i.e., with digits of the form abababab...).

2, 2, 2, 2, 3, 2, 3, 2, 3, 4, 2, 4, 4, 3, 4, 2, 3, 4, 5, 5, 3, 2, 5, 3, 5, 4, 3, 6, 6, 4, 3, 2, 6, 6, 4, 6, 5, 6, 4, 7, 3, 5, 2, 6, 7, 7, 4, 7, 7, 6, 3, 4, 5, 8, 8, 4, 8, 5, 8, 4, 3, 6, 5, 2, 7, 8, 9, 5, 4, 9, 3, 7, 5, 8, 6, 9, 9, 9, 5, 9, 3, 8, 9, 5, 10, 2, 6

Offset

0,1

Links

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

Formula

a(n) <= A000196(n) + 1 for any n > 0.

a(n) <= 10 for any n in A033619.

a(n) = 2 iff n belongs to A000225 or to A000975.

Example

For n = 49:
- we have:
     b  49 in base b  Undulating?
     -  ------------  -----------
     2        110001  No
     3          1211  No
     4           301  No
     5           144  No
     6           121  Yes
- so a(49) = 6.

Prog

(PARI) is(n, base=10) = my (d=digits(n, base)); #d<=2 || d[1..#d-2]==d[3..#d]
a(n) = for (b=2, oo, if (is(n, b), return (b)))
(Python)
def A344888(n):
    b, m = 2, n
    while True:
        m, x = divmod(m, b)
        m, y = divmod(m, b)
        while m > 0:
            m, z = divmod(m, b)
            if z != x:
                break
            if m > 0:
                m, z = divmod(m, b)
                if z != y:
                    break
            else:
                return b
        else:
            return b
        b += 1
        m = n # Chai Wah Wu, Jun 02 2021

Crossrefs

Cf. A000196, A000225, A000975, A033619.

Sequence in context: A331362 A139325 A341829 * A216325 A322868 A240975

Adjacent sequences: A344885 A344886 A344887 * A344889 A344890 A344891

Keyword

nonn,look,base

Author

Rémy Sigrist, Jun 01 2021

Status

approved