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

Author

N. J. A. Sloane

Keywords

• base
• nonn
• easy

Comments

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

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

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

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

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

Programs

Magma

[n: n in [1..500] |  Set(Intseq(n)) subset [1, 2, 3, 5, 7]]; // Vincenzo Librandi, Dec 17 2018

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.

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

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