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A243845 Numbers generated by recursive procedure a(n) = nozero(a(n-1) * 3), in which the function nozero(x) removes all zeros from x, starting with a(1) = 1.

%I A243845 #28 Jun 29 2020 02:26:31
%S A243845 1,3,9,27,81,243,729,2187,6561,19683,5949,17847,53541,16623,49869,
%T A243845 14967,4491,13473,4419,13257,39771,119313,357939,173817,521451,
%U A243845 1564353,469359,14877,44631,133893,41679,12537,37611,112833,338499,115497,346491,139473,418419
%N A243845 Numbers generated by recursive procedure a(n) = nozero(a(n-1) * 3), in which the function nozero(x) removes all zeros from x, starting with a(1) = 1.
%C A243845 Numbers in the following sequence: Let a(1) = 1, then begin the recursive sequence a(n) = nozero(a(n-1) * 3), where the function nozero(x) removes all zeros from x.
%C A243845 The sequence returns standard powers of 3 until step 11, where a(11) = nozero(19683 * 3) = nozero(59049) = 5949.
%C A243845 At step 28, a(28) = nozero(469359 * 3) = nozero(1408077) = 14877. At step 108, a(108) = nozero(4959 * 3) = 14877. Therefore a(28) = a(108) and the sequence repeats. Because this is the first instance where a member of this sequence is repeated one has a(n + L) = a(n) for n >= 28 with the primitive (least) period length L = 108 - 28 = 80.
%H A243845 Anthony Sand, <a href="/A243845/b243845.txt">Table of n, a(n) for n = 1..108</a>
%H A243845 <a href="/index/Rec#order_80">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F A243845 a(n) = A004719(a(n-1) * 3) for n>1, a(1) = 1.
%e A243845 a(2) = nozero(3*a(1)) = nozero(3) = 3.
%t A243845 NestList[FromDigits@ DeleteCases[IntegerDigits[3 #], _?(# == 0 &)] &, 1, 38] (* _Michael De Vlieger_, Jun 27 2020 *)
%o A243845 (Sage)
%o A243845 L=[1]
%o A243845 for i in [1..108]:
%o A243845     T=(3*L[i-1]).digits(base=10)
%o A243845     TT=filter(lambda a: a != 0, T)
%o A243845     L.append(sum(TTi*10^i for i, TTi in enumerate(TT)))
%o A243845 L # - _Tom Edgar_, Jun 17 2014
%Y A243845 Cf. A000244, A004719, A242350, A243846.
%K A243845 nonn,base
%O A243845 1,2
%A A243845 _Anthony Sand_, Jun 12 2014
%E A243845 Edited: Name, comments and formula reformulated. - _Wolfdieter Lang_, Jul 13 2014