A001742 Numbers whose digits contain no loops (version 2).
1, 2, 3, 5, 7, 11, 12, 13, 15, 17, 21, 22, 23, 25, 27, 31, 32, 33, 35, 37, 51, 52, 53, 55, 57, 71, 72, 73, 75, 77, 111, 112, 113, 115, 117, 121, 122, 123, 125, 127, 131, 132, 133, 135, 137, 151, 152, 153, 155, 157, 171, 172, 173, 175, 177, 211, 212, 213, 215
Offset: 1
Examples
From _Hieronymus Fischer_, May 30 2012: (Start) a(10^3) = 12557. a(10^4) = 275557. a(10^5) = 11155557. a(10^6) = 223555557. (End)
Links
- Hieronymus Fischer, Table of n, a(n) for n = 1..10000
- Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
- Index entries for 10-automatic sequences.
Crossrefs
Programs
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Magma
[n: n in [1..500] | Set(Intseq(n)) subset [1, 2, 3, 5, 7]]; // Vincenzo Librandi, Dec 17 2018
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Mathematica
nlQ[n_]:=And@@(MemberQ[{1,2,3,5,7},#]&/@IntegerDigits[n]); Select[Range[ 160],nlQ] (* Harvey P. Dale, Mar 23 2012 *) Table[FromDigits/@Tuples[{1, 2, 3, 5, 7}, n], {n, 3}] // Flatten (* Vincenzo Librandi, Dec 17 2018 *) -
Perl
for (my $k = 1; $k < 1000; $k++) {print "$k, " if ($k =~ m/^[12357]+$/)} # Charles R Greathouse IV, Jun 10 2011
Formula
From Hieronymus Fischer, May 30 2012: (Start)
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)).
a(1*(5^n-1)/4) = 1*(10^n-1)/9.
a(2*(5^n-1)/4) = 2*(10^n-1)/9.
a(3*(5^n-1)/4) = 1*(10^n-1)/3.
a(4*(5^n-1)/4) = 5*(10^n-1)/9.
a(5*(5^n-1)/4) = 7*(10^n-1)/9.
a(n) = (10^log_5(4*n+1)-1)/9 for n=(5^k-1)/4, k > 0.
a(n) < (10^log_5(4*n+1)-1)/9 for (5^k-1)/4 < n < (5^(k+1)-1)/4, k > 0.
a(n) <= A202268(n), equality holds for n=(5^k-1)/4, k > 0.
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.
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.
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)
Sum_{n>=1} 1/a(n) = 3.961674246441345455010500439753914974057344229353697593567607096540565407371... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024
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