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%I #26 Jan 01 2021 11:37:59
%S 1,12,13,124,15,1236,17,1248,139,125,1,12346,1,127,135,1248,1,12369,1,
%T 1245,137,12,1,123468,15,12,139,1247,1,12356,1,1248,13,12,157,123469,
%U 1,12,13,12458,1,12367,1,124,1359,12,1,123468,17,125
%N a(n) is the concatenation in increasing order of all single-digit divisors of n.
%C Sequence has period 2520 = 2^3 * 3^2 * 5 * 7.
%C a(n) = 1 iff n is a 11-rough number: not divisible by 2, 3, 5 or 7 (A008364). - Bernard Schott_, Dec 31 2020
%H Michel Marcus, <a href="/A129476/b129476.txt">Table of n, a(n) for n = 1..5040</a>
%F Let n be the rank and result be the number for this rank let a1...ak be k digits (a1...ak in [0,9]) result=a1*10^(k-1)...ak*10^0 with (i|n) => i in {a1...ak}.
%e a(10)=125 because 1, 2 and 5 divides 10. 10 also divides 10 but it is not a digit so it doesn't appear.
%e a(2520) = 123456789. - Bernard Schott_, Dec 31 2020
%p a:= n-> parse(cat(seq(`if`(irem(n, i)=0, i, [][]), i=1..9))):
%p seq(a(n), n=1..50); # _Alois P. Heinz_, Dec 31 2020
%t Table[FromDigits[Select[Divisors[n],#<10&]],{n,50}] (* _Harvey P. Dale_, Jun 07 2015 *)
%o (PARI) a(n) = fromdigits(select(x->(x<10), divisors(n))); \\ _Michel Marcus_, Dec 31 2020
%o (Python)
%o def a(n): return int('1'+"".join(d for d in "23456789" if n%int(d) == 0))
%o print([a(n) for n in range(1, 51)]) # _Michael S. Branicky_, Dec 31 2020
%Y Cf. A008364, A037278.
%K nonn,base,easy
%O 1,2
%A Colin Pitrat (colin.pitrat(AT)rez-gif.supelec.fr), May 29 2007
%E Editing and comment from _Charles R Greathouse IV_, Nov 02 2009
%E More terms from _Harvey P. Dale_, Jun 07 2015
%E Name edited by _Joerg Arndt_, Jan 01 2021
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