A038565 Number of times digits are repeated in A038564.

1, 1, 2, 1, 3, 3, 3, 3, 3, 5, 5, 3, 3, 4, 3, 3, 3, 5, 3, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Naohiro Nomoto

Keywords

• nonn
• base
• easy

Comments

Next term > 1 is a(221) = 7, corresponding to A038564(221) = 26300344.

Examples

			54023  [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ],
54203  [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ],
55868  [ 1(2),2(2),3(2),4(2),5(2),6(2),7(2),8(2),9(2) ],
500407 [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ].
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A038564.

Programs

Python

from sympy import divisors
from collections import Counter
def okval(n):
    c = Counter()
    for d in divisors(n, generator=True): c.update(str(d))
    return c["1"] if len(set([c[i] for i in "123456789"])) == 1 else False
print([okval(k) for k in range(1, 60000) if okval(k)]) # Michael S. Branicky, Nov 13 2022

Extensions

More terms from Sascha Kurz, Oct 18 2001