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%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
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