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Numbers whose digits contain no loops (version 2).
16

%I #49 Jul 30 2024 01:53:57

%S 1,2,3,5,7,11,12,13,15,17,21,22,23,25,27,31,32,33,35,37,51,52,53,55,

%T 57,71,72,73,75,77,111,112,113,115,117,121,122,123,125,127,131,132,

%U 133,135,137,151,152,153,155,157,171,172,173,175,177,211,212,213,215

%N Numbers whose digits contain no loops (version 2).

%C Numbers all of whose decimal digits are in {1,2,3,5,7}.

%C If n is represented as a zerofree base-5 number (see A084545) according to n = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=1,2,3,5,7 for k=1..5. - _Hieronymus Fischer_, May 30 2012

%H Hieronymus Fischer, <a href="/A001742/b001742.txt">Table of n, a(n) for n = 1..10000</a>

%H Robert Baillie and Thomas Schmelzer, <a href="https://library.wolfram.com/infocenter/MathSource/7166/">Summing Kempner's Curious (Slowly-Convergent) Series</a>, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.

%H <a href="/index/Ar#10-automatic">Index entries for 10-automatic sequences</a>.

%F From _Hieronymus Fischer_, May 30 2012: (Start)

%F a(n) = Sum_{j=0..m-1} ((2*b_j(n)+1) mod 10 + 2*floor(b_j(n)/5) - floor((b_j(n)+3)/5) - floor((b_j(n)+4)/5))*10^j, where b_j(n) = floor((4*n+1-5^m)/(4*5^j)), m = floor(log_5(4*n+1)).

%F a(1*(5^n-1)/4) = 1*(10^n-1)/9.

%F a(2*(5^n-1)/4) = 2*(10^n-1)/9.

%F a(3*(5^n-1)/4) = 1*(10^n-1)/3.

%F a(4*(5^n-1)/4) = 5*(10^n-1)/9.

%F a(5*(5^n-1)/4) = 7*(10^n-1)/9.

%F a(n) = (10^log_5(4*n+1)-1)/9 for n=(5^k-1)/4, k > 0.

%F a(n) < (10^log_5(4*n+1)-1)/9 for (5^k-1)/4 < n < (5^(k+1)-1)/4, k > 0.

%F a(n) <= A202268(n), equality holds for n=(5^k-1)/4, k > 0.

%F a(n) = A084545(n) iff all digits of A084545(n) are <= 3, a(n) > A084545(n), otherwise.

%F G.f.: g(x) = (x^(1/4)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(5/4)*(1 + z(j) + z(j)^2 + 2*z(j)^3 + 2*z(j)^4 - 7*z(j)^5)/(1-z(j)^5), where z(j) = x^5^j.

%F Also g(x) = (x^(1/4)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(5/4)*(1-z(j))*(1 + 2z(j) + 3*z(j)^2 + 5*z(j)^3 + 7*z(j)^4)/(1-z(j)^5), where z(j) = x^5^j.

%F Also: g(x)=(1/(1-x))*(h_(5,0)(x) + h_(5,1)(x) + h_(5,2)(x) + 2*h_(5,3)(x) + 2*h_(5,4)(x) - 7*h_(5,5)(x)), where h_(5,k)(x) = Sum_{j>=0} 10^j*x^((5^(j+1)-1)/4)*(x^5^j)^k/(1-(x^5^j)^5). (End)

%F Sum_{n>=1} 1/a(n) = 3.961674246441345455010500439753914974057344229353697593567607096540565407371... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - _Amiram Eldar_, Feb 15 2024

%e From _Hieronymus Fischer_, May 30 2012: (Start)

%e a(10^3) = 12557.

%e a(10^4) = 275557.

%e a(10^5) = 11155557.

%e a(10^6) = 223555557. (End)

%t nlQ[n_]:=And@@(MemberQ[{1,2,3,5,7},#]&/@IntegerDigits[n]); Select[Range[ 160],nlQ] (* _Harvey P. Dale_, Mar 23 2012 *)

%t Table[FromDigits/@Tuples[{1, 2, 3, 5, 7}, n], {n, 3}] // Flatten (* _Vincenzo Librandi_, Dec 17 2018 *)

%o (Perl) for (my $k = 1; $k < 1000; $k++) {print "$k, " if ($k =~ m/^[12357]+$/)} # _Charles R Greathouse IV_, Jun 10 2011

%o (Magma) [n: n in [1..500] | Set(Intseq(n)) subset [1, 2, 3, 5, 7]]; // _Vincenzo Librandi_, Dec 17 2018

%Y Cf. A001729 (version 1), A190222 (noncomposite terms), A190223 (n with all divisors in this sequence).

%Y Cf. A046034, A084545, A029581, A084984, A001743, A001744, A014261, A014263, A202267, A202268.

%K base,nonn,easy

%O 1,2

%A _N. J. A. Sloane_