A038565 - OEIS

%I #17 Nov 14 2022 00:36:56

%S 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,

%T 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,

%U 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

%N Number of times digits are repeated in A038564.

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

%H Michael S. Branicky, <a href="/A038565/b038565.txt">Table of n, a(n) for n = 1..10000</a>

%e 54023  [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ],

%e 54203  [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ],

%e 55868  [ 1(2),2(2),3(2),4(2),5(2),6(2),7(2),8(2),9(2) ],

%e 500407 [ 1(1),2(1),3(1),4(1),5(1),6(1),7(1),8(1),9(1) ].

%o (Python)

%o from sympy import divisors

%o from collections import Counter

%o def okval(n):

%o     c = Counter()

%o     for d in divisors(n, generator=True): c.update(str(d))

%o     return c["1"] if len(set([c[i] for i in "123456789"])) == 1 else False

%o print([okval(k) for k in range(1, 60000) if okval(k)]) # _Michael S. Branicky_, Nov 13 2022

%Y Cf. A038564.

%K nonn,base,easy

%O 1,3

%A _Naohiro Nomoto_

%E More terms from _Sascha Kurz_, Oct 18 2001