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

Author

Rémy Sigrist, Jun 01 2021

Keywords

• nonn
• look
• base

Examples

			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.
		

Links

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

Crossrefs

Cf. A000196, A000225, A000975, A033619.

Programs

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

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.